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

Answer 1

Answer: (A)

(a)

Answer 2

Let the cell’s emf be $E_1$ and its internal resistance be $r$. When the cell is short‐circuited through a resistor $R = r/2$ the effective resistance is:

$R_{eff}=\frac{r\,({r/2})}{r+(r/2)}=\frac{r^2/2}{3r/2}=\frac{r}{3}$

Thus the terminal (loaded) voltage is

$V=E_1\frac{R}{r+R} = E_1\frac{r/2}{r+(r/2)} = E_1\frac{r/2}{3r/2}=\frac{E_1}{3}$

Since the balancing length is proportional to the voltage of the cell, the new length $l_2$ is one‐third of the original $l_1$.