Question

In the experiment to determine the internal resistance of a cell ($E_1$) using potentiometer, the resistance drawn from the resistance box is $R$. The potential difference across the balancing length of the wire is equal to the terminal potential difference $(V)$ of the cell. The value of internal resistance $(r)$ of the cell is

• A. $R\left(\dfrac{E_1}{V}+1\right)$
• B. $R\left(\dfrac{V}{E_1}-1\right)$
• C. $R\left(\dfrac{V}{E_1}+1\right)$
• D. $R\left(\dfrac{E_1}{V}-1\right)$

Hint:

In potentiometer experiments, EMF is measured without current, while terminal voltage is measured when current flows.

Answer 1

The Correct Option is D

Step 1: Relation between EMF and terminal voltage.
For a cell with internal resistance $r$ and external resistance $R$: \[ V = \frac{E_1 R}{R + r} \]
Step 2: Rearranging the equation.
\[ \frac{E_1}{V} = \frac{R + r}{R} \]
Step 3: Solving for internal resistance.
\[ \frac{E_1}{V} = 1 + \frac{r}{R} \] \[ r = R\left(\frac{E_1}{V} - 1\right) \]
Step 4: Conclusion.
The internal resistance of the cell is $R\left(\dfrac{E_1}{V} - 1\right)$.