A circuit consists of two capacitors, a 24V battery and an AC source connected as shown in figure. The AC voltage is given by $ε = (20 \cos 120 πt) V$ , where t is in second.

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Charge on capacitor $C_{2}$ as function of time $Q_{2} = (30 μ C) \cos (120 π t) + 36 μ C$

Steady state current is 33.9 mA $\sin (120 πt + π)$

Maximum energy stored in capacitor $C_{1} and C_{2} is 36 mJ$

Minimum energy stored in capacitors $C_{1} and C_{2} is 36 μ J$

answer is A, B, D.

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Detailed Solution

$E = 24 + 20 \cos 120 π t C h a r g e o n C_{2} \begin{array}{l} Q_{2} = {EC}_{2} = (24 + 20 \cos 120 πt) 1.5 μC \\ = 36 + 30 \cos 120 πtμC \end{array}$

$\begin{array}{l} i_{2} = \frac{{dQ}_{2}}{dt} = - 30 \times 120 π s i n (120 π t) \\ i_{1} = \frac{dQ}{dt} = \frac{d}{dt} (24 + 20 c o s (120 π t) 3 μ C) \\ = - 60 \times 120 π tsin (120 π t) \\ i = i_{1} + i_{2} = - 90 \times 120 π tsin (120 π t) \\ i = 33.9 s i n (120 π t + π) mA \end{array}$
Minimum energy stored
$= \frac{1}{2} {CV}^{2} = \frac{1}{2} (3 + 1.5) (4)^{2} μJ = 36 μJ$