An LC circuit contains a 20 mH inductor and a `50 muF` capacitor with an initial charge of 10 mC. The resistance of circuit is negligible. Let the instant the circuit is closed be t = 0.
What is the natural frequency of the circuit ?

Explanation

Given Inductance `L = 20 mH = 20xx10^(-3)H`
Capacitance of capacitor
`C=50 mu f=50xx10^(-6)F`
Initial charge on the capacitor,
`Q_(i)=10mc =10xx10^(3)C`
To get the natural frequency of resonant frequency,
`f_(r )=(1)/(2pi sqrt(LC))`
`= (1)/(2xx3.14xx sqrt(20xx10^(-3)xx50xx10^(-6)))`
`= (7xx10^(3))/(44)=159.2 Hz`.
The nature frequency of the circuit
`omega = 2pi V = 2pi xx 159.2`
`=999.78 ~~ 1000 =10^(3)` rad/s