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  • A series LCR circuit containing a resistance of has angular resonance frequency<img alt="\mathrm{4\

A series LCR circuit containing a resistance of120 \Omega has angular resonance frequency\mathrm{4\times 10^{5}\; rads^{-1}} 4 105 rad s−1. At resonance the voltages across resistance and inductance are 60 V and 40 V respectively. Find the values of L and C. At what frequency(in rad/s) the current in the circuit lags the voltage by \mathrm{45^{\circ}} ?

Option: 1

\mathrm{2 \times 10^5}


Option: 2

4 \times 10^5


Option: 3

6 \times 10^5


Option: 4

8 \times 10^5


Answers (1)

best_answer

At resonance as \mathrm{X=0, I=\frac{V}{R}=\frac{60}{120}=\frac{1}{2} A \text { and } V_L=I X_L=I \omega L \text {, }}

\mathrm{ L=\frac{V_L}{I \omega}=\frac{40}{(1 / 2) \times 4 \times 10^5}=2 \times 10^{-4} \text { Henry }}

\mathrm{at \: resonance, \omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}so \mathrm{C}=\frac{1}{\omega^2 \mathrm{~L}} i.e.,}

\mathrm{\mathrm{C}=\frac{1}{0.2 \times 10^{-3} \times\left(4 \times 10^5\right)^2}=\frac{1}{32} \mu \mathrm{F}}

Now in case of series LCR circuit \mathrm{\tan \phi=\frac{X_L-X_c}{R}}

For the current to lag the applied voltage by \mathrm{45\degree}

\mathrm{\tan 45=\frac{\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}}{\mathrm{R}}.}

\mathrm{\text { i.e., } 1 \times 120=\omega \times 2 \times 10^{-4}-\frac{1}{\omega(1 / 32) \times 10^{-6}}}

\mathrm{\text { i.e., } \omega^2-6 \times 10^5 \omega-16 \times 10^{10}=0}

\mathrm{\text { f.e., } \omega=\frac{6 \times 10^5+10 \times 10^5}{2}=8 \times 10^5 \frac{\mathrm{rad}}{\mathrm{s}} \text {. }}

 

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Anam Khan

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