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A meter bridge circuit is powered by an ideal battery of emf 5 V and negligible internal resistance. The bridge wire has a resistance per unit length equal to 0.1 Ω cm - 1 . Unknown resistance X is connected in the left gap and 6 Ω in the right gap. The null point divides the wire in the ratio 2 : 3 . What is the current (in A ) drawn from the battery?

Correct answer

1

Step-by-step solution

l 1 l 2 = 2 3 = v 1 v 2 Now v 1 + v 2 = v = 5   V = I R For potentiometer wire R 1 R 2 = l 1 l 2 = 2 3 x 6 = 2 3 ⇒ x l = 4   Ω m The resistance per unit length of the wire is 0.1 Ω c m = 1 Ω m ∴   x + 1 = R e f f = 5 ∴   i = V R e f f = 5 5 = 1   A

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