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?

Question

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?,

MARKS App · Accepted Answer

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

Search any question & find its solution

Looking for more such questions to practice?