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JEE Mains · Physics · STD 12 -7. Alternating current
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
- A \(10^{-15}\,C\)
- B \(10^{-6}\,C\)
- C \(10^{-10}\,C\)
- D \(10^{-8}\,C\)
Answer & Solution
Correct Answer
(D) \(10^{-8}\,C\)
Step-by-step Solution
Detailed explanation
\(\mathrm{C}=10^{-11} \,\mathrm{F}, \mathrm{L}=10^{-5} \,\mathrm{Henry},\) and \(\mathrm{R}=100\) \(ohm\) \({I_{max}}({\text{wL}} - 1/{\text{wC}} + {\text{R}}) = {V_{max}}.........(1)\) \(\mathrm{wL}=\) inductive reactance \(1 /{wC}=\) capacitive reactance \({R}\) is resistance…
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