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{\bf Question}

In another circuit, the applied voltage is $V=100\sin \omega
t+16\sin3\omega t$ volts.

The corresponding current is $I=5.8\sin\left(\omega
t-\ds\frac{\pi}{3}\right)+1.6\sin\left(3\omega
t-\ds\frac{\pi}{6}\right)$ amps.  Find the mean and rms values of
the power $P$ dissipated over a period.
\medskip

{\bf Answer}
%Question 11 - written over this answer is "long question leave it out"%

$I=5 \cdot 8 \sin\left(\omega t-\ds\frac{\pi}{3}\right)+1 \cdot 6
\sin\left(3 \omega t -\ds\frac{\pi}{6}\right)$

$V=100\sin \omega t+16 \sin 3 \omega t$

$P=IV=\begin{array} {rcl} & & 580 \sin \omega t \sin\left(\omega
t-\ds\frac{\pi}{3}\right)\\ & + & 160 \sin \omega t
\sin\left(3\omega t-\ds\frac{\pi}{6}\right)\\ & + & 92 \cdot 8
\sin 3\omega t \sin\left(\omega t-\ds\frac{\pi}{3}\right)\\ & + &
25 \cdot 6 \sin 3\omega t \sin\left(3\omega
t-\ds\frac{\pi}{6}\right) \end{array}$

What is the period? It's the longest times for any of the series
to repeat. The most slowly varying sine is $\sin \omega t
\Rightarrow$ period is $t=\ds\frac{2\pi}{\omega}$.

$$\bar{P}=\ds\frac{\omega}{2\pi}\ds\int_0^{\frac{2\pi}{\omega}}P
\,dt$$

This is a complicated integral. Expand the sines using the
addition formula

$$\sin(A-B)=\sin A\cos B-\cos A\sin B$$

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