Practice exam-oriented numerical problems with detailed solutions, circuit diagrams, shortcut methods, PDF-inspired patterns, formula strategy and common mistake analysis.
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Contact Kumar SirFirst decide whether the problem is about drift velocity, resistance, power, cells, Kirchhoff law, bridge, meter bridge or potentiometer. Most wrong solutions begin with the wrong chapter label.
Check whether the circuit is open, closed, balanced, unbalanced, steady-state, series, parallel or mixed. Formula selection depends on this condition, not only on the words in the question.
Convert mA to A, cm to m, mm² to m², micro-ohm-metre carefully. JEE and NEET often hide the main trap inside unit conversion.
When all options are numeric, estimate order of magnitude first. Eliminate options with wrong units or impossible trends before doing exact arithmetic.
Replace complicated cell networks by Thevenin equivalent wherever possible: equivalent emf plus equivalent internal resistance.
Do not use P/Q=R/S in an unbalanced bridge. Do not write E=nE for parallel cells. Do not reverse l1 and l2 in potentiometer internal resistance.
| Concept | Formula | Meaning, Units and Use |
|---|---|---|
| Current | I=Q/t or I=dQ/dt | Charge per second. Unit ampere. Use when charge and time are given. Mistake: using minutes without conversion. |
| Drift velocity | I=neAvd | n number density, A area, e charge. Use microscopic current problems. Mistake: forgetting area in m^2. |
| Current density | J=I/A=nevd | Unit A m^-2. Use conductor cross-section questions. Mistake: treating J as scalar only. |
| Ohm's law | V=IR | Use for ohmic conductors at constant temperature. Mistake: applying to diode/filament blindly. |
| Resistance | R=rho L/A | Use wire geometry. Mistake: resistivity is material property, not size property. |
| Conductivity | sigma=1/rho | Use material conduction. Unit S m^-1. Mistake: confusing mobility with conductivity. |
| Series resistors | Req=R1+R2+... | Current same. Use chain networks. Mistake: adding voltage drops without current direction. |
| Parallel resistors | 1/Req=sum(1/Ri) | Voltage same. Use branch networks. Mistake: using arithmetic sum. |
| Cells in series | I=nE/(R+nr) | n identical cells aiding. Mistake: ignoring nr. |
| Cells in parallel | I=E/(R+r/n) | Identical cells same polarity. Mistake: writing equivalent emf nE. |
| Terminal voltage | V=E-Ir | Discharging cell. Open circuit gives V=E. Mistake: using V=E under load. |
| KCL | Sum Iin=Sum Iout | Use junctions. Mistake: saying current is consumed. |
| KVL | Sum V=0 | Use closed loops. Mistake: wrong sign across cells/resistors. |
| Wheatstone bridge | P/Q=R/S | Only at balance. Mistake: applying to unbalanced bridge. |
| Meter bridge | X/R=l/(100-l) | Uniform wire at null point. Mistake: taking length from wrong end. |
| Potentiometer | V=kl | At null point no current drawn from test cell. Mistake: changing rheostat between readings. |
| Internal resistance by potentiometer | r=R(l1/l2-1) | l1 open circuit, l2 with load R. Mistake: reversing l1 and l2. |
| Power | P=VI=I^2R=V^2/R | Choose formula depending on known constant. Mistake: using wrong shared quantity. |
| Heating | H=I^2Rt | Joule heating. Mistake: forgetting square on current. |
A cell of emf 2.2 V is connected to a 4 ohm resistor. Terminal voltage is 2.0 V. Find internal resistance and explain why terminal voltage is less than emf.
Final Answer: 0.4 ohm
Presentation: Current I=V/R=2/4=0.5 A. E=V+Ir, so r=(E-V)/I=0.2/0.5=0.4 ohm. Terminal voltage is less because Ir is lost inside the cell.
Concept Tested: Internal resistance
In a meter bridge, the balance point is 35 cm from the left end when unknown X is in the left gap and 10 ohm is in the right gap. Calculate X. Why should balance point be near the middle?
Final Answer: 5.38 ohm
Presentation: X/10=35/65, so X=5.38 ohm. Near middle, percentage error due to end resistance and scale reading is minimized.
Concept Tested: Meter bridge
A cell balances at 75 cm on a potentiometer wire. Another cell balances at 100 cm. If second emf is 2 V, find first emf. State one advantage of potentiometer over voltmeter.
Final Answer: 1.5 V
Presentation: E1/E2=l1/l2=75/100. E1=1.5 V. Advantage: at balance, potentiometer draws no current from test cell, so it measures emf accurately.
Concept Tested: Potentiometer
Using Kirchhoff rules, write equations for a circuit having two loops with a common resistor R3. Mesh currents I1 and I2 pass through R3 in opposite directions.
Final Answer: Equations
Presentation: Left loop: E1-I1R1-(I1-I2)R3=0. Right loop: E2-I2R2-(I2-I1)R3=0, with signs chosen according to loop direction.
Concept Tested: Kirchhoff rules
A wire has resistance 10 ohm at 27 C and 12 ohm at 127 C. Calculate temperature coefficient of resistance.
Final Answer: 0.002 per C
Presentation: R=R0(1+alpha DeltaT). 12=10(1+100alpha), alpha=0.002 per C.
Concept Tested: Temperature dependence
Derive I=neAvd and calculate vd for I=1.6 A, n=8e28 m^-3, A=2e-6 m^2.
Final Answer: 6.25e-5 m/s
Presentation: vd=I/(neA)=1.6/(8e28 x 1.6e-19 x 2e-6)=6.25e-5 m/s.
Concept Tested: Drift velocity
A cell of emf 2.2 V is connected to a 4 ohm resistor. Terminal voltage is 2.0 V. Find internal resistance and explain why terminal voltage is less than emf.
Final Answer: 0.4 ohm
Presentation: Current I=V/R=2/4=0.5 A. E=V+Ir, so r=(E-V)/I=0.2/0.5=0.4 ohm. Terminal voltage is less because Ir is lost inside the cell.
Concept Tested: Internal resistance
In a meter bridge, the balance point is 35 cm from the left end when unknown X is in the left gap and 10 ohm is in the right gap. Calculate X. Why should balance point be near the middle?
Final Answer: 5.38 ohm
Presentation: X/10=35/65, so X=5.38 ohm. Near middle, percentage error due to end resistance and scale reading is minimized.
Concept Tested: Meter bridge
A cell balances at 75 cm on a potentiometer wire. Another cell balances at 100 cm. If second emf is 2 V, find first emf. State one advantage of potentiometer over voltmeter.
Final Answer: 1.5 V
Presentation: E1/E2=l1/l2=75/100. E1=1.5 V. Advantage: at balance, potentiometer draws no current from test cell, so it measures emf accurately.
Concept Tested: Potentiometer
Using Kirchhoff rules, write equations for a circuit having two loops with a common resistor R3. Mesh currents I1 and I2 pass through R3 in opposite directions.
Final Answer: Equations
Presentation: Left loop: E1-I1R1-(I1-I2)R3=0. Right loop: E2-I2R2-(I2-I1)R3=0, with signs chosen according to loop direction.
Concept Tested: Kirchhoff rules
A wire has resistance 10 ohm at 27 C and 12 ohm at 127 C. Calculate temperature coefficient of resistance.
Final Answer: 0.002 per C
Presentation: R=R0(1+alpha DeltaT). 12=10(1+100alpha), alpha=0.002 per C.
Concept Tested: Temperature dependence
Derive I=neAvd and calculate vd for I=1.6 A, n=8e28 m^-3, A=2e-6 m^2.
Final Answer: 6.25e-5 m/s
Presentation: vd=I/(neA)=1.6/(8e28 x 1.6e-19 x 2e-6)=6.25e-5 m/s.
Concept Tested: Drift velocity
A cell of emf 2.2 V is connected to a 4 ohm resistor. Terminal voltage is 2.0 V. Find internal resistance and explain why terminal voltage is less than emf.
Final Answer: 0.4 ohm
Presentation: Current I=V/R=2/4=0.5 A. E=V+Ir, so r=(E-V)/I=0.2/0.5=0.4 ohm. Terminal voltage is less because Ir is lost inside the cell.
Concept Tested: Internal resistance
In a meter bridge, the balance point is 35 cm from the left end when unknown X is in the left gap and 10 ohm is in the right gap. Calculate X. Why should balance point be near the middle?
Final Answer: 5.38 ohm
Presentation: X/10=35/65, so X=5.38 ohm. Near middle, percentage error due to end resistance and scale reading is minimized.
Concept Tested: Meter bridge
A cell balances at 75 cm on a potentiometer wire. Another cell balances at 100 cm. If second emf is 2 V, find first emf. State one advantage of potentiometer over voltmeter.
Final Answer: 1.5 V
Presentation: E1/E2=l1/l2=75/100. E1=1.5 V. Advantage: at balance, potentiometer draws no current from test cell, so it measures emf accurately.
Concept Tested: Potentiometer
Using Kirchhoff rules, write equations for a circuit having two loops with a common resistor R3. Mesh currents I1 and I2 pass through R3 in opposite directions.
Final Answer: Equations
Presentation: Left loop: E1-I1R1-(I1-I2)R3=0. Right loop: E2-I2R2-(I2-I1)R3=0, with signs chosen according to loop direction.
Concept Tested: Kirchhoff rules
A wire has resistance 10 ohm at 27 C and 12 ohm at 127 C. Calculate temperature coefficient of resistance.
Final Answer: 0.002 per C
Presentation: R=R0(1+alpha DeltaT). 12=10(1+100alpha), alpha=0.002 per C.
Concept Tested: Temperature dependence
Derive I=neAvd and calculate vd for I=1.6 A, n=8e28 m^-3, A=2e-6 m^2.
Final Answer: 6.25e-5 m/s
Presentation: vd=I/(neA)=1.6/(8e28 x 1.6e-19 x 2e-6)=6.25e-5 m/s.
Concept Tested: Drift velocity
A cell of emf 2.2 V is connected to a 4 ohm resistor. Terminal voltage is 2.0 V. Find internal resistance and explain why terminal voltage is less than emf.
Final Answer: 0.4 ohm
Presentation: Current I=V/R=2/4=0.5 A. E=V+Ir, so r=(E-V)/I=0.2/0.5=0.4 ohm. Terminal voltage is less because Ir is lost inside the cell.
Concept Tested: Internal resistance
In a meter bridge, the balance point is 35 cm from the left end when unknown X is in the left gap and 10 ohm is in the right gap. Calculate X. Why should balance point be near the middle?
Final Answer: 5.38 ohm
Presentation: X/10=35/65, so X=5.38 ohm. Near middle, percentage error due to end resistance and scale reading is minimized.
Concept Tested: Meter bridge
A cell balances at 75 cm on a potentiometer wire. Another cell balances at 100 cm. If second emf is 2 V, find first emf. State one advantage of potentiometer over voltmeter.
Final Answer: 1.5 V
Presentation: E1/E2=l1/l2=75/100. E1=1.5 V. Advantage: at balance, potentiometer draws no current from test cell, so it measures emf accurately.
Concept Tested: Potentiometer
Using Kirchhoff rules, write equations for a circuit having two loops with a common resistor R3. Mesh currents I1 and I2 pass through R3 in opposite directions.
Final Answer: Equations
Presentation: Left loop: E1-I1R1-(I1-I2)R3=0. Right loop: E2-I2R2-(I2-I1)R3=0, with signs chosen according to loop direction.
Concept Tested: Kirchhoff rules
A wire has resistance 10 ohm at 27 C and 12 ohm at 127 C. Calculate temperature coefficient of resistance.
Final Answer: 0.002 per C
Presentation: R=R0(1+alpha DeltaT). 12=10(1+100alpha), alpha=0.002 per C.
Concept Tested: Temperature dependence
Derive I=neAvd and calculate vd for I=1.6 A, n=8e28 m^-3, A=2e-6 m^2.
Final Answer: 6.25e-5 m/s
Presentation: vd=I/(neA)=1.6/(8e28 x 1.6e-19 x 2e-6)=6.25e-5 m/s.
Concept Tested: Drift velocity
A cell of emf 2.2 V is connected to a 4 ohm resistor. Terminal voltage is 2.0 V. Find internal resistance and explain why terminal voltage is less than emf.
Final Answer: 0.4 ohm
Presentation: Current I=V/R=2/4=0.5 A. E=V+Ir, so r=(E-V)/I=0.2/0.5=0.4 ohm. Terminal voltage is less because Ir is lost inside the cell.
Concept Tested: Internal resistance
In a meter bridge, the balance point is 35 cm from the left end when unknown X is in the left gap and 10 ohm is in the right gap. Calculate X. Why should balance point be near the middle?
Final Answer: 5.38 ohm
Presentation: X/10=35/65, so X=5.38 ohm. Near middle, percentage error due to end resistance and scale reading is minimized.
Concept Tested: Meter bridge
A cell balances at 75 cm on a potentiometer wire. Another cell balances at 100 cm. If second emf is 2 V, find first emf. State one advantage of potentiometer over voltmeter.
Final Answer: 1.5 V
Presentation: E1/E2=l1/l2=75/100. E1=1.5 V. Advantage: at balance, potentiometer draws no current from test cell, so it measures emf accurately.
Concept Tested: Potentiometer
Using Kirchhoff rules, write equations for a circuit having two loops with a common resistor R3. Mesh currents I1 and I2 pass through R3 in opposite directions.
Final Answer: Equations
Presentation: Left loop: E1-I1R1-(I1-I2)R3=0. Right loop: E2-I2R2-(I2-I1)R3=0, with signs chosen according to loop direction.
Concept Tested: Kirchhoff rules
A wire has resistance 10 ohm at 27 C and 12 ohm at 127 C. Calculate temperature coefficient of resistance.
Final Answer: 0.002 per C
Presentation: R=R0(1+alpha DeltaT). 12=10(1+100alpha), alpha=0.002 per C.
Concept Tested: Temperature dependence
Derive I=neAvd and calculate vd for I=1.6 A, n=8e28 m^-3, A=2e-6 m^2.
Final Answer: 6.25e-5 m/s
Presentation: vd=I/(neA)=1.6/(8e28 x 1.6e-19 x 2e-6)=6.25e-5 m/s.
Concept Tested: Drift velocity
A cell of emf 6 V and internal resistance 1 ohm is connected to a 5 ohm resistor. A voltmeter of resistance 10 ohm is connected across the 5 ohm resistor. What is the voltmeter reading?
Correct Answer: A
Detailed Solution: The voltmeter is not ideal. The external load is 5 parallel 10 = 10/3 ohm. Total resistance = 1+10/3=13/3 ohm. Current from cell = 6/(13/3)=18/13 A. Terminal/load voltage = I x 10/3 = 60/13 = 4.62 V. If options are adjusted for exact finite-voltmeter data, choose the nearest consistent value; here the concept is finite voltmeter loading.
Difficulty: Difficult | Concept Tested: Internal resistance + non-ideal meter
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a Wheatstone bridge, arms are 2 ohm, 3 ohm, 4 ohm and 6 ohm. A 10 ohm galvanometer is connected between the middle points. The equivalent resistance between supply terminals is closest to:
Correct Answer: B
Detailed Solution: Since 2/3 = 4/6, the bridge is balanced and no current flows through the galvanometer. Equivalent = (2+3) parallel (4+6) = 5 parallel 10 = 10/3 ohm. Nearest option is 3 ohm if exact option appears; concept: ignore galvanometer only at balance.
Difficulty: Medium | Concept Tested: Wheatstone bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A meter bridge gives balance point 39.5 cm for X in left gap and 60.5 cm after interchanging X and R. If R=10 ohm, find X approximately ignoring end correction.
Correct Answer: C
Detailed Solution: For first setting X/R=l/(100-l)=39.5/60.5. X=10 x 39.5/60.5 = 6.53 ohm. Interchange reading checks consistency and end error.
Difficulty: Medium | Concept Tested: Meter bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A metallic wire carries current I. If both length and radius are doubled while the same potential difference is applied, drift velocity becomes:
Correct Answer: A
Detailed Solution: R=rho L/A. New R proportional 2L/(4A)=R/2, so current doubles. vd=I/(neA); area becomes 4A, so vd becomes 2/4 = 1/2. If length also changes field E=V/L, microscopic vd proportional E, so vd becomes half. Correct conceptual answer: one-half. This tests consistency of macro and micro pictures.
Difficulty: Difficult | Concept Tested: Drift velocity + geometry
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Twelve identical cells each emf E and internal resistance r are arranged as 3 in series per row and 4 rows in parallel. The equivalent source is:
Correct Answer: B
Detailed Solution: Each row has emf 3E and internal resistance 3r. Four identical rows in parallel keep emf 3E and reduce internal resistance to 3r/4.
Difficulty: Medium | Concept Tested: Mixed cell combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two resistors R and 2R are connected first in series and then in parallel across the same ideal battery. Ratio of total power in parallel to total power in series is:
Correct Answer: C
Detailed Solution: Series equivalent = 3R, P_s=V^2/(3R). Parallel equivalent = 2R/3, P_p=V^2/(2R/3)=3V^2/(2R). Ratio = (3/2)/(1/3)=9/2.
Difficulty: Medium | Concept Tested: Power and combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A 10 ohm wire with alpha=0.004/C is in series with a 20 ohm carbon resistor with alpha=-0.0005/C. Equivalent temperature coefficient is:
Correct Answer: B
Detailed Solution: For series, alpha_eq=(R1 alpha1 + R2 alpha2)/(R1+R2)=(10x.004+20x-.0005)/30=(.04-.01)/30=.001/C.
Difficulty: Difficult | Concept Tested: Temperature coefficient
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 5 ohm resistor. A voltmeter of resistance 10 ohm is connected across the 5 ohm resistor. What is the voltmeter reading?
Correct Answer: A
Detailed Solution: The voltmeter is not ideal. The external load is 5 parallel 10 = 10/3 ohm. Total resistance = 1+10/3=13/3 ohm. Current from cell = 6/(13/3)=18/13 A. Terminal/load voltage = I x 10/3 = 60/13 = 4.62 V. If options are adjusted for exact finite-voltmeter data, choose the nearest consistent value; here the concept is finite voltmeter loading.
Difficulty: Difficult | Concept Tested: Internal resistance + non-ideal meter
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a Wheatstone bridge, arms are 2 ohm, 3 ohm, 4 ohm and 6 ohm. A 10 ohm galvanometer is connected between the middle points. The equivalent resistance between supply terminals is closest to:
Correct Answer: B
Detailed Solution: Since 2/3 = 4/6, the bridge is balanced and no current flows through the galvanometer. Equivalent = (2+3) parallel (4+6) = 5 parallel 10 = 10/3 ohm. Nearest option is 3 ohm if exact option appears; concept: ignore galvanometer only at balance.
Difficulty: Medium | Concept Tested: Wheatstone bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A meter bridge gives balance point 39.5 cm for X in left gap and 60.5 cm after interchanging X and R. If R=10 ohm, find X approximately ignoring end correction.
Correct Answer: C
Detailed Solution: For first setting X/R=l/(100-l)=39.5/60.5. X=10 x 39.5/60.5 = 6.53 ohm. Interchange reading checks consistency and end error.
Difficulty: Medium | Concept Tested: Meter bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A metallic wire carries current I. If both length and radius are doubled while the same potential difference is applied, drift velocity becomes:
Correct Answer: A
Detailed Solution: R=rho L/A. New R proportional 2L/(4A)=R/2, so current doubles. vd=I/(neA); area becomes 4A, so vd becomes 2/4 = 1/2. If length also changes field E=V/L, microscopic vd proportional E, so vd becomes half. Correct conceptual answer: one-half. This tests consistency of macro and micro pictures.
Difficulty: Difficult | Concept Tested: Drift velocity + geometry
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Twelve identical cells each emf E and internal resistance r are arranged as 3 in series per row and 4 rows in parallel. The equivalent source is:
Correct Answer: B
Detailed Solution: Each row has emf 3E and internal resistance 3r. Four identical rows in parallel keep emf 3E and reduce internal resistance to 3r/4.
Difficulty: Medium | Concept Tested: Mixed cell combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two resistors R and 2R are connected first in series and then in parallel across the same ideal battery. Ratio of total power in parallel to total power in series is:
Correct Answer: C
Detailed Solution: Series equivalent = 3R, P_s=V^2/(3R). Parallel equivalent = 2R/3, P_p=V^2/(2R/3)=3V^2/(2R). Ratio = (3/2)/(1/3)=9/2.
Difficulty: Medium | Concept Tested: Power and combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A 10 ohm wire with alpha=0.004/C is in series with a 20 ohm carbon resistor with alpha=-0.0005/C. Equivalent temperature coefficient is:
Correct Answer: B
Detailed Solution: For series, alpha_eq=(R1 alpha1 + R2 alpha2)/(R1+R2)=(10x.004+20x-.0005)/30=(.04-.01)/30=.001/C.
Difficulty: Difficult | Concept Tested: Temperature coefficient
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 5 ohm resistor. A voltmeter of resistance 10 ohm is connected across the 5 ohm resistor. What is the voltmeter reading?
Correct Answer: A
Detailed Solution: The voltmeter is not ideal. The external load is 5 parallel 10 = 10/3 ohm. Total resistance = 1+10/3=13/3 ohm. Current from cell = 6/(13/3)=18/13 A. Terminal/load voltage = I x 10/3 = 60/13 = 4.62 V. If options are adjusted for exact finite-voltmeter data, choose the nearest consistent value; here the concept is finite voltmeter loading.
Difficulty: Difficult | Concept Tested: Internal resistance + non-ideal meter
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a Wheatstone bridge, arms are 2 ohm, 3 ohm, 4 ohm and 6 ohm. A 10 ohm galvanometer is connected between the middle points. The equivalent resistance between supply terminals is closest to:
Correct Answer: B
Detailed Solution: Since 2/3 = 4/6, the bridge is balanced and no current flows through the galvanometer. Equivalent = (2+3) parallel (4+6) = 5 parallel 10 = 10/3 ohm. Nearest option is 3 ohm if exact option appears; concept: ignore galvanometer only at balance.
Difficulty: Medium | Concept Tested: Wheatstone bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A meter bridge gives balance point 39.5 cm for X in left gap and 60.5 cm after interchanging X and R. If R=10 ohm, find X approximately ignoring end correction.
Correct Answer: C
Detailed Solution: For first setting X/R=l/(100-l)=39.5/60.5. X=10 x 39.5/60.5 = 6.53 ohm. Interchange reading checks consistency and end error.
Difficulty: Medium | Concept Tested: Meter bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A metallic wire carries current I. If both length and radius are doubled while the same potential difference is applied, drift velocity becomes:
Correct Answer: A
Detailed Solution: R=rho L/A. New R proportional 2L/(4A)=R/2, so current doubles. vd=I/(neA); area becomes 4A, so vd becomes 2/4 = 1/2. If length also changes field E=V/L, microscopic vd proportional E, so vd becomes half. Correct conceptual answer: one-half. This tests consistency of macro and micro pictures.
Difficulty: Difficult | Concept Tested: Drift velocity + geometry
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Twelve identical cells each emf E and internal resistance r are arranged as 3 in series per row and 4 rows in parallel. The equivalent source is:
Correct Answer: B
Detailed Solution: Each row has emf 3E and internal resistance 3r. Four identical rows in parallel keep emf 3E and reduce internal resistance to 3r/4.
Difficulty: Medium | Concept Tested: Mixed cell combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two resistors R and 2R are connected first in series and then in parallel across the same ideal battery. Ratio of total power in parallel to total power in series is:
Correct Answer: C
Detailed Solution: Series equivalent = 3R, P_s=V^2/(3R). Parallel equivalent = 2R/3, P_p=V^2/(2R/3)=3V^2/(2R). Ratio = (3/2)/(1/3)=9/2.
Difficulty: Medium | Concept Tested: Power and combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A 10 ohm wire with alpha=0.004/C is in series with a 20 ohm carbon resistor with alpha=-0.0005/C. Equivalent temperature coefficient is:
Correct Answer: B
Detailed Solution: For series, alpha_eq=(R1 alpha1 + R2 alpha2)/(R1+R2)=(10x.004+20x-.0005)/30=(.04-.01)/30=.001/C.
Difficulty: Difficult | Concept Tested: Temperature coefficient
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 5 ohm resistor. A voltmeter of resistance 10 ohm is connected across the 5 ohm resistor. What is the voltmeter reading?
Correct Answer: A
Detailed Solution: The voltmeter is not ideal. The external load is 5 parallel 10 = 10/3 ohm. Total resistance = 1+10/3=13/3 ohm. Current from cell = 6/(13/3)=18/13 A. Terminal/load voltage = I x 10/3 = 60/13 = 4.62 V. If options are adjusted for exact finite-voltmeter data, choose the nearest consistent value; here the concept is finite voltmeter loading.
Difficulty: Difficult | Concept Tested: Internal resistance + non-ideal meter
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a Wheatstone bridge, arms are 2 ohm, 3 ohm, 4 ohm and 6 ohm. A 10 ohm galvanometer is connected between the middle points. The equivalent resistance between supply terminals is closest to:
Correct Answer: B
Detailed Solution: Since 2/3 = 4/6, the bridge is balanced and no current flows through the galvanometer. Equivalent = (2+3) parallel (4+6) = 5 parallel 10 = 10/3 ohm. Nearest option is 3 ohm if exact option appears; concept: ignore galvanometer only at balance.
Difficulty: Medium | Concept Tested: Wheatstone bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A meter bridge gives balance point 39.5 cm for X in left gap and 60.5 cm after interchanging X and R. If R=10 ohm, find X approximately ignoring end correction.
Correct Answer: C
Detailed Solution: For first setting X/R=l/(100-l)=39.5/60.5. X=10 x 39.5/60.5 = 6.53 ohm. Interchange reading checks consistency and end error.
Difficulty: Medium | Concept Tested: Meter bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A metallic wire carries current I. If both length and radius are doubled while the same potential difference is applied, drift velocity becomes:
Correct Answer: A
Detailed Solution: R=rho L/A. New R proportional 2L/(4A)=R/2, so current doubles. vd=I/(neA); area becomes 4A, so vd becomes 2/4 = 1/2. If length also changes field E=V/L, microscopic vd proportional E, so vd becomes half. Correct conceptual answer: one-half. This tests consistency of macro and micro pictures.
Difficulty: Difficult | Concept Tested: Drift velocity + geometry
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Twelve identical cells each emf E and internal resistance r are arranged as 3 in series per row and 4 rows in parallel. The equivalent source is:
Correct Answer: B
Detailed Solution: Each row has emf 3E and internal resistance 3r. Four identical rows in parallel keep emf 3E and reduce internal resistance to 3r/4.
Difficulty: Medium | Concept Tested: Mixed cell combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two resistors R and 2R are connected first in series and then in parallel across the same ideal battery. Ratio of total power in parallel to total power in series is:
Correct Answer: C
Detailed Solution: Series equivalent = 3R, P_s=V^2/(3R). Parallel equivalent = 2R/3, P_p=V^2/(2R/3)=3V^2/(2R). Ratio = (3/2)/(1/3)=9/2.
Difficulty: Medium | Concept Tested: Power and combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A 10 ohm wire with alpha=0.004/C is in series with a 20 ohm carbon resistor with alpha=-0.0005/C. Equivalent temperature coefficient is:
Correct Answer: B
Detailed Solution: For series, alpha_eq=(R1 alpha1 + R2 alpha2)/(R1+R2)=(10x.004+20x-.0005)/30=(.04-.01)/30=.001/C.
Difficulty: Difficult | Concept Tested: Temperature coefficient
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 5 ohm resistor. A voltmeter of resistance 10 ohm is connected across the 5 ohm resistor. What is the voltmeter reading?
Correct Answer: A
Detailed Solution: The voltmeter is not ideal. The external load is 5 parallel 10 = 10/3 ohm. Total resistance = 1+10/3=13/3 ohm. Current from cell = 6/(13/3)=18/13 A. Terminal/load voltage = I x 10/3 = 60/13 = 4.62 V. If options are adjusted for exact finite-voltmeter data, choose the nearest consistent value; here the concept is finite voltmeter loading.
Difficulty: Difficult | Concept Tested: Internal resistance + non-ideal meter
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a Wheatstone bridge, arms are 2 ohm, 3 ohm, 4 ohm and 6 ohm. A 10 ohm galvanometer is connected between the middle points. The equivalent resistance between supply terminals is closest to:
Correct Answer: B
Detailed Solution: Since 2/3 = 4/6, the bridge is balanced and no current flows through the galvanometer. Equivalent = (2+3) parallel (4+6) = 5 parallel 10 = 10/3 ohm. Nearest option is 3 ohm if exact option appears; concept: ignore galvanometer only at balance.
Difficulty: Medium | Concept Tested: Wheatstone bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A meter bridge gives balance point 39.5 cm for X in left gap and 60.5 cm after interchanging X and R. If R=10 ohm, find X approximately ignoring end correction.
Correct Answer: C
Detailed Solution: For first setting X/R=l/(100-l)=39.5/60.5. X=10 x 39.5/60.5 = 6.53 ohm. Interchange reading checks consistency and end error.
Difficulty: Medium | Concept Tested: Meter bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A metallic wire carries current I. If both length and radius are doubled while the same potential difference is applied, drift velocity becomes:
Correct Answer: A
Detailed Solution: R=rho L/A. New R proportional 2L/(4A)=R/2, so current doubles. vd=I/(neA); area becomes 4A, so vd becomes 2/4 = 1/2. If length also changes field E=V/L, microscopic vd proportional E, so vd becomes half. Correct conceptual answer: one-half. This tests consistency of macro and micro pictures.
Difficulty: Difficult | Concept Tested: Drift velocity + geometry
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Twelve identical cells each emf E and internal resistance r are arranged as 3 in series per row and 4 rows in parallel. The equivalent source is:
Correct Answer: B
Detailed Solution: Each row has emf 3E and internal resistance 3r. Four identical rows in parallel keep emf 3E and reduce internal resistance to 3r/4.
Difficulty: Medium | Concept Tested: Mixed cell combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two resistors R and 2R are connected first in series and then in parallel across the same ideal battery. Ratio of total power in parallel to total power in series is:
Correct Answer: C
Detailed Solution: Series equivalent = 3R, P_s=V^2/(3R). Parallel equivalent = 2R/3, P_p=V^2/(2R/3)=3V^2/(2R). Ratio = (3/2)/(1/3)=9/2.
Difficulty: Medium | Concept Tested: Power and combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A 10 ohm wire with alpha=0.004/C is in series with a 20 ohm carbon resistor with alpha=-0.0005/C. Equivalent temperature coefficient is:
Correct Answer: B
Detailed Solution: For series, alpha_eq=(R1 alpha1 + R2 alpha2)/(R1+R2)=(10x.004+20x-.0005)/30=(.04-.01)/30=.001/C.
Difficulty: Difficult | Concept Tested: Temperature coefficient
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 5 ohm resistor. A voltmeter of resistance 10 ohm is connected across the 5 ohm resistor. What is the voltmeter reading?
Correct Answer: A
Detailed Solution: The voltmeter is not ideal. The external load is 5 parallel 10 = 10/3 ohm. Total resistance = 1+10/3=13/3 ohm. Current from cell = 6/(13/3)=18/13 A. Terminal/load voltage = I x 10/3 = 60/13 = 4.62 V. If options are adjusted for exact finite-voltmeter data, choose the nearest consistent value; here the concept is finite voltmeter loading.
Difficulty: Difficult | Concept Tested: Internal resistance + non-ideal meter
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a Wheatstone bridge, arms are 2 ohm, 3 ohm, 4 ohm and 6 ohm. A 10 ohm galvanometer is connected between the middle points. The equivalent resistance between supply terminals is closest to:
Correct Answer: B
Detailed Solution: Since 2/3 = 4/6, the bridge is balanced and no current flows through the galvanometer. Equivalent = (2+3) parallel (4+6) = 5 parallel 10 = 10/3 ohm. Nearest option is 3 ohm if exact option appears; concept: ignore galvanometer only at balance.
Difficulty: Medium | Concept Tested: Wheatstone bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A meter bridge gives balance point 39.5 cm for X in left gap and 60.5 cm after interchanging X and R. If R=10 ohm, find X approximately ignoring end correction.
Correct Answer: C
Detailed Solution: For first setting X/R=l/(100-l)=39.5/60.5. X=10 x 39.5/60.5 = 6.53 ohm. Interchange reading checks consistency and end error.
Difficulty: Medium | Concept Tested: Meter bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A metallic wire carries current I. If both length and radius are doubled while the same potential difference is applied, drift velocity becomes:
Correct Answer: A
Detailed Solution: R=rho L/A. New R proportional 2L/(4A)=R/2, so current doubles. vd=I/(neA); area becomes 4A, so vd becomes 2/4 = 1/2. If length also changes field E=V/L, microscopic vd proportional E, so vd becomes half. Correct conceptual answer: one-half. This tests consistency of macro and micro pictures.
Difficulty: Difficult | Concept Tested: Drift velocity + geometry
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Twelve identical cells each emf E and internal resistance r are arranged as 3 in series per row and 4 rows in parallel. The equivalent source is:
Correct Answer: B
Detailed Solution: Each row has emf 3E and internal resistance 3r. Four identical rows in parallel keep emf 3E and reduce internal resistance to 3r/4.
Difficulty: Medium | Concept Tested: Mixed cell combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two resistors R and 2R are connected first in series and then in parallel across the same ideal battery. Ratio of total power in parallel to total power in series is:
Correct Answer: C
Detailed Solution: Series equivalent = 3R, P_s=V^2/(3R). Parallel equivalent = 2R/3, P_p=V^2/(2R/3)=3V^2/(2R). Ratio = (3/2)/(1/3)=9/2.
Difficulty: Medium | Concept Tested: Power and combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A 10 ohm wire with alpha=0.004/C is in series with a 20 ohm carbon resistor with alpha=-0.0005/C. Equivalent temperature coefficient is:
Correct Answer: B
Detailed Solution: For series, alpha_eq=(R1 alpha1 + R2 alpha2)/(R1+R2)=(10x.004+20x-.0005)/30=(.04-.01)/30=.001/C.
Difficulty: Difficult | Concept Tested: Temperature coefficient
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 5 ohm resistor. A voltmeter of resistance 10 ohm is connected across the 5 ohm resistor. What is the voltmeter reading?
Correct Answer: A
Detailed Solution: The voltmeter is not ideal. The external load is 5 parallel 10 = 10/3 ohm. Total resistance = 1+10/3=13/3 ohm. Current from cell = 6/(13/3)=18/13 A. Terminal/load voltage = I x 10/3 = 60/13 = 4.62 V. If options are adjusted for exact finite-voltmeter data, choose the nearest consistent value; here the concept is finite voltmeter loading.
Difficulty: Difficult | Concept Tested: Internal resistance + non-ideal meter
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a Wheatstone bridge, arms are 2 ohm, 3 ohm, 4 ohm and 6 ohm. A 10 ohm galvanometer is connected between the middle points. The equivalent resistance between supply terminals is closest to:
Correct Answer: B
Detailed Solution: Since 2/3 = 4/6, the bridge is balanced and no current flows through the galvanometer. Equivalent = (2+3) parallel (4+6) = 5 parallel 10 = 10/3 ohm. Nearest option is 3 ohm if exact option appears; concept: ignore galvanometer only at balance.
Difficulty: Medium | Concept Tested: Wheatstone bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A square has four 4 ohm sides and a 4 ohm diagonal between opposite corners. Equivalent resistance between the other two opposite corners is:
Correct Answer: C
Detailed Solution: The diagonal connects points at equal potential by symmetry for terminals at the other diagonal, so no current flows in it. Equivalent is two paths of 8 ohm in parallel = 4 ohm.
Difficulty: Medium | Concept Tested: Symmetry bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A source of emf 24 V and internal resistance 6 ohm supplies a variable load R. At what R is load power maximum, and what is maximum power?
Correct Answer: A
Detailed Solution: Maximum power transfer occurs at R=r=6 ohm. Pmax=E^2/(4r)=576/24=24 W.
Difficulty: Medium | Concept Tested: Maximum power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of resistance 99 ohm gives full scale at 1 mA. What shunt makes it an ammeter of range 1 A?
Correct Answer: D
Detailed Solution: At full scale, Ig=0.001 A through galvanometer and 0.999 A through shunt. Same voltage: Ig G = Is S. S=0.001x99/0.999=0.099 ohm.
Difficulty: Medium | Concept Tested: Shunt
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of 50 ohm resistance reads full scale at 2 mA. Series resistance needed for 10 V voltmeter is:
Correct Answer: C
Detailed Solution: Total resistance needed V/Ig=10/0.002=5000 ohm. Series resistance = 5000-50=4950 ohm.
Difficulty: Medium | Concept Tested: Meter conversion
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Terminal voltage of a cell falls from 1.5 V at zero current to 1.1 V at 2 A. Internal resistance is:
Correct Answer: A
Detailed Solution: V=E-Ir. Slope magnitude of V-I graph is r=(1.5-1.1)/2=0.2 ohm.
Difficulty: Medium | Concept Tested: V-I graph
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A wire connected to constant voltage is stretched to three times its length, volume constant. The heat produced per second becomes:
Correct Answer: B
Detailed Solution: R proportional L^2 for constant volume, so R becomes 9R. At constant V, P=V^2/R becomes P/9.
Difficulty: Medium | Concept Tested: Resistance geometry + power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A square has four 4 ohm sides and a 4 ohm diagonal between opposite corners. Equivalent resistance between the other two opposite corners is:
Correct Answer: C
Detailed Solution: The diagonal connects points at equal potential by symmetry for terminals at the other diagonal, so no current flows in it. Equivalent is two paths of 8 ohm in parallel = 4 ohm.
Difficulty: Medium | Concept Tested: Symmetry bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A source of emf 24 V and internal resistance 6 ohm supplies a variable load R. At what R is load power maximum, and what is maximum power?
Correct Answer: A
Detailed Solution: Maximum power transfer occurs at R=r=6 ohm. Pmax=E^2/(4r)=576/24=24 W.
Difficulty: Medium | Concept Tested: Maximum power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of resistance 99 ohm gives full scale at 1 mA. What shunt makes it an ammeter of range 1 A?
Correct Answer: D
Detailed Solution: At full scale, Ig=0.001 A through galvanometer and 0.999 A through shunt. Same voltage: Ig G = Is S. S=0.001x99/0.999=0.099 ohm.
Difficulty: Medium | Concept Tested: Shunt
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of 50 ohm resistance reads full scale at 2 mA. Series resistance needed for 10 V voltmeter is:
Correct Answer: C
Detailed Solution: Total resistance needed V/Ig=10/0.002=5000 ohm. Series resistance = 5000-50=4950 ohm.
Difficulty: Medium | Concept Tested: Meter conversion
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Terminal voltage of a cell falls from 1.5 V at zero current to 1.1 V at 2 A. Internal resistance is:
Correct Answer: A
Detailed Solution: V=E-Ir. Slope magnitude of V-I graph is r=(1.5-1.1)/2=0.2 ohm.
Difficulty: Medium | Concept Tested: V-I graph
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A wire connected to constant voltage is stretched to three times its length, volume constant. The heat produced per second becomes:
Correct Answer: B
Detailed Solution: R proportional L^2 for constant volume, so R becomes 9R. At constant V, P=V^2/R becomes P/9.
Difficulty: Medium | Concept Tested: Resistance geometry + power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A square has four 4 ohm sides and a 4 ohm diagonal between opposite corners. Equivalent resistance between the other two opposite corners is:
Correct Answer: C
Detailed Solution: The diagonal connects points at equal potential by symmetry for terminals at the other diagonal, so no current flows in it. Equivalent is two paths of 8 ohm in parallel = 4 ohm.
Difficulty: Medium | Concept Tested: Symmetry bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A source of emf 24 V and internal resistance 6 ohm supplies a variable load R. At what R is load power maximum, and what is maximum power?
Correct Answer: A
Detailed Solution: Maximum power transfer occurs at R=r=6 ohm. Pmax=E^2/(4r)=576/24=24 W.
Difficulty: Medium | Concept Tested: Maximum power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of resistance 99 ohm gives full scale at 1 mA. What shunt makes it an ammeter of range 1 A?
Correct Answer: D
Detailed Solution: At full scale, Ig=0.001 A through galvanometer and 0.999 A through shunt. Same voltage: Ig G = Is S. S=0.001x99/0.999=0.099 ohm.
Difficulty: Medium | Concept Tested: Shunt
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of 50 ohm resistance reads full scale at 2 mA. Series resistance needed for 10 V voltmeter is:
Correct Answer: C
Detailed Solution: Total resistance needed V/Ig=10/0.002=5000 ohm. Series resistance = 5000-50=4950 ohm.
Difficulty: Medium | Concept Tested: Meter conversion
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Terminal voltage of a cell falls from 1.5 V at zero current to 1.1 V at 2 A. Internal resistance is:
Correct Answer: A
Detailed Solution: V=E-Ir. Slope magnitude of V-I graph is r=(1.5-1.1)/2=0.2 ohm.
Difficulty: Medium | Concept Tested: V-I graph
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A wire connected to constant voltage is stretched to three times its length, volume constant. The heat produced per second becomes:
Correct Answer: B
Detailed Solution: R proportional L^2 for constant volume, so R becomes 9R. At constant V, P=V^2/R becomes P/9.
Difficulty: Medium | Concept Tested: Resistance geometry + power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A square has four 4 ohm sides and a 4 ohm diagonal between opposite corners. Equivalent resistance between the other two opposite corners is:
Correct Answer: C
Detailed Solution: The diagonal connects points at equal potential by symmetry for terminals at the other diagonal, so no current flows in it. Equivalent is two paths of 8 ohm in parallel = 4 ohm.
Difficulty: Medium | Concept Tested: Symmetry bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A source of emf 24 V and internal resistance 6 ohm supplies a variable load R. At what R is load power maximum, and what is maximum power?
Correct Answer: A
Detailed Solution: Maximum power transfer occurs at R=r=6 ohm. Pmax=E^2/(4r)=576/24=24 W.
Difficulty: Medium | Concept Tested: Maximum power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of resistance 99 ohm gives full scale at 1 mA. What shunt makes it an ammeter of range 1 A?
Correct Answer: D
Detailed Solution: At full scale, Ig=0.001 A through galvanometer and 0.999 A through shunt. Same voltage: Ig G = Is S. S=0.001x99/0.999=0.099 ohm.
Difficulty: Medium | Concept Tested: Shunt
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of 50 ohm resistance reads full scale at 2 mA. Series resistance needed for 10 V voltmeter is:
Correct Answer: C
Detailed Solution: Total resistance needed V/Ig=10/0.002=5000 ohm. Series resistance = 5000-50=4950 ohm.
Difficulty: Medium | Concept Tested: Meter conversion
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Terminal voltage of a cell falls from 1.5 V at zero current to 1.1 V at 2 A. Internal resistance is:
Correct Answer: A
Detailed Solution: V=E-Ir. Slope magnitude of V-I graph is r=(1.5-1.1)/2=0.2 ohm.
Difficulty: Medium | Concept Tested: V-I graph
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A wire connected to constant voltage is stretched to three times its length, volume constant. The heat produced per second becomes:
Correct Answer: B
Detailed Solution: R proportional L^2 for constant volume, so R becomes 9R. At constant V, P=V^2/R becomes P/9.
Difficulty: Medium | Concept Tested: Resistance geometry + power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A square has four 4 ohm sides and a 4 ohm diagonal between opposite corners. Equivalent resistance between the other two opposite corners is:
Correct Answer: C
Detailed Solution: The diagonal connects points at equal potential by symmetry for terminals at the other diagonal, so no current flows in it. Equivalent is two paths of 8 ohm in parallel = 4 ohm.
Difficulty: Medium | Concept Tested: Symmetry bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A source of emf 24 V and internal resistance 6 ohm supplies a variable load R. At what R is load power maximum, and what is maximum power?
Correct Answer: A
Detailed Solution: Maximum power transfer occurs at R=r=6 ohm. Pmax=E^2/(4r)=576/24=24 W.
Difficulty: Medium | Concept Tested: Maximum power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of resistance 99 ohm gives full scale at 1 mA. What shunt makes it an ammeter of range 1 A?
Correct Answer: D
Detailed Solution: At full scale, Ig=0.001 A through galvanometer and 0.999 A through shunt. Same voltage: Ig G = Is S. S=0.001x99/0.999=0.099 ohm.
Difficulty: Medium | Concept Tested: Shunt
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of 50 ohm resistance reads full scale at 2 mA. Series resistance needed for 10 V voltmeter is:
Correct Answer: C
Detailed Solution: Total resistance needed V/Ig=10/0.002=5000 ohm. Series resistance = 5000-50=4950 ohm.
Difficulty: Medium | Concept Tested: Meter conversion
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Terminal voltage of a cell falls from 1.5 V at zero current to 1.1 V at 2 A. Internal resistance is:
Correct Answer: A
Detailed Solution: V=E-Ir. Slope magnitude of V-I graph is r=(1.5-1.1)/2=0.2 ohm.
Difficulty: Medium | Concept Tested: V-I graph
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A wire connected to constant voltage is stretched to three times its length, volume constant. The heat produced per second becomes:
Correct Answer: B
Detailed Solution: R proportional L^2 for constant volume, so R becomes 9R. At constant V, P=V^2/R becomes P/9.
Difficulty: Medium | Concept Tested: Resistance geometry + power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A square has four 4 ohm sides and a 4 ohm diagonal between opposite corners. Equivalent resistance between the other two opposite corners is:
Correct Answer: C
Detailed Solution: The diagonal connects points at equal potential by symmetry for terminals at the other diagonal, so no current flows in it. Equivalent is two paths of 8 ohm in parallel = 4 ohm.
Difficulty: Medium | Concept Tested: Symmetry bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A source of emf 24 V and internal resistance 6 ohm supplies a variable load R. At what R is load power maximum, and what is maximum power?
Correct Answer: A
Detailed Solution: Maximum power transfer occurs at R=r=6 ohm. Pmax=E^2/(4r)=576/24=24 W.
Difficulty: Medium | Concept Tested: Maximum power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of resistance 99 ohm gives full scale at 1 mA. What shunt makes it an ammeter of range 1 A?
Correct Answer: D
Detailed Solution: At full scale, Ig=0.001 A through galvanometer and 0.999 A through shunt. Same voltage: Ig G = Is S. S=0.001x99/0.999=0.099 ohm.
Difficulty: Medium | Concept Tested: Shunt
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of 50 ohm resistance reads full scale at 2 mA. Series resistance needed for 10 V voltmeter is:
Correct Answer: C
Detailed Solution: Total resistance needed V/Ig=10/0.002=5000 ohm. Series resistance = 5000-50=4950 ohm.
Difficulty: Medium | Concept Tested: Meter conversion
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Terminal voltage of a cell falls from 1.5 V at zero current to 1.1 V at 2 A. Internal resistance is:
Correct Answer: A
Detailed Solution: V=E-Ir. Slope magnitude of V-I graph is r=(1.5-1.1)/2=0.2 ohm.
Difficulty: Medium | Concept Tested: V-I graph
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A wire connected to constant voltage is stretched to three times its length, volume constant. The heat produced per second becomes:
Correct Answer: B
Detailed Solution: R proportional L^2 for constant volume, so R becomes 9R. At constant V, P=V^2/R becomes P/9.
Difficulty: Medium | Concept Tested: Resistance geometry + power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
An infinite ladder has each section: a 2 ohm series resistor followed by a 2 ohm shunt resistor to return. Find equivalent resistance in ohm from the input.
Correct Answer: A
Detailed Solution: Let equivalent be X. After first series 2 ohm, remaining network X is in parallel with 2 ohm shunt: X=2+(2X)/(2+X). This gives X^2-2X-4=0, so X=1+sqrt(5).
Difficulty: JEE Advanced | Concept Tested: Infinite network
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A battery has n identical cells. Match condition R much greater than r and R much smaller than r with best grouping for maximum current.
Correct Answer: B
Detailed Solution: Current for m rows each with s cells: I=sE/(R+sr/m), sm=n. For high external R, increase emf: series. For low external R, reduce internal resistance: parallel.
Difficulty: JEE Advanced | Concept Tested: Cell grouping optimization
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a two-cell circuit, a 12 V cell with 2 ohm internal resistance is connected opposite to a 6 V cell with 1 ohm internal resistance and external 3 ohm. Current direction is:
Correct Answer: A
Detailed Solution: Net emf=12-6=6 V, total resistance=2+1+3=6 ohm, current=1 A in direction of stronger source.
Difficulty: JEE Advanced | Concept Tested: Opposing cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Mesh currents I1=3 A and I2=1 A pass in opposite directions through a common 4 ohm resistor. Power in common resistor is:
Correct Answer: D
Detailed Solution: Actual current in shared resistor = |I1-I2|=2 A. Power=I^2R=4x4=16 W.
Difficulty: JEE Advanced | Concept Tested: Mesh current
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
If relaxation time doubles and number density halves, conductivity becomes:
Correct Answer: C
Detailed Solution: sigma=ne^2 tau/m. n becomes n/2 and tau becomes 2tau, product remains same.
Difficulty: JEE Advanced | Concept Tested: Microscopic Ohm law
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
An infinite ladder has each section: a 2 ohm series resistor followed by a 2 ohm shunt resistor to return. Find equivalent resistance in ohm from the input.
Correct Answer: A
Detailed Solution: Let equivalent be X. After first series 2 ohm, remaining network X is in parallel with 2 ohm shunt: X=2+(2X)/(2+X). This gives X^2-2X-4=0, so X=1+sqrt(5).
Difficulty: JEE Advanced | Concept Tested: Infinite network
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A battery has n identical cells. Match condition R much greater than r and R much smaller than r with best grouping for maximum current.
Correct Answer: B
Detailed Solution: Current for m rows each with s cells: I=sE/(R+sr/m), sm=n. For high external R, increase emf: series. For low external R, reduce internal resistance: parallel.
Difficulty: JEE Advanced | Concept Tested: Cell grouping optimization
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a two-cell circuit, a 12 V cell with 2 ohm internal resistance is connected opposite to a 6 V cell with 1 ohm internal resistance and external 3 ohm. Current direction is:
Correct Answer: A
Detailed Solution: Net emf=12-6=6 V, total resistance=2+1+3=6 ohm, current=1 A in direction of stronger source.
Difficulty: JEE Advanced | Concept Tested: Opposing cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Mesh currents I1=3 A and I2=1 A pass in opposite directions through a common 4 ohm resistor. Power in common resistor is:
Correct Answer: D
Detailed Solution: Actual current in shared resistor = |I1-I2|=2 A. Power=I^2R=4x4=16 W.
Difficulty: JEE Advanced | Concept Tested: Mesh current
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
If relaxation time doubles and number density halves, conductivity becomes:
Correct Answer: C
Detailed Solution: sigma=ne^2 tau/m. n becomes n/2 and tau becomes 2tau, product remains same.
Difficulty: JEE Advanced | Concept Tested: Microscopic Ohm law
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
An infinite ladder has each section: a 2 ohm series resistor followed by a 2 ohm shunt resistor to return. Find equivalent resistance in ohm from the input.
Correct Answer: A
Detailed Solution: Let equivalent be X. After first series 2 ohm, remaining network X is in parallel with 2 ohm shunt: X=2+(2X)/(2+X). This gives X^2-2X-4=0, so X=1+sqrt(5).
Difficulty: JEE Advanced | Concept Tested: Infinite network
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A battery has n identical cells. Match condition R much greater than r and R much smaller than r with best grouping for maximum current.
Correct Answer: B
Detailed Solution: Current for m rows each with s cells: I=sE/(R+sr/m), sm=n. For high external R, increase emf: series. For low external R, reduce internal resistance: parallel.
Difficulty: JEE Advanced | Concept Tested: Cell grouping optimization
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a two-cell circuit, a 12 V cell with 2 ohm internal resistance is connected opposite to a 6 V cell with 1 ohm internal resistance and external 3 ohm. Current direction is:
Correct Answer: A
Detailed Solution: Net emf=12-6=6 V, total resistance=2+1+3=6 ohm, current=1 A in direction of stronger source.
Difficulty: JEE Advanced | Concept Tested: Opposing cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Mesh currents I1=3 A and I2=1 A pass in opposite directions through a common 4 ohm resistor. Power in common resistor is:
Correct Answer: D
Detailed Solution: Actual current in shared resistor = |I1-I2|=2 A. Power=I^2R=4x4=16 W.
Difficulty: JEE Advanced | Concept Tested: Mesh current
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
If relaxation time doubles and number density halves, conductivity becomes:
Correct Answer: C
Detailed Solution: sigma=ne^2 tau/m. n becomes n/2 and tau becomes 2tau, product remains same.
Difficulty: JEE Advanced | Concept Tested: Microscopic Ohm law
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
An infinite ladder has each section: a 2 ohm series resistor followed by a 2 ohm shunt resistor to return. Find equivalent resistance in ohm from the input.
Correct Answer: A
Detailed Solution: Let equivalent be X. After first series 2 ohm, remaining network X is in parallel with 2 ohm shunt: X=2+(2X)/(2+X). This gives X^2-2X-4=0, so X=1+sqrt(5).
Difficulty: JEE Advanced | Concept Tested: Infinite network
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A battery has n identical cells. Match condition R much greater than r and R much smaller than r with best grouping for maximum current.
Correct Answer: B
Detailed Solution: Current for m rows each with s cells: I=sE/(R+sr/m), sm=n. For high external R, increase emf: series. For low external R, reduce internal resistance: parallel.
Difficulty: JEE Advanced | Concept Tested: Cell grouping optimization
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a two-cell circuit, a 12 V cell with 2 ohm internal resistance is connected opposite to a 6 V cell with 1 ohm internal resistance and external 3 ohm. Current direction is:
Correct Answer: A
Detailed Solution: Net emf=12-6=6 V, total resistance=2+1+3=6 ohm, current=1 A in direction of stronger source.
Difficulty: JEE Advanced | Concept Tested: Opposing cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Mesh currents I1=3 A and I2=1 A pass in opposite directions through a common 4 ohm resistor. Power in common resistor is:
Correct Answer: D
Detailed Solution: Actual current in shared resistor = |I1-I2|=2 A. Power=I^2R=4x4=16 W.
Difficulty: JEE Advanced | Concept Tested: Mesh current
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
If relaxation time doubles and number density halves, conductivity becomes:
Correct Answer: C
Detailed Solution: sigma=ne^2 tau/m. n becomes n/2 and tau becomes 2tau, product remains same.
Difficulty: JEE Advanced | Concept Tested: Microscopic Ohm law
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
An infinite ladder has each section: a 2 ohm series resistor followed by a 2 ohm shunt resistor to return. Find equivalent resistance in ohm from the input.
Correct Answer: A
Detailed Solution: Let equivalent be X. After first series 2 ohm, remaining network X is in parallel with 2 ohm shunt: X=2+(2X)/(2+X). This gives X^2-2X-4=0, so X=1+sqrt(5).
Difficulty: JEE Advanced | Concept Tested: Infinite network
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A battery has n identical cells. Match condition R much greater than r and R much smaller than r with best grouping for maximum current.
Correct Answer: B
Detailed Solution: Current for m rows each with s cells: I=sE/(R+sr/m), sm=n. For high external R, increase emf: series. For low external R, reduce internal resistance: parallel.
Difficulty: JEE Advanced | Concept Tested: Cell grouping optimization
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a two-cell circuit, a 12 V cell with 2 ohm internal resistance is connected opposite to a 6 V cell with 1 ohm internal resistance and external 3 ohm. Current direction is:
Correct Answer: A
Detailed Solution: Net emf=12-6=6 V, total resistance=2+1+3=6 ohm, current=1 A in direction of stronger source.
Difficulty: JEE Advanced | Concept Tested: Opposing cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Mesh currents I1=3 A and I2=1 A pass in opposite directions through a common 4 ohm resistor. Power in common resistor is:
Correct Answer: D
Detailed Solution: Actual current in shared resistor = |I1-I2|=2 A. Power=I^2R=4x4=16 W.
Difficulty: JEE Advanced | Concept Tested: Mesh current
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
If relaxation time doubles and number density halves, conductivity becomes:
Correct Answer: C
Detailed Solution: sigma=ne^2 tau/m. n becomes n/2 and tau becomes 2tau, product remains same.
Difficulty: JEE Advanced | Concept Tested: Microscopic Ohm law
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
The uploaded PDF is mostly scanned/image-based. I visually used its repeated question styles: resistance variation, wire stretching, combination of resistors, Kirchhoff network diagrams, Wheatstone bridge, meter bridge, potentiometer, cells and temperature coefficient. The following problems are reconstructed in clean English with professional SVG-style circuit logic and full solutions.
A cell of emf 2.2 V is connected to a 4 ohm resistor. Terminal voltage is 2.0 V. Find internal resistance and explain why terminal voltage is less than emf.
Final Answer: 0.4 ohm
Presentation: Current I=V/R=2/4=0.5 A. E=V+Ir, so r=(E-V)/I=0.2/0.5=0.4 ohm. Terminal voltage is less because Ir is lost inside the cell.
Concept Tested: Internal resistance
In a meter bridge, the balance point is 35 cm from the left end when unknown X is in the left gap and 10 ohm is in the right gap. Calculate X. Why should balance point be near the middle?
Final Answer: 5.38 ohm
Presentation: X/10=35/65, so X=5.38 ohm. Near middle, percentage error due to end resistance and scale reading is minimized.
Concept Tested: Meter bridge
A cell balances at 75 cm on a potentiometer wire. Another cell balances at 100 cm. If second emf is 2 V, find first emf. State one advantage of potentiometer over voltmeter.
Final Answer: 1.5 V
Presentation: E1/E2=l1/l2=75/100. E1=1.5 V. Advantage: at balance, potentiometer draws no current from test cell, so it measures emf accurately.
Concept Tested: Potentiometer
Using Kirchhoff rules, write equations for a circuit having two loops with a common resistor R3. Mesh currents I1 and I2 pass through R3 in opposite directions.
Final Answer: Equations
Presentation: Left loop: E1-I1R1-(I1-I2)R3=0. Right loop: E2-I2R2-(I2-I1)R3=0, with signs chosen according to loop direction.
Concept Tested: Kirchhoff rules
A wire has resistance 10 ohm at 27 C and 12 ohm at 127 C. Calculate temperature coefficient of resistance.
Final Answer: 0.002 per C
Presentation: R=R0(1+alpha DeltaT). 12=10(1+100alpha), alpha=0.002 per C.
Concept Tested: Temperature dependence
Derive I=neAvd and calculate vd for I=1.6 A, n=8e28 m^-3, A=2e-6 m^2.
Final Answer: 6.25e-5 m/s
Presentation: vd=I/(neA)=1.6/(8e28 x 1.6e-19 x 2e-6)=6.25e-5 m/s.
Concept Tested: Drift velocity
A cell of emf 2.2 V is connected to a 4 ohm resistor. Terminal voltage is 2.0 V. Find internal resistance and explain why terminal voltage is less than emf.
Final Answer: 0.4 ohm
Presentation: Current I=V/R=2/4=0.5 A. E=V+Ir, so r=(E-V)/I=0.2/0.5=0.4 ohm. Terminal voltage is less because Ir is lost inside the cell.
Concept Tested: Internal resistance
In a meter bridge, the balance point is 35 cm from the left end when unknown X is in the left gap and 10 ohm is in the right gap. Calculate X. Why should balance point be near the middle?
Final Answer: 5.38 ohm
Presentation: X/10=35/65, so X=5.38 ohm. Near middle, percentage error due to end resistance and scale reading is minimized.
Concept Tested: Meter bridge
A cell balances at 75 cm on a potentiometer wire. Another cell balances at 100 cm. If second emf is 2 V, find first emf. State one advantage of potentiometer over voltmeter.
Final Answer: 1.5 V
Presentation: E1/E2=l1/l2=75/100. E1=1.5 V. Advantage: at balance, potentiometer draws no current from test cell, so it measures emf accurately.
Concept Tested: Potentiometer
Using Kirchhoff rules, write equations for a circuit having two loops with a common resistor R3. Mesh currents I1 and I2 pass through R3 in opposite directions.
Final Answer: Equations
Presentation: Left loop: E1-I1R1-(I1-I2)R3=0. Right loop: E2-I2R2-(I2-I1)R3=0, with signs chosen according to loop direction.
Concept Tested: Kirchhoff rules
A wire has resistance 10 ohm at 27 C and 12 ohm at 127 C. Calculate temperature coefficient of resistance.
Final Answer: 0.002 per C
Presentation: R=R0(1+alpha DeltaT). 12=10(1+100alpha), alpha=0.002 per C.
Concept Tested: Temperature dependence
Derive I=neAvd and calculate vd for I=1.6 A, n=8e28 m^-3, A=2e-6 m^2.
Final Answer: 6.25e-5 m/s
Presentation: vd=I/(neA)=1.6/(8e28 x 1.6e-19 x 2e-6)=6.25e-5 m/s.
Concept Tested: Drift velocity
A cell of emf 6 V and internal resistance 1 ohm is connected to a 5 ohm resistor. A voltmeter of resistance 10 ohm is connected across the 5 ohm resistor. What is the voltmeter reading?
Correct Answer: A
Detailed Solution: The voltmeter is not ideal. The external load is 5 parallel 10 = 10/3 ohm. Total resistance = 1+10/3=13/3 ohm. Current from cell = 6/(13/3)=18/13 A. Terminal/load voltage = I x 10/3 = 60/13 = 4.62 V. If options are adjusted for exact finite-voltmeter data, choose the nearest consistent value; here the concept is finite voltmeter loading.
Difficulty: Difficult | Concept Tested: Internal resistance + non-ideal meter
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a Wheatstone bridge, arms are 2 ohm, 3 ohm, 4 ohm and 6 ohm. A 10 ohm galvanometer is connected between the middle points. The equivalent resistance between supply terminals is closest to:
Correct Answer: B
Detailed Solution: Since 2/3 = 4/6, the bridge is balanced and no current flows through the galvanometer. Equivalent = (2+3) parallel (4+6) = 5 parallel 10 = 10/3 ohm. Nearest option is 3 ohm if exact option appears; concept: ignore galvanometer only at balance.
Difficulty: Medium | Concept Tested: Wheatstone bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A meter bridge gives balance point 39.5 cm for X in left gap and 60.5 cm after interchanging X and R. If R=10 ohm, find X approximately ignoring end correction.
Correct Answer: C
Detailed Solution: For first setting X/R=l/(100-l)=39.5/60.5. X=10 x 39.5/60.5 = 6.53 ohm. Interchange reading checks consistency and end error.
Difficulty: Medium | Concept Tested: Meter bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A metallic wire carries current I. If both length and radius are doubled while the same potential difference is applied, drift velocity becomes:
Correct Answer: A
Detailed Solution: R=rho L/A. New R proportional 2L/(4A)=R/2, so current doubles. vd=I/(neA); area becomes 4A, so vd becomes 2/4 = 1/2. If length also changes field E=V/L, microscopic vd proportional E, so vd becomes half. Correct conceptual answer: one-half. This tests consistency of macro and micro pictures.
Difficulty: Difficult | Concept Tested: Drift velocity + geometry
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Twelve identical cells each emf E and internal resistance r are arranged as 3 in series per row and 4 rows in parallel. The equivalent source is:
Correct Answer: B
Detailed Solution: Each row has emf 3E and internal resistance 3r. Four identical rows in parallel keep emf 3E and reduce internal resistance to 3r/4.
Difficulty: Medium | Concept Tested: Mixed cell combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two resistors R and 2R are connected first in series and then in parallel across the same ideal battery. Ratio of total power in parallel to total power in series is:
Correct Answer: C
Detailed Solution: Series equivalent = 3R, P_s=V^2/(3R). Parallel equivalent = 2R/3, P_p=V^2/(2R/3)=3V^2/(2R). Ratio = (3/2)/(1/3)=9/2.
Difficulty: Medium | Concept Tested: Power and combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A 10 ohm wire with alpha=0.004/C is in series with a 20 ohm carbon resistor with alpha=-0.0005/C. Equivalent temperature coefficient is:
Correct Answer: B
Detailed Solution: For series, alpha_eq=(R1 alpha1 + R2 alpha2)/(R1+R2)=(10x.004+20x-.0005)/30=(.04-.01)/30=.001/C.
Difficulty: Difficult | Concept Tested: Temperature coefficient
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A square has four 4 ohm sides and a 4 ohm diagonal between opposite corners. Equivalent resistance between the other two opposite corners is:
Correct Answer: C
Detailed Solution: The diagonal connects points at equal potential by symmetry for terminals at the other diagonal, so no current flows in it. Equivalent is two paths of 8 ohm in parallel = 4 ohm.
Difficulty: Medium | Concept Tested: Symmetry bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A source of emf 24 V and internal resistance 6 ohm supplies a variable load R. At what R is load power maximum, and what is maximum power?
Correct Answer: A
Detailed Solution: Maximum power transfer occurs at R=r=6 ohm. Pmax=E^2/(4r)=576/24=24 W.
Difficulty: Medium | Concept Tested: Maximum power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Power loss, drift velocity, equivalent resistance, terminal voltage, cells in series/parallel, meter bridge length ratio and potentiometer comparison repeat because they test one formula with a hidden condition.
Mixed resistance networks, internal resistance, non-ideal cells, Kirchhoff loop equations and Wheatstone balance appear repeatedly because they mix algebra with physical interpretation.
Unbalanced bridge, current reversal, symmetric networks, potentiometer sensitivity, maximum power and equivalent source methods are high-value multi-concept areas.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 5 ohm resistor. A voltmeter of resistance 10 ohm is connected across the 5 ohm resistor. What is the voltmeter reading?
Correct Answer: A
Detailed Solution: The voltmeter is not ideal. The external load is 5 parallel 10 = 10/3 ohm. Total resistance = 1+10/3=13/3 ohm. Current from cell = 6/(13/3)=18/13 A. Terminal/load voltage = I x 10/3 = 60/13 = 4.62 V. If options are adjusted for exact finite-voltmeter data, choose the nearest consistent value; here the concept is finite voltmeter loading.
Difficulty: Difficult | Concept Tested: Internal resistance + non-ideal meter
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a Wheatstone bridge, arms are 2 ohm, 3 ohm, 4 ohm and 6 ohm. A 10 ohm galvanometer is connected between the middle points. The equivalent resistance between supply terminals is closest to:
Correct Answer: B
Detailed Solution: Since 2/3 = 4/6, the bridge is balanced and no current flows through the galvanometer. Equivalent = (2+3) parallel (4+6) = 5 parallel 10 = 10/3 ohm. Nearest option is 3 ohm if exact option appears; concept: ignore galvanometer only at balance.
Difficulty: Medium | Concept Tested: Wheatstone bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A meter bridge gives balance point 39.5 cm for X in left gap and 60.5 cm after interchanging X and R. If R=10 ohm, find X approximately ignoring end correction.
Correct Answer: C
Detailed Solution: For first setting X/R=l/(100-l)=39.5/60.5. X=10 x 39.5/60.5 = 6.53 ohm. Interchange reading checks consistency and end error.
Difficulty: Medium | Concept Tested: Meter bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer wire has potential gradient 0.015 V/cm. A cell balances at 86 cm. When a 5 ohm resistor is connected across it, balance length becomes 75 cm. Internal resistance is:
Correct Answer: D
Detailed Solution: r=R(l1/l2-1)=5(86/75-1)=5(11/75)=0.733 ohm. Potential gradient cancels; it is a common trap.
Difficulty: Medium | Concept Tested: Potentiometer internal resistance
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A metallic wire carries current I. If both length and radius are doubled while the same potential difference is applied, drift velocity becomes:
Correct Answer: A
Detailed Solution: R=rho L/A. New R proportional 2L/(4A)=R/2, so current doubles. vd=I/(neA); area becomes 4A, so vd becomes 2/4 = 1/2. If length also changes field E=V/L, microscopic vd proportional E, so vd becomes half. Correct conceptual answer: one-half. This tests consistency of macro and micro pictures.
Difficulty: Difficult | Concept Tested: Drift velocity + geometry
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Twelve identical cells each emf E and internal resistance r are arranged as 3 in series per row and 4 rows in parallel. The equivalent source is:
Correct Answer: B
Detailed Solution: Each row has emf 3E and internal resistance 3r. Four identical rows in parallel keep emf 3E and reduce internal resistance to 3r/4.
Difficulty: Medium | Concept Tested: Mixed cell combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two resistors R and 2R are connected first in series and then in parallel across the same ideal battery. Ratio of total power in parallel to total power in series is:
Correct Answer: C
Detailed Solution: Series equivalent = 3R, P_s=V^2/(3R). Parallel equivalent = 2R/3, P_p=V^2/(2R/3)=3V^2/(2R). Ratio = (3/2)/(1/3)=9/2.
Difficulty: Medium | Concept Tested: Power and combination
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A 10 ohm wire with alpha=0.004/C is in series with a 20 ohm carbon resistor with alpha=-0.0005/C. Equivalent temperature coefficient is:
Correct Answer: B
Detailed Solution: For series, alpha_eq=(R1 alpha1 + R2 alpha2)/(R1+R2)=(10x.004+20x-.0005)/30=(.04-.01)/30=.001/C.
Difficulty: Difficult | Concept Tested: Temperature coefficient
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a network, left loop has 12 V source and 4 ohm resistor, right loop has 6 V source and 2 ohm resistor, and a common 3 ohm resistor connects the loops. Sources aid their own clockwise mesh currents. Find mesh currents I1 and I2.
Correct Answer: D
Detailed Solution: KVL gives (4+3)I1-3I2=12 and -3I1+(2+3)I2=6. Solving: multiply second by 7 and first by 3 gives I2=30/26=1.154 A and I1=(12+3I2)/7=2.209 A. This is not a direct formula question; it tests mesh setup.
Difficulty: Difficult | Concept Tested: Kirchhoff mesh
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Two cells E1=12 V,r1=2 ohm and E2=6 V,r2=1 ohm are connected in parallel with same polarity. Find equivalent emf and internal resistance.
Correct Answer: B
Detailed Solution: Equivalent emf for non-identical parallel cells = (E1/r1+E2/r2)/(1/r1+1/r2)=(12/2+6/1)/(1/2+1)=12/1.5=8 V. r_eq=(r1 r2)/(r1+r2)=2/3 ohm.
Difficulty: Difficult | Concept Tested: Non-identical cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A square has four 4 ohm sides and a 4 ohm diagonal between opposite corners. Equivalent resistance between the other two opposite corners is:
Correct Answer: C
Detailed Solution: The diagonal connects points at equal potential by symmetry for terminals at the other diagonal, so no current flows in it. Equivalent is two paths of 8 ohm in parallel = 4 ohm.
Difficulty: Medium | Concept Tested: Symmetry bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A source of emf 24 V and internal resistance 6 ohm supplies a variable load R. At what R is load power maximum, and what is maximum power?
Correct Answer: A
Detailed Solution: Maximum power transfer occurs at R=r=6 ohm. Pmax=E^2/(4r)=576/24=24 W.
Difficulty: Medium | Concept Tested: Maximum power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of resistance 99 ohm gives full scale at 1 mA. What shunt makes it an ammeter of range 1 A?
Correct Answer: D
Detailed Solution: At full scale, Ig=0.001 A through galvanometer and 0.999 A through shunt. Same voltage: Ig G = Is S. S=0.001x99/0.999=0.099 ohm.
Difficulty: Medium | Concept Tested: Shunt
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A galvanometer of 50 ohm resistance reads full scale at 2 mA. Series resistance needed for 10 V voltmeter is:
Correct Answer: C
Detailed Solution: Total resistance needed V/Ig=10/0.002=5000 ohm. Series resistance = 5000-50=4950 ohm.
Difficulty: Medium | Concept Tested: Meter conversion
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Terminal voltage of a cell falls from 1.5 V at zero current to 1.1 V at 2 A. Internal resistance is:
Correct Answer: A
Detailed Solution: V=E-Ir. Slope magnitude of V-I graph is r=(1.5-1.1)/2=0.2 ohm.
Difficulty: Medium | Concept Tested: V-I graph
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A wire connected to constant voltage is stretched to three times its length, volume constant. The heat produced per second becomes:
Correct Answer: B
Detailed Solution: R proportional L^2 for constant volume, so R becomes 9R. At constant V, P=V^2/R becomes P/9.
Difficulty: Medium | Concept Tested: Resistance geometry + power
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
An infinite ladder has each section: a 2 ohm series resistor followed by a 2 ohm shunt resistor to return. Find equivalent resistance in ohm from the input.
Correct Answer: A
Detailed Solution: Let equivalent be X. After first series 2 ohm, remaining network X is in parallel with 2 ohm shunt: X=2+(2X)/(2+X). This gives X^2-2X-4=0, so X=1+sqrt(5).
Difficulty: JEE Advanced | Concept Tested: Infinite network
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
For a balanced Wheatstone bridge with a galvanometer between middle points, which statements are true?
Correct Answer: A,C
Detailed Solution: At balance, the middle points are equipotential, so galvanometer current is zero independent of galvanometer resistance. Supply current is not zero.
Difficulty: JEE Advanced | Concept Tested: Multiple correct bridge
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A battery has n identical cells. Match condition R much greater than r and R much smaller than r with best grouping for maximum current.
Correct Answer: B
Detailed Solution: Current for m rows each with s cells: I=sE/(R+sr/m), sm=n. For high external R, increase emf: series. For low external R, reduce internal resistance: parallel.
Difficulty: JEE Advanced | Concept Tested: Cell grouping optimization
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A potentiometer balances a cell at 120 cm. With external load 6 ohm, balance length is 90 cm. If the same cell is connected to 3 ohm, terminal voltage is what fraction of emf?
Correct Answer: C
Detailed Solution: First r=6(120/90-1)=2 ohm. With load 3 ohm, V/E=R/(R+r)=3/5.
Difficulty: JEE Advanced | Concept Tested: Potentiometer + terminal voltage
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A delta of three 6 ohm resistors is converted to star. Each star arm is:
Correct Answer: B
Detailed Solution: For equal delta R, each star arm = R/3 = 2 ohm. This is often needed before reducing unbalanced bridge networks.
Difficulty: JEE Advanced | Concept Tested: Star-delta
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
In a two-cell circuit, a 12 V cell with 2 ohm internal resistance is connected opposite to a 6 V cell with 1 ohm internal resistance and external 3 ohm. Current direction is:
Correct Answer: A
Detailed Solution: Net emf=12-6=6 V, total resistance=2+1+3=6 ohm, current=1 A in direction of stronger source.
Difficulty: JEE Advanced | Concept Tested: Opposing cells
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Mesh currents I1=3 A and I2=1 A pass in opposite directions through a common 4 ohm resistor. Power in common resistor is:
Correct Answer: D
Detailed Solution: Actual current in shared resistor = |I1-I2|=2 A. Power=I^2R=4x4=16 W.
Difficulty: JEE Advanced | Concept Tested: Mesh current
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
If relaxation time doubles and number density halves, conductivity becomes:
Correct Answer: C
Detailed Solution: sigma=ne^2 tau/m. n becomes n/2 and tau becomes 2tau, product remains same.
Difficulty: JEE Advanced | Concept Tested: Microscopic Ohm law
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 5 ohm resistor. A voltmeter of resistance 10 ohm is connected across the 5 ohm resistor. What is the voltmeter reading?
Correct Answer: A
Detailed Solution: The voltmeter is not ideal. The external load is 5 parallel 10 = 10/3 ohm. Total resistance = 1+10/3=13/3 ohm. Current from cell = 6/(13/3)=18/13 A. Terminal/load voltage = I x 10/3 = 60/13 = 4.62 V. If options are adjusted for exact finite-voltmeter data, choose the nearest consistent value; here the concept is finite voltmeter loading.
Difficulty: Difficult | Concept Tested: Internal resistance + non-ideal meter
Shortcut Method: Redraw the equivalent circuit first; then decide whether voltage, current, null point, or power is fixed.
Common Student Mistake: Applying a memorized formula without checking the hidden circuit condition.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
Correction: identify the exact condition, write units, redraw the circuit if needed, and only then substitute values.
If Current Electricity numerical problems, potentiometer, Wheatstone bridge, meter bridge, Kirchhoff's rules, internal resistance or JEE/NEET numerical problems are not clear, students may contact Kumar Sir for one-to-one Physics guidance.
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