The current supplied by the battery in the network shown in the figure below in steady state is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.60 A
In the steady state a capacitor is fully charged and no current flows through it, so the branch with the \( 2\,\mu\mathrm{F} \) capacitor and the \( 2\,\Omega \) resistor carries no current.
The battery current then flows only through the \( 4\,\Omega \) and \( 1\,\Omega \) resistors, which are in series.
\( \displaystyle I = \frac{E}{R_{\mathrm{total}}} = \frac{3\,\mathrm{V}}{(4 + 1)\,\Omega} = 0.60\,\mathrm{A} \)