Question

In a potentiometer experiment, the balancing length with a cell is \(240~\text{cm}\). On shunting the cell with \(2~\Omega\), the balancing length becomes \(120~\text{cm}\). The internal resistance of the cell is

• A. \(0.5~\Omega\)
• B. \(1~\Omega\)
• C. \(2~\Omega\)
• D. \(4~\Omega\)

Hint:

Use: \( \frac{l_2}{l_1} = \frac{R}{R + r} \), where \(r\) is internal resistance, \(R\) is external shunt.

Answer 1

The Correct Option is C

Let:
- \(r\) = internal resistance of the cell
- \(R = 2~\Omega\) = external shunt resistance
- \(l_1 = 240~\text{cm}\) = balancing length without shunt
- \(l_2 = 120~\text{cm}\) = balancing length with shunt
Since potential drop \(\propto\) balancing length, we use: \[ \frac{l_2}{l_1} = \frac{V'}{V} = \frac{R}{R + r} \Rightarrow \frac{120}{240} = \frac{2}{2 + r} \Rightarrow \frac{1}{2} = \frac{2}{2 + r} \Rightarrow 2 + r = 4 \Rightarrow r = 2~\Omega \]

Answer 2

Step 1: Understand the given data
- Initial balancing length, \(L_1 = 240~\text{cm}\)
- Balancing length after shunting with resistance, \(L_2 = 120~\text{cm}\)
- Shunt resistance, \(R = 2~\Omega\)

Step 2: Use the potentiometer balancing condition
Balancing length is proportional to the emf of the cell.
When the cell is shunted by resistance \(R\), the effective emf changes due to internal resistance \(r\).

Step 3: Write the relation for emf before and after shunting
Original emf, \(E = k L_1\)
Effective emf after shunt, \(E' = k L_2\)
where \(k\) is a constant related to the potentiometer wire.

Step 4: Calculate emf ratio and relate to internal resistance
The current through the shunt and cell changes the effective emf as:
\[ E' = E \times \frac{R}{R + r} \]
Using balancing lengths:
\[ \frac{L_2}{L_1} = \frac{R}{R + r} \]
Substitute values:
\[ \frac{120}{240} = \frac{2}{2 + r} \]
\[ \frac{1}{2} = \frac{2}{2 + r} \]
Cross-multiplied:
\[ 2 + r = 4 \implies r = 2~\Omega \]

Step 5: Final answer
The internal resistance of the cell is \(2~\Omega\).