A potentiometer wire has length 4 m and resistance 8 . The resistance that must be connected in series with the wire and the driving battery of e.m.f. 2 V, so as to get a potential gradient of 1 mV per cm on the wire is

Step-by-step solution

Potential gradient k = 1 mV/cm = 10⁻³ V 10⁻² m = 0.1 V/m Length of the potentiometer wire L = 4 m Potential difference across the potentiometer wire V = k L = 0.1 4 = 0.4 V Current in the circuit I = V R_p = 0.4 8 = 0.05 A Total resistance of the circuit R_ eq = E I = 2 0.05 = 40 Resistance to be connected in series R = R_ eq - R_p = 40 - 8 = 32 Answer: 32