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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=480nFR=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)
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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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