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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?

Answers (1)

best_answer

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

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

Posted by

Pankaj Sanodiya

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