Q7.6  Obtain the resonant frequency \omega _{r} of a series LCR circuit with L=2.0HC=32\mu F and  R=10\Omega . What is the Q -value of this circuit?

Answers (1)

Given, in a circuit,

Inductance,  L=2H

Capacitance,    C=32\mu F=32*10^{-6}F

Resistance, R=10\Omega

Now,

Resonance frequency (frequency of maximum current OR minimum impedance OR frequency at which inductive reactance cancels out capacitive reactance )

\omega _r=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{2*32*10^{-6}}}=\frac{1}{8*10^{-3}}=125s^{-1}

Hence Resonance frequency is 125 per second.

Q-Value: 

Q=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{1}{10}\sqrt{\frac{2}{32*10^{-6}}}=25

Hence Q - value of the circuit is 25.

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