Answer:
(a) 4190 rad/sec
(b) 4064 rad/sec
(c) Percentage change is 3 %
Explanation:
We have given inductance [tex]L=490mH=490\times 10^{-3}H[/tex]
Capacitance [tex]C=0.116\mu F=0.116\times 10^{-6}F[/tex]
We know that resonance frequency is given by [tex]\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{490\times 10^{-3}\times 0.116\times 10^{-6}}}=4190rad/sec[/tex]
Now resistance is given as R = 1020 ohm '
(b) We know that damped frequency is given by
[tex]\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[/tex]
(c) Percentage change in frequency [tex]=\frac{4190-4064}{4190}\times 100=3[/tex]%
