A cell of e.m.f 4 V and internal resistance \( 0.2\,\Omega \) is connected to \( 1.8\,\Omega \) (ohm) external resistance. The potential difference across the terminal of the cell will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3.6 V
The current in the circuit follows from Ohm’s law for the whole circuit, \( I = \frac{E}{R + r} \). The potential difference across the terminals of the cell is the potential difference across the external resistance, \( V = IR = E - Ir \).
\( \displaystyle I = \frac{4}{1.8 + 0.2} = 2\,\mathrm{A} \)
\( V = IR = 2\times1.8 = 3.6\,\mathrm{V} \)
Check: \( V = E - Ir = 4 - 2\times0.2 = 3.6\,\mathrm{V} \).