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Question

# In the circuit shown in figure, the key k1, was closed for long time. At t=0, key k1 is opened and k2 is closed. What is the charge on capacitor C1, at the instant energy stored in it is three times of energy stored in inductor?

A
103μC
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B
5μC
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C
52μC
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D
10μC
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Solution

## The correct option is C 10√3μCThe voltage drop across C1 will be half of battery voltage as there is no current flowing when key k1 is closed at t=∞ and capacitance of both the capacitors is sameV1=20/2=10VCharge on equivalent capacitor Q0=C1V1=2μF∗10=20μC before k2 is closed charge on capacitor C1isQ0=CV=20μCAs the key k2 is closed and k1 is opened, the energy will oscillate between capacitor and inductor in LC circuit,total energy of system will remain constant.Uc+UL=Q202×C1The question states that energy in capacitor is three times of inductor i.e.Uc=3UL43Q22C1=Q02×C1Q=√32Q0=10√3μc

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