In a series L - C - R circuit, C = 10⁻¹¹ Farad, L = 10⁻⁵ Henry and R… | JEE Main 2013 (22 Apr, Shift 1)

Q21JEE Main 2013 · 22 APR · SHIFT 1Difficult

LC Oscillations

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:

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Answer: (d)

DC case: $Q = CE \Rightarrow E = \frac{10^{-9}}{10^{-11}} = 100\text{ V}$
At resonance $Z = R$ : $I_0 = \frac{E_0}{R} = \frac{100}{100} = 1\text{ A}$
$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-16}}} = 10^{8}\text{ rad/s}$
$Q_0 = \frac{I_0}{\omega_0} = 10^{-8}\text{ C}$
So the answer is (d) $10^{-8}\text{ C}$ .

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