Explain the dissipation of electrical energy when an alternating current passes through a resistor.

(N/A) In an $A.C.$ circuit, the voltage and current vary sinusoidally. The sum of the instantaneous current values over one complete cycle is zero, meaning the average current is zero.
However, the fact that the average current is zero does not mean that the average power consumed is zero or that there is no dissipation of electrical energy. Joule heating is given by $H = I^{2}Rt$, which depends on $I^{2}$. Since $I^{2}$ is always positive regardless of whether $I$ is positive or negative, energy is dissipated.
The instantaneous power dissipated in the resistor is:
$P = I^{2}R = (I_{m} \sin \omega t)^{2} R = I_{m}^{2} R \sin^{2} \omega t$
The average power $\bar{P}$ over a cycle is:
$\bar{P} = \langle P \rangle = \langle I^{2} R \rangle = I_{m}^{2} R \langle \sin^{2} \omega t \rangle$
Using the trigonometric identity $\sin^{2} \omega t = \frac{1 - \cos 2\omega t}{2}$, we find the average value over a cycle:
$\langle \sin^{2} \omega t \rangle = \langle \frac{1}{2} - \frac{1}{2} \cos 2\omega t \rangle = \frac{1}{2} - 0 = \frac{1}{2}$
Therefore, the average power dissipated is:
$\bar{P} = \frac{1}{2} I_{m}^{2} R$