Chapter 6, Problem 57RE

(a)

To determine

To calculate:The average value of $V^2$ over a 1-sec interval.

Answer to Problem 57RE

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

Explanation of Solution

Given information:

Household Electricity we model the voltage V in our homes with the sine function

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

Formula 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 both sides

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

If f is integrated on [a, b] , its average (mean) value on [a, b] is

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

So,

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

Calculator has a numerical integration capability, which will denote as NINT. In particular, use $NINT\left(f\left(x\right),x,a,b\right)$ to denote a calculator approximation of ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ . When write

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=NINT\left(f\left(x\right),x,a,b\right)$

The understanding that the right-hand side of the equation is an approximation of the left-hand side

Now, use above expression with this function

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}\left({\sin }^2\left(120\pi t\right)\right)}dt=NINT\left({\sin }^2\left(120\pi t\right),t,0,1\right)\\ =\frac{1}{2}\end{array}$

So,

  $av\left(V^2\right)=\frac{{\left(V_{\max }\right)}^2}{2}$

  $\begin{array}{l}{V}_{\text{rms}}\text{is root mean square velocity and it is given by,}\\ V_{\text{rms}}=\sqrt{\text{average value of }{V}^2}\\ =\sqrt{av\left(V^2\right)}\\ =\sqrt{\frac{V_{\max }{}^2}{2}}\\ V_{\text{rms}}=\frac{V_{\max }}{\sqrt{2}}\end{array}$

Conclusion:

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

(a)

To determine

To calculate:The peak value of the voltage.

Answer to Problem 57RE

The value is $V_{\max }=339.41\text{volt}$

Explanation of Solution

Given information:

Household Electricity we model the voltage V in our homes with the sine function

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

Formula used:

  $V_{\max }=\sqrt{2}\times V_{\text{rms}}$

Calculation:

Since, it has been that

  $V_{\text{rms}}=240\text{volt}$

Hence,

  $\begin{array}{l}{V}_{\max }=\sqrt{2}\times V_{\text{rms}}\\ =\sqrt{2}\times 240\\ V_{\max }=339.41\text{volt}\end{array}$

Conclusion:

The value is $V_{\max }=339.41\text{volt}$