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Question

# A circuit containing capacitors C1 and C2 , shown in the figure is in the steady state with key K1 closed and K2 opened. At the instant t=0,K1 is opened and K2 is closed. Determine the first instant t, when energy in the inductor becomes one third of that in the capacitor.

A
3.31×106s
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B
8.05×105s
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C
7.05×105s
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D
6.05×105s
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Solution

## The correct option is A 3.31×10−6sSince at t=0, charge is maximum (=q0).Therefore current will be zero.12Li2=13(12q2C)or i=q√3LC=qω√3From the expression i=ω√q20−q2We have, qω√3=ω√q20−q2or q=√32q0Since at t=0, charge in maximum or q0 so we can writeq=q0cosωt or √3q02=q0cosωtω=1√LC1=1√0.2×10−3×2×10−6=158110 rad/sor ωt=π6or t=π6ω=π6×158110=3.31×10−6s

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