In a series \(L-C-R\) circuit, \(C = 10^{-11}\,Farad.\) \(L = 10^{-5}\,Henry\) and \(R =100\,Ohm.\) when a constant \(D.C.\) voltage \(E\) is applied to the circuit, the capacitor acquires a charge \(10^{-9}\,C\) .The \(D.C.\) source is replaced by a sinusoidal voltage source in which the peak voltage \(E_0\) is equal to the constant \(D.C.\) voltage \(E.\) At resonance the peak value of the charge acquired by the capacitor will be

\({I_{max}}({\text{wL}} - 1/{\text{wC}} + {\text{R}}) = {V_{max}}.........(1)\)

\(\mathrm{wL}=\) inductive reactance

\(1 /{wC}=\) capacitive reactance

\({R}\) is resistance

at resonance: \( w |=1 / w C\)

so equation \((1)\) becomes

\({I_{max}}(\mathrm{R})=V_{max}\)

\(I_{max} =\mathrm{E}_{\mathrm{o}} / \mathrm{R} .........(1)\)

Calculation of Eo

initially the circuit is connected with constant battery source:

at steady state potential across capacitor becomes \(=E=E \circ\)

we know that

\(q=C E \quad\left(\text { Given at steady state } q=10^{-9} \,\mathrm{C}\right)\)

put values and get \(E\) as

\(E=E_{0}=10^{-9} / 10^{-11}=100\,\mathrm{volt}.........(2)\)

Put value of \(Eo\) in \(1\)

Get \(1_{max}\) \(=1\,amp\)

now at resonance \(w=(1 / L C)^{1 / 2}=10^{8}\)

\(Q _{max} ={I_{max}} / w=10^{-8}\)