Question

A null point is obtained at $200\text{ cm}$ on potentiometer wire when cell in secondary circuit is shunted by $5\Omega$. When a resistance of $15\Omega$ is used for shunting, null point moves to $300\text{ cm}$ . The internal resistance of the cell is

• A. $3\Omega$
• B. $4\Omega$
• C. $5\Omega$
• D. $6\Omega$

Hint:

If the shunt resistance triples and the null point increases by 1.5x, the internal resistance often equals the initial shunt.

Answer 1

The Correct Option is C

Step 1: Concept
The internal resistance $r$ is given by $r = R \left( \frac{E-V}{V} \right) = R \left( \frac{l_0 - l}{l} \right)$, where $l_0$ is balancing length on open circuit.

Step 2: Meaning
We have two cases: $r = 5 \left( \frac{l_0 - 200}{200} \right)$ and $r = 15 \left( \frac{l_0 - 300}{300} \right)$.

Step 3: Analysis
Equating the two expressions for $r$: $5 \left( \frac{l_0 - 200}{200} \right) = 15 \left( \frac{l_0 - 300}{300} \right)$ $\frac{l_0 - 200}{40} = \frac{l_0 - 300}{20} \implies l_0 - 200 = 2l_0 - 600 \implies l_0 = 400 \text{ cm}$. Now, $r = 5 \left( \frac{400 - 200}{200} \right) = 5 \left( \frac{200}{200} \right) = 5\Omega$.

Step 4: Conclusion
The internal resistance of the cell is $5\Omega$. Final Answer: (C)