A 10 V battery of negligible internal resistance is connected with a 200 V battery and a set of parallel resistances as shown in the figure. The value of current in the circuit is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 15 A
The circuit is a single loop: the two batteries and the resistor combination are all in series. The three \( 38\,\Omega \) resistors are in parallel with each other.
\( \displaystyle R_{\mathrm{eq}} = \frac{38\,\Omega}{3} \)
Going round the loop, the two batteries are connected so that they oppose each other, so the net emf is their difference.
\( \displaystyle E_{\mathrm{net}} = 200\,\mathrm{V} - 10\,\mathrm{V} = 190\,\mathrm{V} \)
\( \displaystyle I = \frac{E_{\mathrm{net}}}{R_{\mathrm{eq}}} = \frac{190 \times 3}{38} = 15\,\mathrm{A} \)