A cell of emf 3 V and internal resistance \( 0.1\,\Omega \) is connected to a \( 3.9\,\Omega \) external resistance. What will be the potential difference across the terminals of the cell ?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2.925 V
The current is set by the total resistance of the circuit, and the terminal voltage is the emf minus the drop across the internal resistance.
\( \displaystyle I = \frac{E}{R + r} = \frac{3}{3.9 + 0.1} = \frac{3}{4} = 0.75\,\mathrm{A} \)
\( V = E - I\,r = 3 - 0.75 \times 0.1 = 3 - 0.075 = 2.925\,\mathrm{V} \)
The same value follows from \( V = IR = 0.75 \times 3.9 = 2.925\,\mathrm{V} \).