current electricity wheatstone bridge is explained here with formulas, derivation, proper circuit diagrams, meter bridge, sensitivity, exam questions and numerical practice.
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Current Electricity | Bridge Circuits | Meter Bridge

Current Electricity - Wheatstone Bridge

A complete professional guide to Wheatstone Bridge, balance condition, galvanometer null deflection, meter bridge, sensitivity, unbalanced bridge methods and exam-level circuit numericals.

CBSE Class 12NEETJEE MainJEE AdvancedIBIGCSEA-Level
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1. Formula Sheet First

Balanced BridgeP / Q = R / S

Four arms P, Q, R and S are in balance when their ratios are equal.

Cross MultiplicationPS = QR

This is the most used numerical form for unknown resistance.

Galvanometer CurrentIg = 0

At null point, no current flows through the galvanometer.

Potential ConditionVB = VD

Junctions B and D are at the same potential at balance.

Meter BridgeR / X = l / (100 - l)

Here l is balancing length in centimetres.

Unknown ResistanceX = R(100 - l) / l

Use this when R is placed in the left gap and X in the right gap.

Sensitivity rule: maximum sensitivity is obtained when all four arms are comparable. In a meter bridge, the best balancing length is usually between 40 cm and 60 cm.

2. What Is Wheatstone Bridge?

A Wheatstone Bridge is a resistance network used to find an unknown resistance very accurately by comparing it with known resistances. It is a null method: instead of measuring a large current directly, we adjust the circuit until the galvanometer shows zero deflection. This makes the method highly accurate for CBSE, NEET, JEE, IB, IGCSE and practical Physics.

Why Used?

To measure unknown resistance and detect very small resistance changes.

Key Idea

When two middle junctions have equal potential, the galvanometer current becomes zero.

Applications

Meter bridge, strain gauge, resistance thermometer, sensors and precision measurement circuits.

3. Proper SVG Wheatstone Bridge Circuit

A B C D P Q R S G VBVD I1 I2 Balance: Ig = 0 Cell: long plate is + Key

4-5. Principle And Balanced Bridge Derivation

Principle

The bridge is balanced when the galvanometer shows null deflection. Then no charge flows through the galvanometer branch, so B and D are at the same potential.

VB = VD, Ig = 0

Physical Meaning

The left and right potential dividers create the same potential at the central junctions. Since there is no potential difference across the galvanometer, no current passes through it.

  1. Let current through P and Q be I1. Let current through R and S be I2.
  2. At balance, Ig = 0, so the current in P continues through Q and the current in R continues through S.
  3. Since VB = VD, potential drop from A to B equals potential drop from A to D: I1P = I2R
  4. Potential drop from B to C equals potential drop from D to C: I1Q = I2S
  5. Dividing the two equations gives P / Q = R / S or PS = QR.

6-8. Unbalanced Bridge, Balancing Point And Sensitivity

Unbalanced Bridge

If P/Q is not equal to R/S, then VB is not equal to VD and current flows through the galvanometer. The simple balance formula cannot be applied directly.

How To Solve

Use Kirchhoff's rules, Thevenin method, star-delta transformation or symmetry, depending on the network.

Balancing Point

The point where galvanometer deflection becomes zero. Null methods are accurate because the answer does not depend on galvanometer resistance at balance.

Sensitivity

High sensitivity means a small change in resistance produces a clear galvanometer deflection. Sensitivity is best when all arms are comparable.

Meter Bridge Range

The best balancing length is 40 cm to 60 cm. Near 0 cm or 100 cm, end correction and contact errors become significant.

Example

If P = 2 Ω, Q = 3 Ω, R = 4 Ω, then S = QR/P = 3 x 4 / 2 = 6 Ω.

Unbalanced Wheatstone Bridge: Equivalent Resistance By Star-Delta Method

When the bridge is unbalanced, the galvanometer branch cannot be removed. To find equivalent resistance between A and C, one reliable method is to convert one triangle into an equivalent star.

Step 1: Choose Delta A-B-DStep 2: Replace By Star A B D C P R G Q S Convert P, R, G delta A B D C O xA xB xD Q S
Delta chosenA-B-D: P, R, G

Here G is the galvanometer resistance between B and D. This triangle is converted into a star with central point O.

Star armsxA = PR/(P+R+G)xB = PG/(P+R+G)xD = RG/(P+R+G)
Equivalent after conversionReq = xA + [(xB+Q)(xD+S)] / [(xB+Q)+(xD+S)]

This gives equivalent resistance between A and C for the unbalanced bridge.

Balanced shortcutIf P/Q = R/S, remove G branch

Then Req = (P+Q) parallel (R+S). This shortcut is valid only when the bridge is balanced.

Worked example: Let P=3 Ω, R=6 Ω, G=6 Ω, Q=3 Ω and S=6 Ω. Sum = 15 Ω. Then xA=18/15=1.2 Ω, xB=18/15=1.2 Ω, xD=36/15=2.4 Ω. So Req=1.2+[(1.2+3)(2.4+6)]/[(1.2+3)+(2.4+6)] = 1.2+(4.2x8.4)/(12.6)=1.2+2.8=4.0 Ω.

9. Meter Bridge

ABUniform 100 cm bridge wire Jockey J Resistance BoxR UnknownX G l cm100 - l cm Cell Key K Meter Bridge: Null point at J
DerivationR / X = l / (100 - l)

For a uniform wire, resistance is proportional to length. At null point, the meter bridge is a Wheatstone Bridge.

UnknownX = R(100 - l) / l

Use the correct side placement. If R and X are interchanged, the formula changes accordingly.

10-16. Exam Practice Bank

CBSE Class 12 Questions With Answers
Short: Why is Wheatstone Bridge called a null method?
Answer: Because at balance the galvanometer current is zero, so the unknown resistance is found from a ratio, not from direct current measurement.
Numerical: P=3 Ω, Q=6 Ω, R=5 Ω. Find S.
Solution: P/Q=R/S, so S=QR/P=6x5/3=10 Ω.
Assertion-Reason: Assertion: Battery resistance does not affect balance condition. Reason: At balance, VB=VD.
Answer: Both true; reason explains assertion.
Case: In a meter bridge, null point is 20 cm. Is measurement reliable?
Answer: Not highly reliable; it should be shifted near 40-60 cm by choosing a better known resistance.
NEET MCQs: Four Options, Answer And Explanation
1. In a balanced Wheatstone Bridge, galvanometer current is:
A. MaximumB. ZeroC. Equal to battery currentD. Infinite
Answer: B. At balance, the two galvanometer terminals are at the same potential.
2. The balance condition is:
A. P+Q=R+SB. PQ=RSC. P/Q=R/SD. P/R=Q/S is never valid
Answer: C. The standard condition is P/Q=R/S, equivalently PS=QR.
3. If P=2 Ω, Q=5 Ω, R=4 Ω, S at balance is:
A. 5 ΩB. 10 ΩC. 2.5 ΩD. 8 Ω
Answer: B. S=QR/P=5x4/2=10 Ω.
4. In a meter bridge, R=10 Ω and l=50 cm. X equals:
A. 5 ΩB. 10 ΩC. 20 ΩD. 100 Ω
Answer: B. X=R(100-l)/l=10x50/50=10 Ω.
5. Best meter bridge balance is near:
A. 5 cmB. 15 cmC. 50 cmD. 98 cm
Answer: C. Middle balance reduces percentage error and end correction influence.
6. If galvanometer deflects, the bridge is:
A. BalancedB. UnbalancedC. Open alwaysD. Independent of resistors
Answer: B. Deflection means VB is not equal to VD.
7. At balance, changing battery EMF ideally:
A. Changes balance ratioB. Does not change balance conditionC. Makes Ig infiniteD. Reverses P and Q
Answer: B. Balance depends on resistance ratio, not ideal battery emf.
8. A bridge has P/Q=2 and R/S=2. Galvanometer current is:
A. ZeroB. Non-zeroC. Negative onlyD. Equal to P/Q
Answer: A. Equal ratios imply balanced bridge.
9. If l=40 cm, R=6 Ω, then X is:
A. 4 ΩB. 6 ΩC. 9 ΩD. 15 Ω
Answer: C. X=6x60/40=9 Ω.
10. The meter bridge wire should be:
A. Non-uniformB. UniformC. InsulatingD. Infinite resistance
Answer: B. R is proportional to length only for a uniform wire.
11. Unknown resistance should be adjusted so null point lies:
A. Near 0 cmB. Near 100 cmC. Between 40 cm and 60 cmD. Outside the wire
Answer: C. This gives better sensitivity and lower percentage error.
12. Which quantity is zero at balance?
A. Battery currentB. Current through PC. Current through galvanometerD. Current through Q
Answer: C. Null deflection means Ig=0.
JEE Main And Advanced Practice
JEE Main 1. P=4 Ω, Q=8 Ω, R=6 Ω. For balance, S is:
A. 6 ΩB. 8 ΩC. 12 ΩD. 16 Ω
Answer: C. S=QR/P=8x6/4=12 Ω.
JEE Main 2. If all four arms are doubled, balance condition:
A. ChangesB. Remains sameC. Becomes P+Q=R+SD. Fails always
Answer: B. Ratios remain unchanged.
JEE Advanced Multiple Correct. At balance:
A. VB=VDB. Ig=0C. Battery current is zeroD. P/Q=R/S
Answer: A, B, D. Battery current is not generally zero.
Integer type: In a meter bridge, R=12 Ω and l=60 cm. Find X.
Solution: X=12x40/60=8 Ω. Integer answer: 8.
IB, IGCSE, ICSE, A-Level And Strong Numerical Practice
IB Data: Explain why a null method reduces systematic error due to galvanometer calibration.
Answer: The reading required is zero deflection, so exact current scale calibration is not needed.
IGCSE: State one use of a meter bridge.
Answer: It is used to determine an unknown resistance by comparison.
A-Level: Discuss why sensitivity falls near the ends of a meter bridge wire.
Answer: A small length error creates a large percentage error when l or 100-l is small.
Difficult Numerical: R=15 Ω and null point = 75 cm. X=15x25/75=5 Ω.

17. Common Student Errors

Wrong: Applying P/Q=R/S in an unbalanced bridge.
Correct: Use it only when Ig=0.
Wrong: Forgetting 100-l in meter bridge.
Correct: The two wire arms are l and 100-l.
Wrong: Confusing R/X with X/R.
Correct: Match the physical placement of R and X.
Wrong: Thinking battery resistance affects null point.
Correct: At balance, the ratio condition is independent of battery and galvanometer resistance.
Wrong: Taking null point near the end as reliable.
Correct: Prefer 40-60 cm.
Wrong: Cross multiplying as PR=QS.
Correct: From P/Q=R/S, the result is PS=QR.

18. Quick Revision Sheet

PrincipleIg = 0

Null deflection means balanced bridge.

ConditionP / Q = R / S

Use for balanced bridge only.

PotentialVB = VD

No p.d. across galvanometer.

Meter BridgeX = R(100-l)/l

When R is left and X is right.

Best Length40 cm to 60 cm

Gives good sensitivity.

TrapPS = QR

Do not write PR=QS.

Still Confused In Wheatstone Bridge?

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