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Exam level1 past question

Wheatstone Bridge and Meter Bridge

Learn the principle and application of Wheatstone and meter bridges for measuring unknown resistances.

Why this shows up in the exam

NEET often includes practical circuit questions involving bridge principles.

How NEET tests this

Numerical · 2 QsApplication

Learn the idea

A Wheatstone bridge is balanced when the galvanometer reads zero; then the ratio of opposite arms is equal. In a meter bridge the uniform wire makes resistance directly proportional to its length, so the same ratio can be written with lengths.

🧠 Memory hook: Think of a road: the longer the road, the more resistance you feel while driving. So a larger resistance needs a longer piece of the uniform wire, just like a longer road takes more time.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • At balance the potential difference between the two mid‑points is zero → no current through the galvanometer.
  • For a balanced bridge: R1 : R2 = R3 : R4 (ratio of opposite arms).
  • In a uniform wire R ∝ L, i.e., resistance per unit length is constant.
  • In a balanced meter bridge the ratio of the two resistances equals the ratio of the lengths of the wire on their respective sides.
  • Length corresponding to 1 Ω = total length of the wire ÷ total resistance of the wire.

How to approach it

  1. 11. Check that the bridge is stated to be balanced (galvanometer deflection = 0).
  2. 22. Write the proportion: resistance ratio = length ratio of the wire on the same sides.
  3. 33. Substitute the given numbers and solve for the unknown resistance or total resistance of the wire.
  4. 44. If the question asks for length per ohm, divide the known total length by the total resistance obtained in step 3.

Worked example — watch it click

A resistance wire connected in the left gap of a meter bridge balances a 10 Ω resistance in the right gap at a point which divides the bridge wire in the ratio 3:2. If the length of the resistance wire is 1.5 m, then the length of 1 Ω of the resistance wire is:

  • ✅1.0 × 10⁻¹ m
  • B)1.5 × 10⁻² m
  • C)1.5 × 10⁻¹ m
  • D)1.0 × 10⁻² m

The concept behind this problem

The problem forces you to use the balance condition to find the total resistance of the uniform wire, then convert that total resistance into a length per ohm – a direct application of R ∝ L in a meter bridge.

Step by step

  1. 1Meter bridge balance: R/10 = 3/2, so R = 15Ω.
  2. 2Length of wire = 1.5 m for 15Ω.
  3. 3Length per ohm = 1.5/15 = 0.1 m = 1.0×10⁻¹ m.

Watch out

Students often invert the length ratio (using 2/3 instead of 3/2) and obtain the wrong total resistance of the wire.

Common slip-ups that cost marks

  • •Mixing up which length belongs to the unknown resistance – always pair each resistance with the length on its own side of the bridge.
  • •Using the inverse ratio (Lright/Lleft) by mistake; the ratio must follow the same order as the resistances.
  • •Treating the wire resistance as negligible – in a meter bridge the wire’s resistance is the quantity being used.

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 1

A resistance wire connected in the left gap of a meter bridge balances a 10 Ω resistance in the right gap at a point which divides the bridge wire in the ratio 3:2. If the length of the resistance wire is 1.5 m, then the length of 1 Ω of the resistance wire is:

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