A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are…

Question

A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths l₁ cm and l₂ cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal, respectively, to

Accepted Answer

(2) kl₁ and k(l₂-l₁). The left end A of the wire is joined to the left end of R. The jockey side goes through G to terminal 1. Key 1–2: G meets the junction between R and X, so the balance measures the p.d. across R: IR=kl₁. Key 1–3: G meets the far end of X, so the balance measures the p.d. across R and X: I(R+X)=kl₂. At balance no current flows through G, so I=1.0 A throughout. R=kl₁ and X=kl₂-kl₁=k(l₂-l₁).

Step-by-step solution

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