9.2 Determination of Internal resistance of a cell using potentiometer

Initially, key $K'$ is kept open and balancing length ${l_1}$ is obtained.

Since cell $\xi $ is in open circuit so its emf balances on the length ${l_1}$. So, $$\xi = k{l_1}$$

With the help of resistance box, resistance $R$ is introduced. So, key $K'$ is closed.

Terminal potential difference $(V)$ balances on length ${l_2}$. So, $$V = k{l_2}$$

By ohm's law, $$\xi = I\left( {R + r} \right)$$

And $V = IR$, $$\begin{equation} \begin{aligned} \Rightarrow \frac{\xi }{V} = \frac{{R + r}}{R} = \frac{{{l_1}}}{{{l_2}}} \\ \Rightarrow r = R\left[ {\frac{{{l_1} - {l_2}}}{{{l_2}}}} \right] \\\end{aligned} \end{equation} $$

The sensitivity of a potentiometer can be increased by increasing the length of potentiometer wire.

Question 29. A potentiometer wire is supplied a constant voltage 3 $V$ . A cell of emf 1.08 $V$ is balanced by the voltage drop across 216 $cm$ of the wire. Find the total length of the potentiometer wire.

Given $\xi = 3V$, ${\xi _1} = 1.08V$, ${l_1} = 216cm$

Using, $$\begin{equation} \begin{aligned} \Rightarrow \frac{\xi }{{{\xi _1}}} = \frac{l}{{{l_1}}} \\ \Rightarrow l = \frac{\xi }{{{\xi _1}}} \times {l_1} \\ \Rightarrow l = \frac{{3 \times 216}}{{1.08}} \\ \Rightarrow l = 600cm \\\end{aligned} \end{equation} $$

Question 30. A potentiometer wire of length 100 $cm$ has a resistance of $10\Omega $. It is connected in series with a resistance and a battery of emf $2V$ and of negligible internal resistance. A source of emf 10 $mV$ is balanced against a length of 40 $cm$ of the potentiometer wire. What is the value of the external resistance?

Solution: Figure shows a potentiometer wire of length 100 $cm$ connected in series to a cell of emf 2 $V$ and an unknown resistance $R$.

The cell of emf 10 $mV$ balances length $AJ=40\ cm$.

Resistance per unit length of potentiometer wire = $\frac{{10}}{{100}}$

Resistance of wire $AJ$ = $$R_{AJ}=\frac{{10}}{{100}} \times 40 = 4\Omega $$

Current through the wire $AJ$, $$I = \frac{{10mV}}{{4\Omega }} = \frac{{10 \times {{10}^{ - 3}}V}}{{4\Omega }} = 2.5 \times {10^{ - 3}}A$$

The same current flows through the potentiometer wire and through the external resistance $R$.

Total resistance = $\left( {R + 10} \right)\Omega $

Therefore, $$2.5 \times {10^{ - 3}} = \frac{{2V}}{{\left( {R + 10} \right)\Omega }}$$

or, $$R + 10 = \frac{2}{{2.5 \times {{10}^{ - 3}}}} = 800$$

$$R = 800 - 10 =790\Omega $$

Question 31. The length of wire of a potentiometer is 100 $cm$ and the emf of its standard cell is $\xi $volt. It is employed to measure the emf of a battery, whose internal resistance is $0.5\Omega $. If the balance point is obtained at $l = 30 cm$ from the positive end. Find the emf of the battery.

Solution: For the standard cell $\xi $ , balancing length is ${l_1} = 100cm$

For the battery of unknown emf $\xi' $, balancing length is ${l_2} = 30cm$

$$\frac{{\xi '}}{\xi } = \frac{{{l_2}}}{{{l_1}}} = \frac{{30}}{{100}}$$

$$\xi ' = \frac{{30}}{{100}}\xi $$

Question 32. When a resistor of $5\Omega $ is connected across the cell it is balanced at 150 $cm$ of potentiometer wire and when a resistor of $10\Omega $ resistance is connected across the cell, it is balanced at 175 $cm$ of the potentiometer wire. Find the internal resistance of the cell.

Solution: In first case, $$r = {R_1}\left( {\frac{{l - {l_1}}}{{{l_1}}}} \right) \Rightarrow r\frac{{{l_1}}}{{{R_1}}} = l - {l_1}$$

In second case, $$r = {R_2}\left( {\frac{{l - {l_2}}}{{{l_2}}}} \right) \Rightarrow r\frac{{{l_2}}}{{{R_2}}} = l - {l_2}$$

Substracting above equations, $$r\left( {\frac{{{l_1}}}{{{R_1}}} - \frac{{{l_2}}}{{{R_2}}}} \right) = l - {l_1} - l + {l_2}$$

Thus, $$r = \frac{{{l_2} - {l_1}}}{{\frac{{{l_1}}}{{{R_1}}} - \frac{{{l_2}}}{{{R_2}}}}} = \frac{{175 - 150}}{{\frac{{150}}{5} - \frac{{175}}{{10}}}} = \frac{{25}}{{12.5}} = 2\Omega $$