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 length l1 cm and l2 cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal, respectively, to [figure]
- k(l2 - l1) and kl2
- kl1 and k(l2 - l1)
- k(l2 - l1) and kl1
- kl1 and kl2
Answer
(2) kl1 and k(l2 - l1)
Solution
At balance no current flows through the galvanometer, so the current through R and X stays I = 1.0 A.
(i) Key between 1 and 2: the galvanometer branch is across R.
IR = kl1 R = kl1 (as I = 1 A)
(ii) Key between 1 and 3: the galvanometer branch is across (R + X).
I(R + X) = kl2 R + X = kl2
X = kl2 - kl1 = k(l2 - l1)
Answer: (2)