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  • In a series L-C-R circuit,  <img alt="L=10^{-5}\, Henry" src="ht

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 : 

Option: 1

10^{-15}C
 


Option: 2

10^{-6}C


Option: 3

10^{-10}C

 


Option: 4

10^{-8}C


Answers (1)

best_answer

C= 10-11F, L = 10-5Henry, and R= 100 ohm
 \omega L= inductive reactance
\frac{1}{\omega C} = capacitive reactance
R is resistance

at resonance:   \omega L=\frac{1}{\omega C}
i.e Imax(R) = Vmax
Imax = Eo/R ......(1)

Calculation of Eo
initially, the circuit is connected with a constant battery source:
at steady state potential across the capacitor becomes = E = Eo
we know that 
q = C E   ( Given at steady state q= 10-9 C)
put values and get E as
E = Eo = 10-9/10-11  = 100 volt.............(2)
Put the value of Eo in equation (1)
Get Imax = 1 amp

The maximum value of potential difference across the capacitor is
\begin{aligned} & \mathrm{V}_{\max }=I_{\max } \mathrm{X}_{\mathrm{C}}=I_{\max } \times \frac{1}{\omega C} \\ & \mathrm{~V}_{\max }=\frac{I_{\max }}{\omega C}\end{aligned}

Therefore, the peak value of the charge acquired by the capacitor is
\begin{aligned} & Q_P=C V_{\max } \\ & Q_P=C \times \frac{I_{\max }}{\omega C}=\frac{I_{\max }}{\omega} \end{aligned}
 \begin{aligned} & \text { At resonance } \omega=\frac{1}{\sqrt{\mathrm{LC}}} \\ & \Rightarrow Q_P=I_{\max } \sqrt{\mathrm{LC}} \\ & Q_P=1 \times \sqrt{10^{-5} \times 10^{-11}} \\ & \therefore Q_P=10^{-8} \mathrm{C} \end{aligned}

Posted by

Divya Prakash Singh

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