Determination of the internal resistance of a cell using the potentiometer

Let,

• $E$ = electromotive force (e.m.f.) of the cell
• $r$ = internal resistance of the cell
• $R$ = external resistance connected to the cell
• $l_1$ = balancing length when the key is open
• $l_2$ = balancing length when the key is closed

When the key is open:

$$E \propto l_1$$

When the key is closed:

$$V \propto l_2$$

Also,

$$V = \frac{ER}{R+r}$$

Therefore,

$$\frac{l_2}{l_1} = \frac{R}{R+r}$$

Rearranging,

$$r = R\left(\frac{l_1-l_2}{l_2}\right)$$

Or,

$$\frac{1}{l_2} = \frac{r}{K}\frac{1}{R} + \frac{1}{K}$$

where

$$K = \frac{l_1}{E}$$

If $\dfrac{1}{l_2}$ is plotted against $\dfrac{1}{R}$, a straight line is obtained.

From the graph,

• Gradient $= \dfrac{Kr}{E}$ (or proportional to $r$)
• Intercept $= \dfrac{K}{E}$

Hence,

$$r = \frac{\text{Gradient}}{\text{Intercept}}$$

1. Set up the circuit as shown in the figure.
2. Check the circuit as done in the potentiometer experiment.
3. Close key $K_1$ and key $K_2$ open.
4. Find the balancing length $l_1$ and record it.
5. Close key $K_2$ to connect the resistor box $R$ across the cell.
6. Measure the new balancing length $l_2$.
7. Repeat the experiment for several values of $R$.
8. Record all values in Table 34.1.

$R$ (Ω)
$l_2$ (cm)
$\dfrac{1}{R}$ (Ω$^{-1}$)
$\dfrac{1}{l_2}$ (cm$^{-1}$)

1. Plot the graph of:

$$\frac{1}{l_2} \quad \text{against} \quad \frac{1}{R}$$

2. Determine the gradient of the graph.
3. Determine the intercept of the graph.
4. Calculate the internal resistance using:

$$r = \frac{\text{Gradient}}{\text{Intercept}}$$

Calculate and state the value of the internal resistance $r$ of the cell.