Explain electrical energy when an ac current passes through a resistor. 

It is seen from the graph that in $A.\;C$. circuit voltage and current varies with sine curve and obtained positive and negative values corresponds to it.

Thus, the sum of the instantaneous current values over one complete cycle is zero and the

average current is zero.

The fact that the average current is zero, however does not mean that the average power

consumed is zero and that there is no dissipation of electrical energy.

Joule heating is given by $\mathrm{I}^{2} \mathrm{R} t$ and depends on $\mathrm{I}^{2}$ which is always positive whether I is positive

or negative and not on $\mathrm{I}$.

The instantaneous power dissipated in the resistor is,

$\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$

$=\mathrm{I}_{\mathrm{m}}^{2}$ Rsin $^{2} \omega t$

The average value of $[\because \mathrm{p}$ over a cycle is,

$\bar{p}=<\mathrm{I}^{2} \mathrm{R}>$ $=\mathrm{I}_{\mathrm{m}}^{2}<\mathrm{R} \sin ^{2} \omega t>=\mathrm{I}_{\mathrm{m}}^{2} \mathrm{R}<\sin ^{2} \omega t>$

here the bar over a $\bar{p}$ denotes its average value and $<>$ denotes average of the quantity inside

the bracket.

$\therefore \bar{p}=\mathrm{I}_{\mathrm{m}}^{2} \mathrm{R}<\sin ^{2} \omega t>\left[\because \mathrm{I}_{\mathrm{m}}^{2}\right.$ and $\mathrm{R}$ is constant $]$

Using the trigonometric,

=$\frac{1}{2}(1-\cos 2 \omega t)$

$=\left(\frac{1}{2}-\frac{1}{2} \cos \omega t\right)$

but $\langle\cos 2 \omega t\rangle=0$, so, $\left\langle\sin ^{2} \omega t\right\rangle=\frac{1}{2}$ $\therefore \bar{p}=\frac{1}{2} \mathrm{I}_{\mathrm{m}}^{2} \mathrm{R}$