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For the circuit shown in the diagram, considering steady-state :

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The current I₁ is 2.4 A

The current I is 4.8 A

Power supplied by 18V battery is 49.2

Charge on $7 μF$ capacitor is $126 μC$

answer is A, C.

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Detailed Solution

In steady state, $I_{1} = I = \frac{18 - 6}{5} = 2.4 A$
${ΔV}_{across} 7 μF capacitance = 7 \times 18 = 126 μC$
Power supplied by battery = 18 x 2.4 = 43.2 W

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