In the steady state, the charge on the capacitor capacity $0 .2 μF is (in μC)$

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answer is 0.36.

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Detailed Solution

$R_{eq} = \frac{2 \times 3}{2 + 3} + 2 .8 = 4 \Rightarrow i = \frac{6}{4} = 1 .5 A;$

$\begin{array}{l} V_{R} = 2 .8 \times 1 .5 = 4 .2 V \Rightarrow V_{C} = 6 V - 4.2 V = 1 .8 V \\ q = {CV}_{C} = 0 .2 μ \times 1 .8 = 0 .36 μC \end{array}$

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