Question

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

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

Hint:

Use \(r=R\left(\dfrac{l_1}{l_2}-1\right)\) for a cell shunted by resistance R.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Without the shunt the balance length measures the emf: \(E\propto l_1=240\) cm. With a shunt \(R\) the balance gives the terminal voltage: \(V\propto l_2=120\) cm.

Step 2: Relation
\(\dfrac{E}{V}=\dfrac{l_1}{l_2}=\dfrac{R+r}{R}\).

Step 3: Calculation
\(\dfrac{240}{120}=2=\dfrac{2+r}{2}\), so \(2+r=4\).

Step 4: Result
\(r=2\ \Omega\).

Final Answer:
The internal resistance of the cell is 2 ohm. \[ \boxed{\text{(D) }2\ \Omega} \]