Question

A voltage $V=V_{0}sin\omega t$ is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is
(i) no power dissipated even though the current flows through the circuit,
(ii) maximum power dissipated in the circuit?

Answer 1

Voltage $V=V_{0}sin\omega t$ is applied to a series LCR circuit.

Current is $I=I_{0}sin(\omega t+\Phi )$

$I_0=\frac{V_0}{Z}$

$\Phi =ta{n}^{-1}\left(\frac{X_C-X_L}{R}\right)$

Instantaneous power supplied by the source is

$P=VI=\left(V_{0}sin\omega t\right)\times \left(I_{0}sin\right(\omega t+\Phi \left)\right)$

= $\frac{V_0{I}_0}{2}[cos\Phi -cos(2\omega t+\Phi \left)\right]$

The average power over a cycle is average of the two terms on the R.H.S. of the above equation.

The second term is time dependent, so, its average is zero.

So, $P=\frac{V_0{I}_0}{2}cos\Phi$

= $\frac{V_0{I}_0}{\sqrt{2}\sqrt{2}}cos\Phi$

= $VIcos\Phi$

$P=I^{2}Zcos\Phi$

$cos\Phi$ is called the power factor.

Case 1.

For pure inductive circuit or pure capacitive circuit, the phase difference between current and voltage i.e., $\Phi$ is $\frac{\pi }{2}$.

$\therefore \Phi =\frac{\pi }{2}$

So $Cos\Phi =0$

Therefore, no power is dissipated.

Case 2.

For power dissipated at resonance in an LCR circuit,

$X_C-X_L=0,\Phi =0$

$\therefore Cos\Phi =1$

So, maximum power is dissipated.