Chapter 9, Problem 1P

a.

To determine

Find the maximum amplitude of the voltage.

Answer to Problem 1P

The maximum amplitude of the voltage is $\underset{\_}{40\text{V}}$.

Explanation of Solution

Given data:

The given sinusoidal voltage is,

$v\left(t\right)=40\cos \left(100\pi t+60°\right)\text{V}$ (1)

Formula used:

Consider the general expression for sinusoidal voltage.

$v\left(t\right)=V_m\cos \left(\omega t+\varphi \right)\text{V}$ (2)

Here,

$V_m$ is the maximum amplitude.

$\omega$ is the frequency.

$\varphi$ is the phase angle.

Calculation:

Compare equation (1) with equation (2), that is

$V_m=40\text{V}$

Conclusion:

Thus, the maximum amplitude of the voltage is $\underset{\_}{40\text{V}}$.

b.

To determine

Find the value of frequency in hertz.

Answer to Problem 1P

The value of frequency in hertz is $\underset{\_}{50\text{Hz}}$.

Explanation of Solution

Calculation:

Compare equation (1) with equation (2), that is

$\begin{array}{l}\omega =100\pi \text{rad}/\text{s}\\ 2\pi f=100\pi \left\{\because \omega =2\pi f\right\}\\ f=\frac{100\pi }{2\pi }\\ f=50\text{Hz}\end{array}$

Conclusion:

Thus, the value of frequency in hertz is $\underset{\_}{50\text{Hz}}$.

c.

To determine

Find the value of frequency in radians per second.

Answer to Problem 1P

The value of frequency in radians per second is $\underset{\_}{314.159\text{rad}/\text{s}}$.

Explanation of Solution

Calculation:

Compare equation (1) with equation (2), that is

$\begin{array}{c}\omega =100\pi \text{rad}/\text{s}\\ =314.159\text{rad}/\text{s}\end{array}$

Conclusion:

Thus, the value of frequency in radians per second is $\underset{\_}{314.159\text{rad}/\text{s}}$.

d.

To determine

Find the value of phase angle in radians.

Answer to Problem 1P

The value of phase angle in radians is $\underset{\_}{1.05\text{rad}}$.

Explanation of Solution

Calculation:

Compare equation (1) with equation (2), that is

$\begin{array}{c}\varphi =60°\\ =\frac{2\pi }{360°}\left(60°\right)\text{rad}\\ =1.047\text{rad}\\ \cong 1.05\text{rad}\end{array}$

Conclusion:

Thus, the value of phase angle in radians is $\underset{\_}{1.05\text{rad}}$.

e.

To determine

Find the value of phase angle in degrees.

Answer to Problem 1P

The value of phase angle in degrees is $\underset{\_}{60°}$.

Explanation of Solution

Calculation:

Compare equation (1) with equation (2), that is

$\varphi =60°$

Conclusion:

Thus, the value of phase angle in degrees is $\underset{\_}{60°}$.

f.

To determine

Find the value of period in milliseconds.

Answer to Problem 1P

The value of period in milliseconds is $\underset{\_}{20\text{ms}}$.

Explanation of Solution

Calculation:

Consider the general expression for time period.

$T=\frac{1}{f}$

Substitute $50\text{Hz}$ for $f$ as follows.

$\begin{array}{c}T=\frac{1}{50\text{Hz}}\\ =0.02\text{s}\\ =\text{20}\text{ms}\end{array}$

Conclusion:

Thus, the value of period in milliseconds is $\underset{\_}{20\text{ms}}$.

g.

To determine

Find the value of first time that obtained after $t=0$ that $v=-40\text{V}$.

Answer to Problem 1P

The value of first time that obtained after $t=0$ that $v=-40\text{V}$ is $\underset{\_}{6.67\text{ms}}$.

Explanation of Solution

Calculation:

Substitute $-40\text{V}$ for $v$ in equation (1) as follows.

$\begin{array}{l}-40=40\cos \left(100\pi t+60°\right)\\ \frac{-40}{40}=\cos \left(100\pi t+60°\right)\\ {\cos }^{-1}\left(-1\right)=\left(100\pi t+60°\right)\\ \pi =\left(100\pi t+60°\right)\end{array}$

Simplify the equation as follows.

$\begin{array}{l}100\pi t=\pi -60°\\ t=\frac{180°-60°}{100\left(180°\right)}\left\{\because \pi =180°\right\}\\ t=\frac{120°}{100\left(180°\right)}\\ t=6.67\text{ms}\end{array}$

Conclusion:

Thus, the value of first time that obtained after $t=0$ that $v=-40\text{V}$ is $\underset{\_}{6.67\text{ms}}$.

h.

To determine

Find the expression of $v\left(t\right)$, when the sinusoidal function is shifted $10/3\text{ms}$ right.

Answer to Problem 1P

The expression of $v\left(t\right)$, when the sinusoidal function is shifted $10/3\text{ms}$ right is $\underset{\_}{40\cos 100\pi t\text{V}}$.

Explanation of Solution

Calculation:

As the sinusoidal function is shifted $10/3\text{ms}$ right, modify equation (1) as follows.

$\begin{array}{c}v\left(t\right)=40\cos \left(100\pi \left(t-\frac{0.01}{3}\right)+60°\right)\\ =40\cos \left(100\pi \left(t-\frac{0.01}{3}\right)+\frac{\pi }{3}\right)\left\{\because \frac{\pi }{3}=60°\right\}\\ =40\cos \left(100\pi t-\frac{\pi }{3}+\frac{\pi }{3}\right)\\ =40\cos 100\pi t\text{V}\end{array}$

Conclusion:

Thus, the expression of $v\left(t\right)$, when the sinusoidal function is shifted $10/3\text{ms}$ right is $\underset{\_}{40\cos 100\pi t\text{V}}$.

i.

To determine

Find the minimum number of milliseconds taken by a function to shift towards left, when $40\sin 100\pi t\text{V}$is used as $v\left(t\right)$.

Answer to Problem 1P

The minimum number of milliseconds taken by a function to shift towards left, when $40\sin 100\pi t\text{V}$is used as $v\left(t\right)$ is $\underset{\_}{11.67\text{ms}}$.

Explanation of Solution

Calculation:

Substitute $40\sin 100\pi t\text{V}$ for $v\left(t\right)$ in equation (1) and modify the equation as follows.

$\begin{array}{l}40\sin 100\pi t=40\cos \left(100\pi \left(t+t_o\right)+60°\right)\\ \cos \left(100\pi t-\frac{\pi }{2}\right)=\cos \left(100\pi \left(t+t_o\right)+\frac{\pi }{3}\right)\left\{\because \frac{\pi }{3}=60°\right\}\\ 100\pi t+\frac{3\pi }{2}=100\pi \left(t+t_o\right)+\frac{\pi }{3}\\ 100\pi t+\frac{3\pi }{2}=100\pi t+100\pi t_o+\frac{\pi }{3}\end{array}$

Simplify the equation as follows.

$\begin{array}{l}\frac{3\pi }{2}=100\pi t_o+\frac{\pi }{3}\\ 100\pi t_o=\frac{3\pi }{2}-\frac{\pi }{3}\\ 100\pi t_o=\frac{9\pi -2\pi }{6}\\ 100\pi t_o=\frac{7\pi }{6}\end{array}$

Rearrange the equation as follows.

$\begin{array}{c}{t}_o=\frac{7\pi }{6\left(100\pi \right)}\\ =11.66\text{ms}\end{array}$

Conclusion:

Thus, the minimum number of milliseconds taken by a function to shift towards left, when $40\sin 100\pi t\text{V}$is used as $v\left(t\right)$ is $\underset{\_}{11.67\text{ms}}$.