Question

In potentiometer experiment, the balancing length with a cell 'E₁' is 'l₁' cm. By shunting the cell with a resistance 'R' equal to half the internal resistance of the cell, the balancing length 'l2' will be (E.M.F. of driver cell E > E₁)

• A. $l_2 = \frac{1}{3} l_1$
• B. $l_2 = \frac{1}{4} l_1$
• C. $l_2 = l_1$
• D. $l_2 = \frac{1}{2} l_1$

Answer 1

Answer: (A)

Option (a) $l_2=\frac{1}{3}l_1$.

The terminal voltage becomes

$$V = E_1 \times \frac{R}{R+r} = E_1 \times \frac{r/2}{r/2 + r} = E_1 \times \frac{1}{3}.$$

Since the potentiometer wire is uniform, the balancing length is proportional to the potential difference. Thus,

$$l_2=\frac{1}{3}l_1.$$

Answer 2

When the cell is shunted by a resistance $R = \frac{r}{2}$, its terminal voltage becomes

$$V = E_1 \times \frac{R}{R+r} = E_1 \times \frac{r/2}{r/2 + r} = E_1 \times \frac{1}{3}.$$

Since the potentiometer wire is uniform, the balancing length is proportional to the potential difference. Thus,

$$l_2=\frac{1}{3}l_1.$$