Question

A potentiometer wire of length 100 cm and resistance \(3\,\Omega\) is connected in series with resistance of \(8\,\Omega\) and an accumulator of 4 volt whose internal resistance is \(1\,\Omega\). A cell of e.m.f. \(E\) is balanced at 50 cm length of the wire. The e.m.f. of the cell is

• A. \(1.00 \, \text{volt}\)
• B. \(0.75 \, \text{volt}\)
• C. \(0.50 \, \text{volt}\)
• D. \(0.25 \, \text{volt}\)

Hint:

In potentiometer problems, always find the potential gradient first.

Answer 1

The Correct Option is C

Step 1: Total resistance of the circuit.
\[ R_{\text{total}} = 3 + 8 + 1 = 12\,\Omega. \]
Step 2: Current in the circuit.
\[ I = \frac{4}{12} = \frac{1}{3}\,\text{A}. \]
Step 3: Potential drop across potentiometer wire.
\[ V_{\text{wire}} = I \times 3 = 1\,\text{V}. \]
Step 4: Potential gradient.
\[ k = \frac{1}{100} = 0.01\,\text{V/cm}. \]
Step 5: Calculating e.m.f. of the cell.
\[ E = k \times 50 = 0.5\,\text{V}. \]
Step 6: Conclusion.
The e.m.f. of the cell is \(0.50 \, \text{V}\).