Get Answers to all your Questions

header-bg qa

Q7.12 (b)    An LC circuit contains a 20mH inductor and a 50\mu F capacitor with an initial charge of 10mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0.

 What is the natural frequency of the circuit?

             

Answers (1)

best_answer

Given,

The inductance of the inductor:

 L=20mH=20*10^{-3}H

The capacitance of the capacitor :

 C=50\mu F=50*10^{-6}F

The initial charge on the capacitor:

Q=10mC=10*10^{-3}C

The natural angular  frequency of the circuit:

w_{natural}=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(20*10^{-3}*50*10^{-6})}}=10^3rad/sec

Hence the natural angular frequency of the circuit is 10^3rad/sec.

The natural frequency of the circuit:

f_{natural}=\frac{w_{natural}}{2\pi}=\frac{10^3}{2\pi}=159Hz

Hence natural frequency of the circuit is 159Hz.

 

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads