In a given L - C circuit, switch S is closed at t = 0, then
A
maximum charge on capacitor can be 2Cε.
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B
when charge on capacitor is half of its maximum charge, current is maximum
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C
Charge on capacitor is εC2 at t=π3√LC
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D
Maximum charge on capacitor can be εC
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Solution
The correct options are A maximum charge on capacitor can be 2Cε. B when charge on capacitor is half of its maximum charge, current is maximum C Charge on capacitor is εC2 at t=π3√LC
ε=qC+Ldidt ⇒ε−qC=Ld2qdt2 εC−qLC=d2qdt2=didt⇒d2idt2=−iLC d2idt2=−w2i=w=1√LC&T=2πw i=i0sinwt or didt=i0wcoswt At t = 0, Ldidt=ε⇒ε=Li0w⇒i0=εLw=ε√CL&i=dqdt=i0sinwt ∫q0dq=∫t0i0sinwtdt⇒q=−i0w[coswt]t0 q=i0w(1−coswt)⇒q=q0(1−coswt) where q0=i0w=εC qmax=2q0=2εC When charge on the capacitor is half of its maximum charge i.e., q0, q=q0(1−coswt) ⇒q0=q0(1−coswt)⇒coswt=0 Therefore, i=i0sinwt=i0×1=i0, so current takes its maximum value. At t=π3√LC, q=q0(1−coswt)=εC(1−cosπ3)=εC2