The internal resistance of a cell of emf 2 V is \( 0.1\,\Omega \). It is connected to an external resistance of \( 3.9\,\Omega \). The terminal potential difference across the cell will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.95\,\mathrm{V} \)
The current drawn from a cell is the emf divided by the total resistance, and the terminal potential difference is the emf less the drop across the internal resistance.
\( \displaystyle I = \frac{\varepsilon}{R + r} = \frac{2}{3.9 + 0.1} = 0.5\,\mathrm{A} \)
\( \displaystyle V = \varepsilon - I r = 2 - 0.5 \times 0.1 = 1.95\,\mathrm{V} \)
The same value follows from the external resistance: \( V = IR = 0.5 \times 3.9 = 1.95\,\mathrm{V} \).