Question

Consider an LC circuit in which L = 490 mH and C = 0.116 �F.

Answer 1

Answer:

(a) 4190 rad/sec

(b) 4064 rad/sec

(c) Percentage change is 3 %  

Explanation:

We have given inductance $L=490mH=490\times 10^{-3}H$

Capacitance $C=0.116\mu F=0.116\times 10^{-6}F$

We know that resonance frequency is given by $\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{490\times 10^{-3}\times 0.116\times 10^{-6}}}=4190rad/sec$

Now resistance is given as R = 1020 ohm '

(b) We know that damped frequency is given by

$\omega =\sqrt{\frac{1}{LC}-(\frac{R}{2L})^2}=\sqrt{\frac{1}{490\times 10^{-3}\times 0.116\times 10^{-3}}-(\frac{1020}{2\times 490\times 10^{-3}})^2}=4064rad/sec$

(c) Percentage change in frequency $=\frac{4190-4064}{4190}\times 100=3$%