Question

To make a potentiometer, a driver cell of e.m.f. 4.0 V is connected across a 1.00 m length of resistance wire. (a) What is the potential difference across each 1 cm length of the wire? What length of wire has a p.d. of 1.0 V across it? [3] (b) A cell of unknown e.m.f. E is connected to the potentiometer and the balance point is found at a distance of 37.0 cm from the end of the wire to which the galvanometer is connected. Estimate the value of E. [3] (c) A standard cell of e.m.f. 1.230 V gives a balance length of 31.2 cm. Use this value to obtain a more accurate value for E. 

Answer 1

(a) The potential difference across each 1 cm length of the wire

$PD =Voltage \times Length = 4.0 V \times 0.01 m = 0.04 Vm$

Length of wire has a p.d. of 1.0 V across 1 cm

$PD =Voltage \times Length \implies 1V = 4.0 V/m \times L \implies L = \frac{1}{4}=0.25 m =25 cm$

(b) $1 cm \to 4.0 V$

$37.0 cm \to 4.0 v \times 37.0 = 148 V$

$E= 148 V$

(c) $31.2 cm \to 1.230 V$

$37.0 cm \to 1.230 v \times \frac{37.0}{31.21.4} = 1.45865384615V$

$E= 1.45865384615V$