Question

e.m.f. of a cell is 1.5 volt and internal resistance is \( 0.1 \, \Omega \). On connecting the cell with an external resistance of \( 2.9 \, \Omega \), what will be the potential difference at the terminals of the cell?

Hint:

The potential difference at the terminals of the cell is calculated by subtracting the voltage drop across the internal resistance from the e.m.f. of the cell.

Answer 1

Step 1: Total Resistance in the Circuit.
The total resistance \( R_{\text{total}} \) is the sum of the internal resistance of the cell and the external resistance: \[ R_{\text{total}} = 0.1 \, \Omega + 2.9 \, \Omega = 3 \, \Omega \]
Step 2: Current in the Circuit.
Using Ohm's law, the current \( I \) in the circuit is given by: \[ I = \frac{V}{R_{\text{total}}} = \frac{1.5 \, \text{V}}{3 \, \Omega} = 0.5 \, \text{A} \]
Step 3: Potential Difference Across the External Resistance.
The potential difference across the external resistance is given by: \[ V_{\text{external}} = I \times R_{\text{external}} = 0.5 \, \text{A} \times 2.9 \, \Omega = 1.45 \, \text{V} \]
Step 4: Conclusion.
The potential difference at the terminals of the cell is: \[ V_{\text{terminals}} = 1.5 \, \text{V} - (0.5 \, \text{A} \times 0.1 \, \Omega) = 1.5 \, \text{V} - 0.05 \, \text{V} = 1.45 \, \text{V} \]