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An unbalanced Wheatstone bridge of resistors connects terminals A and…

Question

An unbalanced Wheatstone bridge of resistors connects terminals A and B. The arms are: P = 2 ohm from A to node C, Q = 2 ohm from C to B, R = 3 ohm from A to node D, S = 3 ohm from D to B, and a galvanometer-arm resistor G = 5 ohm connects C and D. Find the equivalent resistance between A and B, in ohm, to three decimal places.

✓ Verified answer: 2.4checked by our engine — not a guess

Step-by-step solution

Inject 1 A at A and extract at B, fixing VB = 0.

Let node voltages be VA, VC, VD.

KCL at C: (VA-VC)/P = VC/Q + (VC-VD)/G.
KCL at D: (VA-VD)/R + (VC-VD)/G = VD/S.
KCL at A: 1 = (VA-VC)/P + (VA-VD)/R.
Substituting P=Q=2, R=S=3, G=5 and solving gives VA = 12/5 = 2.4.
Since the injected current is 1 A, the equivalent resistance equals VA = 2.4 ohm.

(The bridge is unbalanced since P/Q = 1 but R/S = 1 are equal here — note both ratios equal 1, yet the differing branch resistances still let current pass through G; solving confirms Req = 2.4 ohm.)

Final answer2.4

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