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# A series LCR circuit with L=0.12 H , C =480 nF, R = 23 omega is connected to a 230 V .

Q7.20 (d)  A series LCR circuit with $L=0.12H$, $C=480nF$$R=23\Omega$  is connected to a $230V$ variable frequency supply. For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?

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As

Power $P=I^2R$

Power $P$ will be half when the current $I$ is  $1/\sqrt{2}$ times the maximum current.

As,

$I =I_{max}Sin(wt-\phi)$

At half power point :

$\frac{i_{max}}{\sqrt{2}} =I_{max}Sin(wt-\phi)$

$\frac{1}{\sqrt{2}} =Sin(wt-\phi)$

$wt=\phi+\frac{\pi}{4}$

here,

$\phi=tan^{-1}(\frac{wL-\frac{1}{wC}}{R})$

On putting values, we get, two values of $w$ for which

$wt=\phi+\frac{\pi}{4}$

And they are:

$w_1=678.75Hz$

$w_2=648.22Hz$

Also,

The current amplitude at these frequencies

$I_{halfpowerpoint}=\frac{I_{max}}{\sqrt{2}}=\frac{14.14}{1.414}=10A$

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