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

Power factor of the circuit is given by:

$cos\varphi =\frac{R}{Z}=\frac{R}{\sqrt{R^2+(\omega L-\frac{1}{\omega C}{)}^2}}$

Current flowing in the circuit is given by:

$I=\frac{V}{Z}$

$I=\frac{V_o\sin \left(\omega t\right)}{Z}$

Instantaneous real power dissipated in the circuit is:

$P=I^{2}R$

$P=\frac{V_o^2{\sin }^2\left(\omega t\right)}{Z^2}R$

Average power dissipated in a cycle is given by:

$<P>=\frac{{\int }_0^{2\pi /\omega }Pdt}{{\int }_0^{2\pi /\omega }dt}=\frac{V_o^{2}R}{2Z^2\times 2\pi /\omega }{\int }_0^{2\pi /\omega }(1-cos(2\omega t\left)\right)dt$

$<P>=V_{rms}{I}_{rms}cos\left(\varphi \right)$

(i)

No power is dissipated when P = 0

This implies $cos\varphi =0$

$\varphi =\pi /2$

That is the circuit is purely inductive or capacitive.

(ii)

Maximum power is dissipated when P is maximum.

This implies $cos\varphi =1$

$\varphi =0$

Circuit is purely resistive.