A capacitor of capacitance 0.1 μF is connected to a battery of emf 8V as shown in the figure. Under steady state condition

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Charge on the capacitor is 0.4 μC

Charge on the capacitor is 0.2 μC

Current in the resistor (R) between points A and B is 0.1 A

Current in the resistor (R) between points A and B is 0.4 A

answer is A.

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Detailed Solution

The given circuit can be simplified as
$\begin{array}{l} R_{eq} = 20 Ω \\ i = \frac{8}{20} = 0 .4 A \\ I_{1} = 0 .2 A \\ V_{AB} = 4 V \\ charge on capacitor q = CV = 0 .1 \times 4 \\ = 0 .4 μC \end{array}$

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