Question

In an experiment with potentiometer , to measure the internal resistance of a cell. When it is shunted by $5\Omega$ the null point obtained is at $2\text{m}$ from one end. When cell is shunted by $20\Omega$, the null point is obtained at $3\text{m}$ from the same end. The internal resistance of cell is

• A. $8\Omega$
• B. $6\Omega$
• C. $4\Omega$
• D. $2\Omega$

Answer 1

The correct option is C $4\Omega$

Let the internal resistance of the cell be $r$.
$$\begin{array}{cc}{V}_1=I_1{R}_1.....\left(1\right)& \\ V_2=I_2{R}_2.....\left(2\right)& E=I_2(R_2+r)=I_1(R_1+r).....\left(3\right)\end{array}$$
Using these equation we get,

$\frac{I_1}{I_2}=\frac{R_2+r}{R_1+r}$

$\frac{V_1}{V_2}=\frac{I_1}{I_2}\times \left(\frac{R_1}{R_2}\right)=\frac{I\rho l_1}{I\rho l_2}=\frac{R_1}{R_2}\times \left[\frac{R_2+r}{R_1+r}\right]$

$\frac{l_1}{l_2}=\frac{R_1}{R_2}\left[\frac{R_2+r}{R_1+r}\right]\Rightarrow \frac{2}{3}=\frac{5}{20}\left[\frac{20+r}{5+r}\right]$

$8(5+r)=60+3r$

$8r-3r=60-40=20$

$5r=20$

$r=\frac{20}{5}=4\Omega$