\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

In a certain AC circuit, the applied voltage is $V=V_0 \sin \omega
t$ volts. The corresponding current is $I=V_0 \omega C \cos \omega
t$ amps. Given that the power dissipated $P$ is given by $P=IV$.
Find the mean value of $P$ over a period. Find the corresponding
rms value.
\medskip

{\bf Answer}

$P+IV+V_0\sin \omega t \times V_0 \omega c \cos \omega t$

(The period will be from $t=0$ to $t=\ds\frac{2\pi}{\omega}$)

So

$\begin{array} {rcl} P & = &  V_0^2 \omega c \sin \omega t \cos
\omega t\\ & = & \ds\frac{V_0^2 \omega c}{2}\sin 2 \omega t
\end{array}$

Mean value of $P$ over a period is

$\begin{array} {rcl} & = &
\ds\frac{1}{(\frac{2\pi}{\omega}-0)}\ds\int_0^{\frac{2\pi}{\omega}}
\ds\frac{V_0^2 \omega c}{2} \sin 2 \omega t \,dt\\ & = &
\ds\frac{\omega}{2\pi}\ds\frac{V_0^2 \omega
c}{2}\left[-\ds\frac{cos 2 \omega
t}{2\omega}\right]_0^{\frac{2\pi}{\omega}}\\ & = & \un{0}
\end{array}$

rms value over a period

$\begin{array}{rcl} & = &
\sqrt{ds\frac{\omega}{2\pi}\ds\int_0^{\frac{2\pi}{\omega}}P^2
\,dt}\\ & = & \left[\ds\frac{\omega}{2\pi} \ds\frac{V_0^4 \omega^2
c^2}{4}\ds\int_0^{\frac{2\pi}{\omega}} \sin^2 2\omega t
\,dt\right]^{\frac{1}{2}}\\ & = & \ds\frac{V_0^2
\omega^{\frac{3}{2}}c}{2\sqrt{2\pi}}\left\{\ds\int_0^{\frac{2\pi}{\omega}}\sin^2
2\omega t \,dt\right\}^{\frac{1}{2}}\\ & & 1-2\sin^2 2\omega t =
\cos 4\omega t\\ & = & \ds\frac{V_0^2
\omega^{\frac{3}{2}}c}{2\sqrt{2\pi}}\left\{\ds\frac{1}{2}\ds\int_0^{\frac{2\pi}{\omega}}(1-\cos
4\omega t) \,dt\right\}^{\frac{1}{2}}\\ & = & \ds\frac{V_0^2
\omega^{\frac{3}{2}}c}{4\sqrt{\pi}}\left\{\left[t-\ds\frac{\sin
4\omega
t}{4\omega}\right]_0^{\frac{2\pi}{\omega}}\right\}^{\frac{1}{2}}\\
& = & \un{\ds\frac{V_0^2\omega c}{2\sqrt{2}}}\end{array}$

\end{document}