Two capacitors of capacitances 3C and 2C are connected in series with an inductor of inductance L. Potential differences across the capacitors are VP−VQ=V0,VR−VQ=72V0. Initial current in the circuit is zero. Energy in capacitor with capacitance 3C when current in the inductor is maximum is

Current in the inductor flows till the potentials of the capacitors become equal. So common potential V=3C×V0+2C×7V023C+2C=2V0Now energy stored on the capacitor of 3C U=12(3C)(2V0)2 =6CV02

Two capacitors of capacitances 3C and 2C are connected in series with an inductor of inductance L. Potential differences across the capacitors are VP−VQ=V0,VR−VQ=72V0. Initial current in the circuit is zero. Energy in capacitor with capacitance 3C when current in the inductor is maximum is

Current in the inductor flows till the potentials of the capacitors become equal. So common potential V=3C×V0+2C×7V023C+2C=2V0Now energy stored on the capacitor of 3C U=12(3C)(2V0)2 =6CV02

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Two capacitors of capacitances 3C and 2C are connected in series with an inductor of inductance L. Potential differences across the capacitors are $V_{P} - V_{Q} = V_{0}, V_{R} - V_{Q} = \frac{7}{2} V_{0} .$ Initial current in the circuit is zero. Energy in capacitor with capacitance 3C when current in the inductor is maximum is

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zero

$6 C V_{0}^{2}$

$\frac{3}{2} C V_{0}^{2}$

$3 C V_{0}^{2}$

answer is C.

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Detailed Solution

Current in the inductor flows till the potentials of the capacitors become equal. So common potential
$V = \frac{3 C \times V_{0} + 2 C \times \frac{7 V_{0}}{2}}{3 C + 2 C} = 2 V_{0}$
Now energy stored on the capacitor of 3C
$U = \frac{1}{2} (3 C) {(2 V_{0})}^{2}$
$= 6 C V_{0}^{2}$