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

Concept: In a potentiometer: \[ \text{Balance length} \propto \text{terminal voltage} \] For a cell of emf \(E\) and internal resistance \(r\), connected to an external resistance \(R\): \[ V = E\frac{R}{R+r} \]
Step 1: Write expressions for balance lengths For \(R_1=4\,\Omega\), balance length \(l_1=120\) cm: \[ l_1 \propto \frac{4}{4+r} \] For \(R_2=12\,\Omega\), balance length \(l_2=180\) cm: \[ l_2 \propto \frac{12}{12+r} \]
Step 2: Take ratio of balance lengths \[ \frac{120}{180}=\frac{\frac{4}{4+r}}{\frac{12}{12+r}} \Rightarrow \frac{2}{3}=\frac{4(12+r)}{12(4+r)} \]
Step 3: Solve for \(r\) \[ 2 \times 12(4+r)=3 \times 4(12+r) \] \[ 24(4+r)=12(12+r) \] \[ 96+24r=144+12r \] \[ 12r=48 \Rightarrow r=4 \]