Question

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 1

Answer: (D)

Given $C=10^{-11} F ; L=10^{-5}$ Henry; $R=100 \Omega$
$I_{\max }\left(\omega L-\frac{1}{\omega C}+R\right)=V_{\max } \rightarrow$ (1)
$W L=$ inductive reactance; $\frac{1}{W C}=$ capacitive reactance
' $R$ ' is resistance
At resonance $\omega L=\frac{1}{\omega C}$;
so (1) becomes
$I_{\max }(R)=V_{\max }=\frac{E_{0}}{R} \rightarrow 0$
$q=C E$
(Given at steady state $q=10^{-9} C$ )
Put values of $\epsilon_{0}$ we get
$E=E_{0}=\frac{10^{-9}}{10^{-11}}=100$ volt $\rightarrow$ (2)
put $E_{0}$ in $(1)$
$ \Rightarrow I_{\max }=1 A$;
Resonance $W=\left(\frac{1}{L C}\right)^{Y_{2}}=10^{8}$
$Q_{\max }=\frac{I_{\max }}{\omega}=10^{-8}$