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(a) Obtain the expression for the average power dissipated in series LCR circuit driven by an AC source of voltage $V = V_{m}\sin (\omega t+\phi )$

(b) Define the terms: (i) Wattless current and (ii) Q-factor of LCR circuit

Answers (1)

(a) Given,

Voltage, $V = V_{m}\sin \omega t$

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

where $\phi$ is the phase difference between current and applied voltage,

where,

$i_{m}= \frac{V_{m}}{Z}$

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

Hence, the instantaneous power is

$p= Vi$

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

$=\frac{V_{m}i_{m}}{2}\left ( 2\sin \omega t.\sin \left ( \omega t+\phi \right ) \right )$

$p=\frac{V_{m}i_{m}}{2}\left [ \cos \phi -\cos \left ( 2\omega t +\phi \right ) \right ]$ ................(1)

The average of the second term of the equation is zero

Therefore,

Average power dissipated over a complete cycle

$p=\frac{V_{m}i_{m}}{2} \cos \phi =\frac{V_{m}}{\sqrt{2}}.\frac{i_{m}}{\sqrt{2}} \cos \phi$

$p=VI \cos \phi$

This can be also be written as

$P = I^{2}Z\cos \theta$

(b) (i) Wattless current

In a purely inductive or capacitive circuit, the phase difference between voltage and current is $\frac{\pi}{2}$ . Then the power factor $\cos \phi = 0$ which means power dissipated in the circuit is zero even though the current is flowing in the circuit. Such type of current is known as Wattles's current.

(ii) Q - factor of LCR circuit

Q - factor of LCR circuit is the sharpness of resonance in a circuit it is given by,

$Q = \frac{w_{o}L}{R}= \frac{1}{w_{0}RC}= \frac{1}{R}\sqrt{\frac{L}{C}}$

where $w_{0}$ is the resonant frequency.

Posted by

Safeer PP

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