Chapter 6.4, Problem 91AYU

a.

To determine

Show that $P\left(t\right)=\frac{V_0^2}{R}{\sin }^2\left(2\pi ft\right)$ .

Answer to Problem 91AYU

  $\boxed{P\left(t\right)=\frac{V_0{}^2}{R}{\sin }^2\left(2\pi ft\right)}$ Proved.

Explanation of Solution

Given information:

The voltage $V$ in volts produced by an ac generator is sinusoidal. As a function of time $t$ , the voltage $V$ is,

  $V\left(t\right)=V_0\sin \left(2\pi ft\right)$ where $f$ is the frequency, the number of complete oscillations(cycles)per second $\left(f=60Hz\right)$ . The power $P$ delivered to a resistance $R$ at any time $t$ is defined as,

  $P\left(t\right)=\frac{{\left[V\left(t\right)\right]}^2}{R}$

Show that $P\left(t\right)=\frac{V_0^2}{R}{\sin }^2\left(2\pi ft\right)$ .

Calculation:

Consider the given function.

  $V\left(t\right)=V_0\sin \left(2\pi ft\right)$ and $P\left(t\right)=\frac{{\left[V\left(t\right)\right]}^2}{R}$

  $P\left(t\right)=\frac{{\left[V\left(t\right)\right]}^2}{R}$

  $\begin{array}{l}P\left(t\right)=\frac{{\left[V_0\sin \left(2\pi ft\right)\right]}^2}{R}\\ P\left(t\right)=\frac{V_0{}^2}{R}{\sin }^2\left(2\pi ft\right)\end{array}$

Hence, proved $\boxed{P\left(t\right)=\frac{V_0{}^2}{R}{\sin }^2\left(2\pi ft\right)}$ .

b.

To determine

Express $P$ as a sinusoidal function.

Answer to Problem 91AYU

  $\boxed{P\left(t\right)=\frac{V_0{}^2}{2R}\sin \left(2\pi ft\right)+\frac{V_0^2}{2R}}$

Explanation of Solution

Given information:

The voltage $V$ in volts produced by an ac generator is sinusoidal. As a function of time $t$ , the voltage $V$ is,

  $V\left(t\right)=V_0\sin \left(2\pi ft\right)$ where $f$ is the frequency, the number of complete oscillations(cycles)per second $\left(f=60Hz\right)$ . The power $P$ delivered to a resistance $R$ at any time $t$ is defined as,

  $P\left(t\right)=\frac{{\left[V\left(t\right)\right]}^2}{R}$

The graph of $P$ is shown in the figure. Express $P$ as a sinusoidal function.

  

Calculation:

Consider the given graph, there is vertical shift up by $\frac{V_0^2}{2R}$ units. If we shift graph down by $\frac{V_0^2}{2R}$ unit, the given graph has characteristics of a sine function as the graph will pass through origin. So, we view the equation as a sine function $y=A\sin \left(\omega x\right)with\left|A\right|=\frac{V_0^2}{2R}$ as the curve lies between $-\frac{V_0^2}{2R}$ and $\frac{V_0^2}{2R}$ on $y-axis$ and time period $T=\frac{1}{t}$ as one cycle in the graph begins at $t=0$ and ends at $t=\frac{1}{f}$ .

The given sine function will not be negative as power in an ac generator is always positive.

Now $\begin{array}{l}\omega =\frac{2\pi }{T}\\ =\frac{2\pi }{1/f}\\ \omega =2\pi f\end{array}$

Hence, the sine function of graph is, $\boxed{P\left(t\right)=\frac{V_0{}^2}{2R}\sin \left(2\pi ft\right)+\frac{V_0^2}{2R}}$ .

c.

To determine

Deduce that ${\sin }^2\left(2\pi ft\right)=\frac{1}{2}\left[1-\cos \left(4\pi ft\right)\right]$ .

Answer to Problem 91AYU

  $\boxed{{\sin }^2\left(2\pi ft\right)=\frac{1}{2}\left[1-\cos \left(4\pi ft\right)\right]}$ proved.

Explanation of Solution

Given information:

The voltage $V$ in volts produced by an ac generator is sinusoidal. As a function of time $t$ , the voltage $V$ is,

  $V\left(t\right)=V_0\sin \left(2\pi ft\right)$ where $f$ is the frequency, the number of complete oscillations(cycles)per second $\left(f=60Hz\right)$ . The power $P$ delivered to a resistance $R$ at any time $t$ is defined as,

  $P\left(t\right)=\frac{{\left[V\left(t\right)\right]}^2}{R}$

Deduce that ${\sin }^2\left(2\pi ft\right)=\frac{1}{2}\left[1-\cos \left(4\pi ft\right)\right]$ .

Calculation:

Consider trigonometric identities ${\cos }^{2}x+{\sin }^{2}x=1$ and double angle formula $\cos 2x=1-2{\cos }^{2}x$

  $\begin{array}{l}{\sin }^2\left(2\pi ft\right)=1-{\cos }^2\left(2\pi ft\right)\\ =\frac{1}{2}\left(2-2{\cos }^2\left(2\pi ft\right)\right)\\ =\frac{1}{2}\left[1+\left(1-2{\cos }^2\left(2\pi ft\right)\right)\right]\\ =\frac{1}{2}\left[1-\cos \left(2\pi ft\right)\right]\end{array}$

Hence, proved $\boxed{{\sin }^2\left(2\pi ft\right)=\frac{1}{2}\left[1-\cos \left(4\pi ft\right)\right]}$ .