current electricity potentiometer is explained here with potential gradient, balance point, comparison of EMF, internal resistance, practical writing format, SVG diagrams and exam questions.
If Potentiometer, Potential Gradient, Balance Point, Comparison of EMF, Internal Resistance or CBSE practical numericals are not clear, students may contact Kumar Sir for one-to-one Physics guidance. Contact: +91-9958461445 | Website: KumarPhysicsClasses.com
Current Electricity | Null Method | Practical Physics

Current Electricity - Potentiometer

A complete premium guide to potentiometer principle, potential gradient, balance length, comparison of EMF, internal resistance, sensitivity, practical writing format and exam-level numerical problems.

CBSE PracticalNEETJEE MainJEE AdvancedIBIGCSEA-Level
Contact Kumar Sir

1. Formula Sheet First

Potential Gradientk = V / L

V is potential difference across full wire and L is total wire length.

Potential DifferenceV = kl

Across length l of a uniform potentiometer wire.

Balance PointE = kl

At null deflection, cell EMF equals potential drop along balancing length.

Comparison Of EMFE1/E2 = l1/l2

Valid when the same potential gradient is used.

Internal Resistancer = R(l1/l2 - 1)

Equivalent form: r = R[(l1-l2)/l2]. R is external resistance connected across the cell.

SensitivitySensitivity ∝ 1/k

Smaller potential gradient gives higher sensitivity.

2-4. What Is Potentiometer And Potential Gradient?

A potentiometer is a device used to measure EMF or potential difference accurately by a null deflection method. At balance, it draws no current from the cell under test, so it measures true EMF more accurately than a voltmeter. For a uniform wire carrying steady current, potential drop is directly proportional to length.

Principle

V ∝ l, V = kl

The wire must be uniform, current steady and temperature constant.

Material

Manganin and constantan are preferred because they have high resistivity and low temperature coefficient.

Gradient Derivation

k = V/L = IRp/L

For uniform wire, Rp=ρL/A, hence k=Iρ/A.

Dependency: potential gradient increases with current and resistivity, decreases with larger area, and is controlled by rheostat setting and primary battery voltage.

5. Proper SVG Potentiometer Diagram

Potentiometer Wire AB A (+) B (-) Jockey JBalance length l Primary current I Rheostat Primary Cell Key K Secondary Cell E G Positive terminals connected to same end A

6-7. Primary Circuit, Secondary Circuit And Balance Point

Primary Circuit

Main battery, rheostat, key and potentiometer wire AB. It creates a steady potential gradient along AB.

Secondary Circuit

Cell whose EMF is measured, galvanometer and jockey. At balance, this circuit carries zero current.

Correct Polarity

Positive terminal of primary battery and positive terminal of secondary cell must be connected to the same end A.

At balance: galvanometer deflection is zero and E = kl. On the two sides of the balance point, opposite galvanometer deflections are observed.

8-10. Sensitivity, Valid Conditions And Potentiometer Vs Voltmeter

Sensitivity ∝ 1/kkSensitivity
k ∝ current IIk
Increase SensitivityDecrease k

Use longer wire, reduce current using rheostat, or use smaller driving voltage when possible.

Valid Experiment

Primary cell EMF must be greater than unknown EMF. Otherwise no balance point is obtained.

Why Better Than Voltmeter?

At balance, potentiometer draws no current from the cell and measures true EMF. Voltmeter draws current and reads terminal potential difference.

PotentiometerVoltmeter
Null method, no current drawn at balanceDraws some current
Measures true EMF accuratelyMeasures terminal potential difference
More sensitiveLess sensitive

11. Comparison Of EMF Of Two Cells

Comparison Of EMF: Separate Cells E1 And E2 A (+) B J1, l1 J2, l2 E1long plate + E2 Two-way keyPosition 1: E1Position 2: E2 G Single jockey is moved to J1 or J2 Positive terminals of E1 and E2 connected to same end A
  1. For cell 1 at balance: E1 = kl1
  2. For cell 2 at balance: E2 = kl2
  3. Dividing both equations: E1/E2 = l1/l2
Example: l1=72 cm, l2=48 cm. E1/E2=72/48=3/2. If E2=1.5 V, E1=2.25 V.

12. Internal Resistance Using Potentiometer

Internal Resistance By Potentiometer: Primary Source + Correct Shunt Branch A (+) B Potentiometer wire AB Jockey J l1: K2 openl2: K2 closed Rheostat Primary battery Primary key K1 supplies steady current through AB P (+) N (-)Cell E, r G R K2 between R and N Correct shunt branch across cell: P -> R -> K2 -> N K2 open: E = k l1K2 closed: V = k l2V = ER/(R+r) Secondary path: Cell -> G -> Jockey
  1. With external resistance disconnected, balance length is l1: E = kl1
  2. With external resistance R connected, terminal voltage balance length is l2: V = kl2
  3. Therefore, E/V = l1/l2
  4. For a cell delivering current through R, E/V = (R+r)/R
  5. Equate both ratios: l1/l2 = (R+r)/R
  6. Multiply by R: R l1/l2 = R + r
  7. Subtract R from both sides: r = R(l1/l2) - R
  8. Take R common: r = R(l1/l2 - 1)
  9. Equivalent final form: r = R[(l1-l2)/l2]
Example: R=5 Ω, l1=80 cm, l2=64 cm. Using the preferred formula, r=R(l1/l2 - 1)=5(80/64 - 1)=5(1.25 - 1)=1.25 Ω. Same result: r=5[(80-64)/64]=1.25 Ω.

13-15. CBSE Practical Format, Limitations And Mistakes

Practical File Format

Aim, apparatus, theory, circuit diagram, procedure, observation table, calculations, result, precautions and sources of error.

Experiments

1. Compare EMF of two cells. 2. Determine internal resistance of a cell using potentiometer.

Limitations

Requires long uniform wire, stable current, balance point within wire and primary voltage greater than unknown EMF.

Wrong polarity: positive terminals must go to same end A.
Forgetting E=kl: balance length represents EMF only at null point.
Confusing l1 and l2: l1 is open-circuit EMF length, l2 is terminal voltage length.
Assuming balance always exists: no balance if primary drop is smaller than unknown EMF.
Hard jockey pressure: it changes contact and may damage wire.
Voltmeter logic: potentiometer is a null method, not a current-drawing meter.

16-21. Exam Question Bank

NEET MCQs With Four Options
1. Potentiometer works on:
A. Heating effectB. Null deflection methodC. Magnetic induction onlyD. Capacitance only
Answer: B. At balance, galvanometer current is zero.
2. Potential gradient is:
A. V/LB. VLC. L/VD. I/V
Answer: A. k=V/L.
3. At balance point:
A. E=klB. E=k/lC. E=l/kD. E=0 always
Answer: A. EMF equals potential drop along balancing length.
4. Why is potentiometer more accurate than voltmeter?
A. It draws no current at balanceB. It has low resistanceC. It heats the cellD. It measures current
Answer: A. It measures true EMF under null condition.
5. If k=0.02 V/cm and l=75 cm, E is:
A. 0.75 VB. 1.5 VC. 2.0 VD. 3.0 V
Answer: B. E=kl=0.02x75=1.5 V.
6. If l1=60 cm and l2=40 cm, E1/E2 is:
A. 2/3B. 3/2C. 1/2D. 2
Answer: B. E1/E2=l1/l2=60/40.
7. Sensitivity increases when:
A. k increasesB. k decreasesC. wire is heated stronglyD. balance disappears
Answer: B. Sensitivity is inversely proportional to k.
8. Suitable wire material is:
A. copper onlyB. manganinC. aluminium foilD. iron nail
Answer: B. Manganin has low temperature coefficient.
9. No balance point is obtained if:
A. primary drop is less than unknown EMFB. wire is longC. rheostat is usedD. galvanometer is sensitive
Answer: A. The wire cannot provide the required matching potential drop.
10. Internal resistance formula is:
A. r=R(l1-l2)/l2B. r=R(l2-l1)/l1C. r=Rl1l2D. r=R+l1+l2
Answer: A. r=R[(l1-l2)/l2].
11. l1 is measured when the cell is:
A. short-circuitedB. open-circuitedC. removed from circuitD. reversed always
Answer: B. l1 corresponds to EMF.
12. l2 corresponds to:
A. terminal voltageB. zero EMFC. wire resistance onlyD. primary battery EMF
Answer: A. With R connected, balance length gives terminal p.d.
13. If R=4 Ω, l1=100 cm, l2=80 cm, r is:
A. 1 ΩB. 2 ΩC. 4 ΩD. 5 Ω
Answer: A. r=4(20/80)=1 Ω.
14. Potential gradient is directly proportional to:
A. current in wireB. reciprocal currentC. galvanometer zeroD. secondary EMF only
Answer: A. k=Iρ/A.
15. A potentiometer measures:
A. current onlyB. EMF/potential differenceC. magnetic pole strengthD. capacitance only
Answer: B. It balances potential drops.
16. If primary current is reduced, k:
A. decreasesB. increasesC. becomes infiniteD. becomes negative
Answer: A. k is proportional to current.
17. Balance point shifts to larger length when unknown EMF:
A. increasesB. decreasesC. is zeroD. is removed
Answer: A. l=E/k.
18. The galvanometer is connected in:
A. primary circuit onlyB. secondary circuitC. across rheostat onlyD. nowhere
Answer: B. It detects null condition for the test cell.
19. If l1/l2=5/4 and R=8 Ω, r is:
A. 1 ΩB. 2 ΩC. 4 ΩD. 8 Ω
Answer: B. r=R[(l1-l2)/l2]=8(1/4)=2 Ω.
20. Positive terminal condition is:
A. both positives at same endB. positives at opposite endsC. no polarity neededD. galvanometer reversed only
Answer: A. Wrong polarity prevents proper null balance.
JEE Main, JEE Advanced, IB, IGCSE, CBSE Case Studies And Numericals
JEE Main: k=0.015 V/cm. A cell balances at 120 cm. Find E.
Solution: E=1.8 V.
JEE Main: Reducing current in potentiometer wire increases balance length for the same cell because k decreases.
JEE Advanced Multiple Correct: Sensitivity increases by increasing wire length and decreasing current. Both reduce k.
Integer: R=6 Ω, l1=90 cm, l2=72 cm. r=6(18/72)=1.5 Ω.
CBSE Case: A student connects positive terminals to opposite ends and gets no null point. Reason: wrong polarity; correct by connecting both positives to A.
IB Data: If uncertainty in l is ±1 mm, percentage uncertainty is lower for larger balance lengths.
IGCSE: State one advantage over voltmeter: no current is drawn from the cell at balance.
A-Level: Explain why manganin is preferred: resistance changes little with temperature.

22. Final Revision Sheet

PointFormula / Result
PrincipleV ∝ l for uniform wire
Potential gradientk = V/L = IRp/L = Iρ/A
Balance conditionE = kl
Compare EMFE1/E2 = l1/l2
Internal resistancer = R[(l1-l2)/l2]
SensitivitySensitivity ∝ 1/k
PrincipleV = kl
Gradientk = V/L
EMFE1/E2=l1/l2
Internal rr=R(l1-l2)/l2
AccuracyNull method
PrecautionSame positive end

Still Confused In Potentiometer?

Still confused in Potentiometer, Potential Gradient, Sensitivity, Comparison of EMF, Internal Resistance or CBSE practical questions? Learn Physics step-by-step with Kumar Sir.

Phone: +91-9958461445
Website: KumarPhysicsClasses.com

Contact Kumar Sir
Scroll to Top