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In a series LCR circuit connected to an a.c. source of voltage $v=v_{m}$ $\sin \omega t$ , use phasor diagram to derive an expression for the current in the circuit. Hence obtain the expression for the power dissipated in the circuit. Show that power dissipated at resonance is maximum.

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Phasor diagram

$\left | V \right |=\sqrt{V{_{R}}^{2}+(V_{L}-V_{C})^{2}}$

$=\sqrt{(IR)^{2}+(IX_{L}-IX_{C})^{2}}$

$V=I\sqrt{R^{2}+\left ( X_{L}-X_{C} \right )^{2}}$

$\left | I \right |=\frac{\left | V \right |}{\sqrt{R^{2}+\left ( X_{L}-X_{C} \right )^{2}}}$

$\phi =\tan ^{-1}\left ( \frac{V_{L}-V_{C}}{R} \right )=\tan ^{-1}\left ( \frac{X_{L}-X_{_{C}}}{R} \right )$

$I = i_{m}\sin \left ( \omega t-\phi \right )$

For $V_{C}>V_{L}$

$I=i_{m}\sin \left ( \omega t+\phi \right )$

Power dissipated

$P= VI$

$= V_{m}i_{m}\sin \omega t\; \sin (\omega t-\phi )$

$=\frac{V_{m}i_{m}}{2}\left [ \cos \phi -\cos (2\omega t-\phi ) \right ]$

The average power over a cycle

$P_{av}=\frac{V_{m}i_{m}}{2}\cos \phi$

$=VI\cos \phi$

Resonance $X_{L}=X_{C}$

$\therefore \phi =O$

$\Rightarrow \cos \phi =1$

Power $P_{av}=VI\cos \phi$ has maximum value at resonance.

Posted by

Safeer PP

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