Question

In the potentiometer, when the cell in the secondary circuit is shunted with \(4\,\Omega\) resistance, the balance is obtained at a length \(120\) cm of wire. Now when the same cell is shunted with \(12\,\Omega\) resistance, the balance is shifted to a length of \(180\) cm. The internal resistance of the cell is ________ \( \Omega \).

• A. \(12\)
• B. \(4\)
• C. \(6\)
• D. \(3\)

Hint:

In potentiometer problems, emf cancels out—always compare balance lengths using terminal voltage expressions.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A potentiometer measures the terminal voltage of a cell when shunted. The internal resistance \(r\) is related to the open-circuit balance length (\(l_0\)) and the shunted balance length (\(l\)).
Step 2: Key Formula or Approach:
\[ r = R \left( \frac{l_0}{l} - 1 \right) \]
where \(R\) is the shunt resistance and \(l\) is the corresponding balance length.
Step 3: Detailed Explanation:
Let \(l_0\) be the balance length for the EMF of the cell.
Case 1: \(R_1 = 4 \, \Omega\), \(l_1 = 120 \text{ cm}\).
\[ r = 4 \left( \frac{l_0}{120} - 1 \right) \quad \dots (i) \]
Case 2: \(R_2 = 12 \, \Omega\), \(l_2 = 180 \text{ cm}\).
\[ r = 12 \left( \frac{l_0}{180} - 1 \right) \quad \dots (ii) \]
Equating (i) and (ii):
\[ 4 \left( \frac{l_0 - 120}{120} \right) = 12 \left( \frac{l_0 - 180}{180} \right) \]
\[ \frac{l_0 - 120}{30} = 12 \left( \frac{l_0 - 180}{180} \right) \]
\[ \frac{l_0 - 120}{30} = \frac{l_0 - 180}{15} \]
\[ l_0 - 120 = 2(l_0 - 180) \]
\[ l_0 - 120 = 2l_0 - 360 \Rightarrow l_0 = 240 \text{ cm} \]
Substitute \(l_0 = 240\) into equation (i):
\[ r = 4 \left( \frac{240}{120} - 1 \right) = 4(2 - 1) = 4 \, \Omega \]
Step 4: Final Answer:
The internal resistance of the cell is \(4 \, \Omega\).