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  • The frequencies at which the current amplitude in an LCR series circuit becomes <img alt="\frac{1}{\sqrt{2}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D

The frequencies at which the current amplitude in an LCR series circuit becomes \frac{1}{\sqrt{2}} times its maximum value, are 212 \: \: \mathrm{rad} \mathrm{s}^{-1}$ and $232\: \: \mathrm{rad} \mathrm{s}^{-1}. The value of resistance in the circuit is \mathrm{R=5 \Omega}. The self inductance in the circuit is _________\mathrm{mH}.

Option: 1

250


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{I_0=\frac{1}{\sqrt{2}}\: I_{\max } ,at}

\mathrm{\omega=212 \: \frac{\mathrm{rad}}{\mathrm{s}} \: or\: 232 \: \mathrm{rad}}

\mathrm{Let \: \: \omega_1=212 \: \: \frac{\mathrm{rad}}{\mathrm{s}}}

        \mathrm{\omega_2=232\: \frac{\mathrm{rad}}{\mathrm{s}}}
         \mathrm{R=5 \Omega}
We know that,
Resonant frequency \mathrm{=\omega_0=\sqrt{\omega_1 \omega_2}}

\mathrm{ \frac{1}{\sqrt{L C}}=\omega_0=\sqrt{(212)(232)} \rightarrow \text { (1) } }

\mathrm{At \: resonance, \quad z=R}

\mathrm{\omega _{2}-\omega _{1}=\frac{R}{L}}

\mathrm{20=\frac{5}{L}}

\mathrm{L=\frac{1}{4}=0.25H}

\mathrm{L=250mH}

The self inductance in the circuit is \mathrm{250mH}

 

Posted by

HARSH KANKARIA

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