Question

To determine the internal resistance of a cell with potentiometer, when the cell is shunted by a resistance of $5\Omega$ the balancing length is $250 \text{ cm}$. When the cell is shunted by $20\Omega$, the balancing length of potentiometer wire is $400 \text{ cm}$. The internal resistance, of the cell is

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

Hint:

Balancing length $l_2$ is always less than $l_1$ because terminal voltage is less than EMF.

Answer 1

The Correct Option is B

Step 1: Concept
Internal resistance $r = R\left(\frac{l_1}{l_2} - 1\right)$, where $l_1$ is the open-circuit balance length and $l_2$ is the shunt balance length.

Step 2: Meaning
Case 1: $r = 5\left(\frac{l_1}{250} - 1\right)$. Case 2: $r = 20\left(\frac{l_1}{400} - 1\right)$.

Step 3: Analysis
$5\left(\frac{l_1 - 250}{250}\right) = 20\left(\frac{l_1 - 400}{400}\right)$.
$\frac{l_1 - 250}{50} = \frac{l_1 - 400}{20} \implies 2(l_1 - 250) = 5(l_1 - 400)$.
$2l_1 - 500 = 5l_1 - 2000 \implies 3l_1 = 1500 \implies l_1 = 500 \text{ cm}$.

Step 4: Conclusion
$r = 5\left(\frac{500}{250} - 1\right) = 5(2-1) = 5\Omega$. Final Answer: (B)