Question

Using $P=V_{rms}{I}_{rms}cos\delta$ discuss the special cases for power consumed in an A.C circuit.

Answer 1

An alternating voltage

$E=V_o\sin \omega t$ is applied and current through the load is $i=i_o\sin (\omega t+\delta )$

Where, $\delta =$ Phase difference.

Instantaneous Power Consumed, $P$

$P=Ei=V_o\sin \left(\omega t\right)\times i_o\sin (\omega t+\varphi )$

or, $\frac{d\omega }{dt}=V_o{i}_o\sin \omega t\times \sin (\omega t+\varphi )$

or, $d\omega =V_o{i}_o\sin \omega t\cdot \sin (\omega t+\varphi )dt$

Now, workdone in one complete cycle is

$W=V_o{i}_o{\int }_0^T\sin \left(\omega t\right)\cdot \sin (\omega t+\varphi )dt$

$W=\frac{V_o{i}_o}{2}\cos \varphi {\int }_0^{T}dt-E_o{i}_o/2{\int }_0^T\cos (2\omega t+\delta )dt$

$W=\frac{V_o{i}_o}{2}\cos \varphi \times T-0$

Now power delivered by an AC

$P_{av}=\frac{W}{T}=\frac{V_o{I}_o}{2}\cos \delta$

$\Rightarrow P_{av}=\frac{V_0}{\sqrt{2}}\times \frac{i_o}{\sqrt{2}}\cos \delta$

$P_{av}=V_{rms}\cdot i_{rms}\cos \delta$

If circuit is purely resistivity, $\delta =0$

$P_{av}=V_{rms}\cdot i_{rms}$ maximum

If circuit only capacitor or inductor purely.

$P_{av}=0$