Chapter 5, Problem 59RE

a.

To determine

the average value of $V^2$ over a I-sec interval. Then find $V{rms}_{}$ and verify Equation I.

Answer to Problem 59RE

The value of $V_{rms}=\frac{V_{\max }}{\sqrt{2}}$

Explanation of Solution

Given information:

The voltage V in our homes with the sine function $V=V_{\max }\sin \left(120\pi t\right)$ , which expresses V in volts as a function of time t in seconds. The function runs through 60 cycles each second. The number $V_{\max }$ is the peak voltage.

To measure the voltage effectively, we use an instrument that measures the square root of the average value of the square of the voltage over a 1-second interval: $V_{\max }=\sqrt{{\left(V^2\right)}_{\pi t}}$ . The subscript “max” stands for “root mean square”. It turns out that $V_{\max }=\frac{V_{\max }}{\sqrt{2}}$ (1).

The familiar phrase “115 volts ac” means that t herms voltage is 115. The peak voltage, obtained from Equation 1 as $V_{\max }=115\sqrt{2}$ , is about 163 volts.

Concept Used:

  $av\left(f\right)=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$

Calculation:

Since,

  $V=V_{\max }\sin \left(120\pi t\right)$

Square on both the sides.

  $V^2={\left(V_{\max }\right)}^2{\sin }^2\left(120\pi t\right)$

If $f$ is integrated on $\left[a,b\right]$ , its average mean value is calculated by

  $av\left(f\right)=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$

So,

  $\begin{array}{c}av\left(V^2\right)=\frac{1}{1-0}{\displaystyle \underset{0}{\overset{1}{\int }}{\left(V_{\max }\right)}^2{\sin }^2\left(120\pi t\right)}dx\\ ={\left(V_{\max }\right)}^{2}NINT\left({\sin }^2\left(120\pi t\right),t,0,1\right)\\ =\frac{{\left(V_{\max }\right)}^2}{2}\\ =\frac{V_{\max }}{\sqrt{2}}\end{array}$