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  • A series LCR circuit containing a resistance of 120  has an angular resonance frequency of 4 * 105&

A series LCR circuit containing a resistance of 120 \Omega has an angular resonance frequency of 4 * 105 rad/s. At resonance, the voltage across the resistance and inductance are 60V and 40 V respectively. At what frequency (105) the current in the circuit lags the voltage by 45°?

Option: 1

1 * 105


Option: 2

4 * 105


Option: 3

8 * 105


Option: 4

16 * 105


Answers (1)

best_answer

i = \frac{V}{R} = \frac{60}{120} = 0.5 A

and

V_{L} = i * X_{L} = i\omega L        

\implies L =\frac{V_{L}}{i \omega} = \frac{40}{\frac{1}{2} * 4 * 10^{5}}

                = \frac{80}{4 * 10^{5}} = 8 *10^{-4} H

                = \frac{0.8 mH}{4} = 0.2

At resonance,

\omega _{0}= \frac{1}{\sqrt{2C}}

\implies C = \frac{1}{\omega _{0}^{2}L} = \frac{4}{0.8 * 10^{-3} * (1*10^{5})^{2}} = \frac{1}{32} \mu F

For the LCR circuit,

tan \phi = \frac{X_{L} - X_{C}} {R}

\implies 1 * 120 = \omega * 2 * 10^{-4} - \frac{1}{\omega \frac{1}{2} * 10^{-6}}

\implies \omega = \frac{6 * 10^{5} \pm \sqrt{(6 * 10^{5})^{2} + 64 * 10^{10}}}{2}

                = 8 * 10^{5} rad/s

Posted by

Gaurav

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