In the given circuit diagram, the current through the battery and the charge on the capacitor respectively in steady state are

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$1 A and 3 μC$

$11 A and 3 μC$

$\frac{6}{7} A and \frac{12}{7} μC$

$17 A and 0 μC$

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{1}{R} = 1 + \frac{1}{2} + \frac{1}{3} \\ \frac{1}{R} = \frac{6 + 3 + 2}{6} \\ R = \frac{6}{11} Ω \\ V = IR \\ 6 = I \frac{6}{11} \\ I = 11 A \\ q = CV \\ q = 0 .5 X 6 \\ q = 3 μc \end{array}$

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