Question

In a series $L-C-R$ circuit, $C=10^{-11}$ Farad, $L=10^{-5}$ Henry and $R=100Ohm$, 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; $R=100\Omega$
$I_{\max }\left(\omega L-\frac{1}{\omega C}+R\right)=V_{\max }\to$ (1)
$WL=$ inductive reactance; $\frac{1}{WC}=$ capacitive reactance
' $R$ ' is resistance
At resonance $\omega L=\frac{1}{\omega C}$;
so (1) becomes
$I_{\max }(R)=V_{\max }=\frac{E_0}{R}\to 0$
$q=CE$
(Given at steady state $q=10^{-9}C$ )
Put values of ${ϵ}_0$ we get
$E=E_0=\frac{10^{-9}}{10^{-11}}=100$ volt $\to$ (2)
put $E_0$ in $(1)$
$\Rightarrow I_{\max }=1A$;
Resonance $W={\left(\frac{1}{LC}\right)}^{Y_2}=10^8$
$Q_{\max }=\frac{I_{\max }}{\omega }=10^{-8}$