Answer
\[
\text { (b) } 169.705 \mathrm{V}=V_{p}
\]
(a) $\frac{V_{p}}{\sqrt{2}}=V_{r m s}$
Work Step by Step
(a) We start with calculating $V^{2}$, then using the expression for the average value of a function.
\[
\begin{aligned}
&V_{p} \sin (2 \pi f t)= V \\
&V_{p}^{2} \sin ^{2}(2 \pi f t)= V^{2}\\
&\frac{\int_{0}^{1 / 5} V^{2} d t}{1 / f}=V_{\text {average}}^{2} \\
&\frac{\int_{0}^{2 \pi} V_{p}^{2} \sin ^{2}(2 \pi f t) d t}{1 / f}=V_{\text {average}}^{2} \\
&\frac{V_{p}^{2}}{1 / f} \int_{0}^{1 / f} \frac{-\cos 4 \pi f t+1}{2} d t= V_{\text {average}}^{2} \quad\left\{\text { Using } =-2 \sin ^{2}\theta+1=\cos 2 \theta\right\} \\
V_{\text {average}}^{2} &=\frac{V_{p}^{2}}{1 / f}\left[-\frac{\sin 4 \pi f t}{2}+\frac{t}{2}\right]_{0}^{1 / f} \\
&\frac{V_{p}^{2}}{1 / f}\left[\frac{1 / f}{2}-\frac{\sin 4 \pi-\sin 0}{2}\right] =V_{\text {average}}^{2} \\
&\frac{V_{p}^{2} f}{2 f}=\frac{V_{p}^{2}}{2}=V_{\text {average}}^{2} \\
&\frac{V_{p}}{\sqrt{2}}=R M S(V)=\sqrt{V_{\text {average}}^{2}}
\end{aligned}
\]
(b) Using the above derivative expression:
\[
169.705 \mathrm{V}=120 \sqrt{2} \mathrm{V}=V_{r m s} \sqrt{2}=V_{p}
\]
