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A series LCR circuit containing 5.0 H inductor $80\mu F$ capacitor and $40\Omega$ resistor is connected to 230 V variable frequency ac source .The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be :
Option: 1
25 rad/s and 75 rad/s

Option: 2
50 rad/s and 25 rad/s

Option: 3
46 rad/s and 54 rad/s

Option: 4
42 rad/s and 58 rad/s

Answers (1)

best_answer

$L=5H$ $P_{0}\rightarrow$ Power at resonance

$C=80\mu F$ $w_{0}\rightarrow$ Resonace angular frequency

$R=40\Omega$ at resonace, $X_{c}=X_{L}$

$P=\frac{1}{2}P_{0}$ $w_{0}C=\frac{1}{w_{0}L}$

$w_{0}=\frac{1}{\sqrt{LC}}=\frac{1}{20\times 10^{-3}}$

$w_{0}=50\: rad/s$

$P=V_{rms}I_{rms}\cos \phi _{1}$

$P_{0}=V_{rms}I_{rms}\cos \phi _{0}$

$\cos \phi _{0}=\frac{1}{2}$

$\cos \phi _{1}=\frac{1}{2}$ $\cos \phi _{o}=\frac{R}{Z}$

$\frac{R}{Z_{1}}=\frac{1}{2}$ $\phi _{o}=0^{o};\cos \phi _{o}=1$

$80=Z_{1}=\sqrt{\left ( X_{L}-X_{c} \right )^{2}+R^{2}}$

$6400=\left ( X_{L}-X_{c} \right )^{2}+1600$

$4800=\left ( X_{L}-X_{c} \right )^{2}$

$4800=\left ( wL-\frac{1}{wc} \right )^{2}$

$40\sqrt{3}=\left ( wL-\frac{1}{wc} \right )$

$w=25rad/s$

$w_{o}=50\: rad/s\: \: \: \: \: w=25\: \: rad/s$

Posted by

HARSH KANKARIA

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