Question

In a series $\text{LCR}$ circuit, $C={10}^{-11}\text{F},L={10}^{-5}\text{H}$ and $R=100\Omega$. When a constant $\text{DC}$ voltage $E$ is applied to the circuit, the capacitor acquires a charge of ${10}^{-9}\text{C}$. The $\text{DC}$ source is replaced by a sinusoidal voltage source in which the peak voltage is $E$. At resonance, the peak value of the charge acquired by the capacitor is ${10}^{-n}\text{C}$. The value of $n$ is :

Answer 1

Given :-

$C={10}^{-11}\text{F};L={10}^{-5}\text{H}$

$R=100\Omega ;Q_0={10}^{-9}\text{C}$

When $\text{DC}$ source is applied across the $\text{LCR}$ circuit

Charge on the capacitor is $Q_0=CE$

$\Rightarrow {10}^{-9}={10}^{-11}E$

$\therefore E={10}^2\text{V}=100\text{V}$

When $\text{AC}$ source is applied across the $\text{LCR}$ circuit

At resonance, the peak value of current flowing in the circuit is given by

$I_{max}=\frac{E}{Z}=\frac{E}{R}$ [$\because Z=R$ at resonance]

$\Rightarrow I_{max}=\frac{100}{100}=1\text{A}$

The maximum value of potential difference across the capacitor is
$V_{max}=I_{max}{X}_C=I_{max}\times \frac{1}{\omega C}$

$V_{max}=\frac{I_{max}}{\omega C}$

Therefore, the peak value of the charge acquired by the capacitor is

$Q_P=C{V}_{max}$

$Q_P=C\times \frac{I_{max}}{\omega C}=\frac{I_{max}}{\omega }$

At resonance $\omega =\frac{1}{\sqrt{LC}}$

$\Rightarrow Q_P=I_{max}\sqrt{LC}$

$Q_P=1\times \sqrt{{10}^{-5}\times {10}^{-11}}$

$\therefore Q_P={10}^{-8}\text{C}$

Comparing with $Q_P={10}^{-n}\text{C}$

Therefore, $n=8$.