Chapter 9, Problem 1P

a.

To determine

Find the value of frequency f in hertz.

Answer to Problem 1P

The value of frequency in hertz is $\underset{\_}{127.32\text{Hz}}$.

Explanation of Solution

Given data:

The given sinusoidal current is,

$i\left(t\right)=125\cos \left(800t+36.87°\right)\text{mA}$ (1)

Formula used:

Consider the general expression for sinusoidal current.

$i\left(t\right)=I_m\cos \left(\omega t+\varphi \right)\text{A}$ (2)

Here,

$I_m$ is the maximum amplitude.

$\omega$ is the frequency.

$\varphi$ is the phase angle.

Calculation:

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

$\begin{array}{l}\omega =800\text{rad}/\text{s}\\ 2\pi f=800\left\{\because \omega =2\pi f\right\}\\ f=\frac{800}{2\pi }\\ f=127.32\text{Hz}\end{array}$

Conclusion:

Thus, the value of frequency in hertz is $\underset{\_}{127.32\text{Hz}}$.

b.

To determine

Find the value of period in milliseconds.

Answer to Problem 1P

The value of period in milliseconds is $\underset{\_}{7.85\text{ms}}$.

Explanation of Solution

Calculation:

Consider the general expression for time period.

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

Substitute $127.32\text{Hz}$ for $f$ as follows.

$\begin{array}{c}T=\frac{1}{127.32\text{Hz}}\\ =7.8542\times {10}^{-3}\text{s}\\ \cong \text{7}\text{.85}\text{ms}\end{array}$

Conclusion:

Thus, the value of period in milliseconds is $\underset{\_}{7.85\text{ms}}$.

c.

To determine

Find the maximum amplitude of the current.

Answer to Problem 1P

The maximum amplitude of the current is $\underset{\_}{125\text{mA}}$.

Explanation of Solution

Calculation:

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

$I_m=125\text{mA}$

Conclusion:

Thus, the maximum amplitude of the current is $\underset{\_}{125\text{mA}}$.

d.

To determine

Find the value of voltage $i\left(0\right)$.

Answer to Problem 1P

The value of voltage $i\left(0\right)$ is $\underset{\_}{100\text{mA}}$.

Explanation of Solution

Calculation:

Substitute 0 for $t$ in equation (1) as follows.

$\begin{array}{c}i\left(0\right)=125\cos \left(800\left(0\right)+36.87°\right)\\ =125\cos \left(36.87°\right)\\ =125\left(0.799\right)\\ =\text{99}\text{.875}\text{mA}\end{array}$

$i\left(0\right)\cong \text{100}\text{mA}$

Conclusion:

Thus, the value of voltage $i\left(0\right)$ is $\underset{\_}{100\text{mA}}$.

e.

To determine

Find the value of phase angle in terms of degrees and radians.

Answer to Problem 1P

The value of phase angle in terms of degrees and radians are $\underset{\_}{36.87°\text{and}0.6435\text{rad}}$, respectively.

Explanation of Solution

Calculation:

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

$\varphi =36.87°$

Convert phase angle in terms of radians.

$\begin{array}{c}\varphi =\frac{2\pi }{360°}\left(36.87°\right)\text{rad}\\ =0.6435\text{rad}\end{array}$

Conclusion:

Thus, the value of phase angle in terms of degrees and radians are $\underset{\_}{36.87°\text{and}0.6435\text{rad}}$, respectively.

f.

To determine

Find the smallest possible value of $t$ at which $i=0$.

Answer to Problem 1P

The smallest possible value of $t$ at which $i=0$ is $\underset{\_}{1.16\text{ms}}$.

Explanation of Solution

Calculation:

Substitute 0 for $i\left(t\right)$ in equation (1) as follows.

$\begin{array}{l}125\cos \left(800t+36.87°\right)=0\\ 800t+36.87°={\cos }^{-1}\left(0\right)\\ 800t+36.87°=90°\\ 800t=90°-36.87°\end{array}$

Simplify the equation as follows.

$\begin{array}{l}\left(800\text{rad}/\text{s}\right)t=\frac{53.13°}{57.3°/\text{rad}}\\ \left(800\text{rad}/\text{s}\right)t=0.927\text{rad}\\ t=\frac{0.927\text{rad}}{800\text{rad}/\text{s}}\\ t\cong 1.16\text{ms}\end{array}$

Conclusion:

Thus, the smallest possible value of $t$ at which $i=0$ is $\underset{\_}{1.16\text{ms}}$.

g.

To determine

Find the smallest possible value of $t$ at which $di/dt=0$.

Answer to Problem 1P

The smallest possible value of $t$ at which $di/dt=0$ is $\underset{\_}{3.12\text{ms}}$.

Explanation of Solution

Calculation:

$i\left(t\right)=125\cos \left(800t+36.87°\right)\text{mA}$

Apply differentiation on both sides of equation (1) as follows.

$\frac{di\left(t\right)}{dt}=-\left(0.125\right)\left(800\right)\sin \left(800t+36.87°\right)$

Substitute 0 for $\frac{di\left(t\right)}{dt}$ as follows.

$\begin{array}{l}0=-\left(0.125\right)\left(800\right)\sin \left(800t+36.87°\right)\\ 800t+36.87°={\sin }^{-1}\left(0\right)\\ 800t+36.87°=180°\\ 800t=180°-36.87°\end{array}$

Simplify the equation as follows.

$\begin{array}{l}\left(800\text{rad}/\text{s}\right)t=\frac{143.13°}{57.3°/\text{rad}}\\ \left(800\text{rad}/\text{s}\right)t=2.4979\text{rad}\\ t=\frac{2.4979\text{rad}}{800\text{rad}/\text{s}}\\ t\cong 3.12\text{ms}\end{array}$

Conclusion:

Thus, the smallest possible value of $t$ at which $di/dt=0$ is $\underset{\_}{3.12\text{ms}}$.