Question

A potentiometer wire, of length 4 m and resistance 8 Ω, is connected in series with a battery of emf 2 V and negligible internal resistance. If the emf of the cell balances against a length of 217 cm of the potentiometer wire, find the emf of the cell. When the cell is shunted by a resistance of 15 Ω, the balancing length is reduced by 17 cm. Find the internal resistance of the cell.

Answer 1

Data: L = 4 m, R = 8 Ω, E = 2V, r = 0,

R(shunt) = 15 Ω, l = 217 cm = 2.17 m,

l₁ = 217 – 17 = 200 cm = 2 m

potential gradient, 

K = \(\cfrac VL\) = \(\cfrac{ER}{(R+r)L}\)

= \(\cfrac{2\times8}{(8+0)\times4}\) = 0.5 V/m

If the emf of the cell is E₁,

E₁ = KI

= 0.5 x 2.17 = 1.085 V

The internal resistance of the cell is

r(cell) = R(shunt)\(\left(\cfrac{I}{I_1}-1 \right)\)

= 15 \(\left(\cfrac{217}{200}-1 \right)\) = 15 x 0.085

= 1.275 Ω