current electricity meter bridge is explained here with theory, derivation, balancing condition, sensitivity, errors, SVG diagrams, numericals and exam questions.
If Meter Bridge derivations, balancing condition, sensitivity, Wheatstone Bridge relation or JEE/NEET numerical problems are not clear, students can contact Kumar Sir for one-to-one Physics guidance. Contact: +91-9958461445 | Website: KumarPhysicsClasses.com
Current Electricity | Practical Bridge | Null Method

Meter Bridge - Complete Theory, Derivation, Numericals & Exam Questions

Master Meter Bridge concepts, derivations, balance condition, sensitivity, error analysis and advanced applications for NEET, IIT-JEE, IB and British Curriculum Physics.

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1-2. Introduction: What Is A Meter Bridge?

A Meter Bridge is a practical form of the Wheatstone Bridge used to determine an unknown resistance by comparing it with a known resistance. It uses a one metre long uniform resistance wire, a resistance box, a galvanometer, a jockey, a cell and a key. Historically, slide-wire bridge methods became important because they allowed accurate resistance comparison without depending on direct current measurement.

Measurement

It finds unknown resistance using a null point, so it is more accurate than direct meter reading.

Comparison

It compares the ratio of two resistances with the ratio of two wire lengths.

Laboratory Use

It is a standard CBSE and practical Physics experiment for resistance and resistivity.

3. Principle Of Meter Bridge

Wheatstone Bridge PrincipleR / X = L / (100 - L)

At balance, the ratio of gap resistances equals the ratio of resistance wire lengths.

Unknown ResistanceX = R(100 - L) / L

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

Null PointIg = 0

The galvanometer shows zero current because its two terminals are at the same potential.

4. Construction Of Meter Bridge: Proper SVG Diagram

ABUniform 1 metre bridge wire Jockey J Resistance BoxR UnknownX G L cm100 - L cm Cell Key K

5. Working Of Meter Bridge

  1. Connect the resistance box R in one gap and the unknown resistance X in the other gap.
  2. Connect the cell and key across the ends of the bridge wire.
  3. Touch the jockey gently on the wire and move it until galvanometer deflection becomes zero.
  4. Record the balancing length L from the end near the known resistance.
  5. Calculate X using X = R(100 - L)/L.

6. Derivation Of Meter Bridge Formula

  1. At null point, the meter bridge behaves like a balanced Wheatstone Bridge.
  2. Therefore, R / X = Rwire1 / Rwire2.
  3. For a uniform wire, resistance is directly proportional to length: Rwire ∝ length.
  4. The two wire lengths are L and 100 - L, so R / X = L / (100 - L).
  5. Rearranging gives X = R[(100 - L)/L].

7-8. Unknown Resistance And Resistivity

Easy: R=5 Ω, L=50 cm. X=5(50/50)=5 Ω.
Medium: R=8 Ω, L=40 cm. X=8(60/40)=12 Ω.
Advanced: R=12 Ω, L=75 cm. X=12(25/75)=4 Ω. The null point is near the end, so choose a better R.
Olympiad: If R and X are interchanged, new length L' should be close to 100-L for a uniform wire.
Resistivityρ = RA / l

Here R is resistance of the test wire, A is cross-sectional area, and l is wire length in metre.

Resistivity Example: R=4 Ω, length=2 m, diameter=0.5 mm. A=π(0.25x10-3)2. Then ρ=RA/l.

9. Sensitivity Of Meter Bridge

Sensitivity is maximum near 50 cm because both wire parts have comparable resistance. When the balance point is near an end, a small scale reading error becomes a large percentage error.

50 cm40 cm60 cmBest sensitivity near middleSensitivityL
Exam rule: The practical balancing length should lie between 40 cm and 60 cm. If it is near 0 cm or 100 cm, change the resistance box value.

10. Slide Wire Bridge

Construction And Working

A slide wire bridge uses a uniform resistance wire and a sliding contact to obtain the balance point. It is essentially the same principle as a meter bridge but can be arranged with different lengths and laboratory layouts.

Comparison

Meter bridge has a fixed one metre wire and is simpler for school labs. Slide wire bridge may allow more flexible experimental arrangements.

11-12. Errors And Common Student Mistakes

End correction: Avoid balance near 0 cm or 100 cm; repeat after interchanging R and X.
Contact resistance: Keep connections tight and clean.
Jockey pressure: Touch gently; do not rub the wire.
Non-uniform wire: Use the middle region and take repeated readings.
Wrong length: Measure L from the end connected to known R if using X=R(100-L)/L.
Heating: Close the key only while taking observations.

13-20. Exam Question Bank

NEET MCQs: 25 Questions With Four Options
1. Meter bridge is based on:
A. Ohm's law onlyB. Wheatstone BridgeC. Joule's lawD. Faraday's law
Answer: B. It is the practical form of Wheatstone Bridge.
2. At balance, galvanometer current is:
A. ZeroB. MaximumC. Equal to cell currentD. Infinite
Answer: A. Null point means no current through galvanometer.
3. If R=10 Ω and L=50 cm, X is:
A. 5 ΩB. 10 ΩC. 20 ΩD. 100 Ω
Answer: B. X=10x50/50=10 Ω.
4. Best balancing length is:
A. 5 cmB. 20 cmC. 50 cmD. 95 cm
Answer: C. Sensitivity is best near the middle.
5. If L=40 cm and R=6 Ω, X equals:
A. 4 ΩB. 6 ΩC. 9 ΩD. 12 Ω
Answer: C. X=6x60/40=9 Ω.
6. A null method is accurate because it measures:
A. Large currentB. Zero deflection conditionC. Heat directlyD. EMF only
Answer: B. Zero deflection is less dependent on meter calibration.
7. The wire used should be:
A. Non-uniformB. UniformC. InsulatingD. Superconducting only
Answer: B. Resistance must be proportional to length.
8. Formula for X when R is left gap:
A. X=RL/(100-L)B. X=R(100-L)/LC. X=R+LD. X=100R/L
Answer: B. From R/X=L/(100-L).
9. If balance point is 90 cm, we should:
A. Accept as bestB. Change R to bring null near centreC. Heat wireD. Remove galvanometer
Answer: B. 90 cm gives high percentage error.
10. Resistance of wire segment is proportional to:
A. LengthB. Square of lengthC. Reciprocal lengthD. EMF
Answer: A. For uniform wire, R=rho L/A.
11. End correction arises due to:
A. finite copper strips and contactsB. gravityC. magnetic field onlyD. sound waves
Answer: A. The effective zero and 100 cm points may not match scale ends.
12. If R and X are interchanged, ideal new length is:
A. LB. 100-LC. 2LD. L/2
Answer: B. Interchanging gaps reverses the length ratio.
13. If L=60 cm and X=8 Ω, R is:
A. 4 ΩB. 8 ΩC. 12 ΩD. 16 Ω
Answer: C. R/X=L/(100-L)=60/40=3/2, so R=12 Ω.
14. Jockey should be pressed:
A. Very hardB. GentlyC. Continuously with rubbingD. With oil
Answer: B. Hard pressure changes contact and may damage wire.
15. The balance condition is independent of:
A. length ratioB. resistance ratioC. galvanometer resistance at nullD. uniformity of wire
Answer: C. At null, no galvanometer current flows.
16. If R=4 Ω and L=20 cm, X is:
A. 4 ΩB. 8 ΩC. 12 ΩD. 16 Ω
Answer: D. X=4x80/20=16 Ω.
17. The main purpose of key is:
A. measure lengthB. open/close currentC. change scaleD. act as unknown resistance
Answer: B. It prevents unnecessary heating by controlling current.
18. Resistivity formula is:
A. rho=RA/lB. rho=Rl/AC. rho=A/RlD. rho=R/A
Answer: A. From R=rho l/A.
19. If diameter doubles, area becomes:
A. doubleB. four timesC. halfD. unchanged
Answer: B. Area is proportional to d squared.
20. Zero galvanometer deflection means:
A. no current in entire circuitB. no current in galvanometerC. no resistance in wireD. no battery
Answer: B. Current still flows in the main circuit.
21. If L increases for same R, calculated X:
A. increasesB. decreasesC. unchanged alwaysD. becomes zero
Answer: B. X=R(100-L)/L decreases as L increases.
22. Practical current should be kept small to reduce:
A. lengthB. heating errorC. bridge principleD. scale marks
Answer: B. Heating changes resistance.
23. Null point is obtained by moving:
A. cellB. jockeyC. resistance wireD. screw gauge
Answer: B. The jockey selects the balancing length.
24. If L=25 cm, R=3 Ω, X is:
A. 1 ΩB. 3 ΩC. 6 ΩD. 9 Ω
Answer: D. X=3x75/25=9 Ω.
25. Meter bridge is most suitable for:
A. resistance comparisonB. measuring accelerationC. measuring frequencyD. measuring optical power
Answer: A. It compares an unknown resistance with a known resistance.
JEE Main, JEE Advanced, CBSE Case Studies, IB, IGCSE And A-Level
JEE Main: R=15 Ω gives null at 75 cm. Find X.
Solution: X=15x25/75=5 Ω.
JEE Main: Why does interchanging R and X reduce error?
Answer: Mean of two readings helps compensate end correction.
JEE Advanced: A wire of unknown resistance is first measured, then its length is doubled keeping material and area same. Its resistance doubles, so the new null point shifts according to R/X=L/(100-L).
CBSE Case Study: A student gets null at 18 cm. Questions: Is it reliable? What should be changed? Why middle range? Answer: not reliable; adjust R; percentage error is lower near centre.
IB: Discuss uncertainty in L and its effect on X. Answer: uncertainty in both L and 100-L propagates; it is larger near wire ends.
IGCSE/A-Level: State two precautions: use low current and touch jockey gently.

21-22. Revision Formula Sheet And Quick Cards

ConceptFormula / Result
Balance conditionR / X = L / (100 - L)
Unknown resistanceX = R(100 - L) / L
Resistivityrho = RA / l
Best balance range40 cm to 60 cm
Null pointIg = 0
PrincipleWheatstone Bridge
FormulaX=R(100-L)/L
SensitivityNear 50 cm
ErrorsEnd correction
ApplicationsFind X, compare R
Exam TipCheck side placement

Still Confused About Meter Bridge?

If Meter Bridge derivations, balancing condition, sensitivity, Wheatstone Bridge relation or JEE/NEET numerical problems are not clear, take one-to-one Physics guidance from Kumar Sir.

Phone: +91-9958461445
Website: KumarPhysicsClasses.com

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