Two cells of emf 2 V and 4 V and internal resistance of \( 0.5\,\Omega \) each are connected with an external resistor of resistance \( 3\,\Omega \) as shown in the figure. The terminal potential difference of 2 V cell is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2.25\,\mathrm{V} \)
With the key K closed there is a single loop. The positive terminals (longer plates) of the two cells face opposite ways, so the cells are in series opposition and the net emf is the difference of the two emfs.
\( \displaystyle I = \frac{E_2 - E_1}{R + r_1 + r_2} = \frac{4 - 2}{3 + 0.5 + 0.5} = \frac{2}{4} = 0.5\,\mathrm{A} \)
The 4 V cell drives this current, so it enters the 2 V cell at its positive terminal: the 2 V cell is being charged. For a cell being charged, the terminal voltage exceeds the emf:
\( V = E_1 + I r_1 = 2 + 0.5 \times 0.5 = 2.25\,\mathrm{V} \)