A 10 V battery having 3 Ω internal resistance is connected with a 200 V battery and a 35 Ω resistance as shown in the figure. The terminal potential drop at 10 V battery will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 25 V
The circuit is a single loop in which the two batteries are connected in opposition, so the net emf is their difference and the larger 200 V battery drives the current.
\( \displaystyle I = \frac{200 - 10}{35 + 3} = \frac{190}{38} = 5\,\mathrm{A} \)
This current enters the 10 V battery at its positive terminal, so the 10 V battery is being charged. For a cell being charged, the terminal voltage is the emf plus the drop across its internal resistance.
\( \displaystyle V = E + I r = 10 + 5 \times 3 = 25\,\mathrm{V} \)