In the potentiometer, when the cell in the secondary circuit is shunted with $4 \Omega$ resistance,

Question

In the potentiometer, when the cell in the secondary circuit is shunted with $4 \Omega$ resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with $12 \Omega$ resistance, the balance is shifted to a length of 180 cm . The internal resistance of cell is $\_\_\_\_$ $\Omega$

Professor PC Thomas Classes · Answer

The balancing length $l$ in a potentiometer shunted with a resistance $R$ is proportional to the terminal voltage $V = \frac{ER}{R+r}$. Thus, $l \propto \frac{R}{R+r}$. We can construct the ratio for the two cases: $\frac{l_1}{l_2} = \frac{R_1 / (R_1 + r)}{R_2 / (R_2 + r)}$. Substituting the given values ($l_1 = 120$, $R_1 = 4$, $l_2 = 180$, $R_2 = 12$): $\frac{120}{180} = \frac{4 / (4+r)}{12 / (12+r)} \implies \frac{2}{3} = \frac{12+r}{3(4+r)}$. Cross-multiplying gives $2(4+r) = 12+r \implies 8 + 2r = 12 + r \implies r = 4 \Omega$.