Chapter 9, Problem 40P

In the circuit of Fig. 9.47, find i_o when:

1. (a) ω = 1 rad/s
2. (b) ω = 5 rad/s
3. (c) ω = 10 rad/s

Figure 9.47

To determine

Find the value of the current $i_o\left(t\right)$ in the circuit of Figure 9.47 when $\omega =1\frac{\text{rad}}{\text{s}}$.

Answer to Problem 40P

The value of the current $i_o\left(t\right)$ in the circuit of Figure 9.47 when $\omega =1\frac{\text{rad}}{\text{s}}$ is $2.542\sin \left(\left(1\right)t+8.83°\right)\text{A}$.

Explanation of Solution

Given data:

Refer to Figure 9.47 in the textbook.

The value of the angular frequency $\left(\omega \right)$ is $1\frac{\text{rad}}{\text{s}}$.

Formula used:

Write the expression to calculate the impedance of the passive elements resistor, inductor and capacitor.

$Z_R=R$        (1)

$Z_L=j\omega L$        (2)

$Z_C=\frac{1}{j\omega C}$        (3)

Here,

$\omega$ is the angular frequency,

$R$ is the value of the resistor,

$L$ is the value of the inductor, and

$C$ is the value of the capacitor.

Calculation:

The given circuit is redrawn as Figure 1.

Refer to Figure 1, the current equation is,

$i_s\left(t\right)=5\sin \omega t\text{A}$

Convert the given current into phasor form.

$I_s=\left(5\angle 0°\right)\text{A}$

Substitute $10\Omega$ for $R_1$ in equation (1) to find $Z_{R_1}$.

$Z_{R_1}=10\Omega$

Substitute $10\Omega$ for $R_2$ in equation (1) to find $Z_{R_2}$.

$Z_{R_2}=10\Omega$

Substitute $2\text{H}$ for $L$ in equation (2) to find $Z_L$.

$Z_L=j\omega \left(2\text{H}\right)$        (4)

Substitute $50\text{mF}$ for $C$ in equation (3) to find $Z_C$.

$Z_C=\frac{1}{j\omega \left(50\text{mF}\right)}$

$Z_C=-\frac{j}{\omega \left(50\text{mF}\right)}$        (5)

The Figure 1 is redrawn as impedance circuit in the following Figure 2.

Write the expression to obtain the source transformation from current to voltage.

$V_s=Z_{R_1}{I}_s$

Substitute $10\Omega$ for $Z_{R_1}$ and $\left(5\angle 0°\right)\text{A}$ for $I_s$ in above equation to find $V_s$.

$\begin{array}{c}{V}_s=\left(10\Omega \right)\left(\left(5\angle 0°\right)\text{A}\right)\\ =\left(10\angle 0°\right)\Omega \left(\left(5\angle 0°\right)\text{A}\right)\\ =\left(50\angle 0°\right)\text{V}\end{array}$

Based upon the source transformation, the Figure 2 is reduced as the following Figure3.

Refer to Figure 3, the series connected impedances $Z_{R_2}$ and $Z_L$ are connected in parallel with the impedance $Z_C$ and this parallel combination is connected in series with the impedance $Z_{R_1}$.

Therefore, the total equivalent impedance is calculated as follows.

$Z_{\text{eq}}=Z_{R_1}+\left(Z_C\parallel \left(Z_{R_2}+Z_L\right)\right)$        (6)

Substitute $1\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (4) to find $Z_L$.

$\begin{array}{c}{Z}_L=j\left(1\frac{\text{rad}}{\text{s}}\right)\left(2\text{H}\right)\\ =j\left(1\frac{\text{1}}{\text{s}}\right)\left(2\Omega \cdot \text{s}\right)\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =j2\Omega \end{array}$

Substitute $1\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (5) to find $Z_C$.

$\begin{array}{c}{Z}_C=-\frac{j}{\left(1\frac{\text{rad}}{\text{s}}\right)\left(50\text{mF}\right)}\\ =-\frac{j}{\left(1\right)\left(50\times {10}^{-3}\right)\frac{\text{rad}}{\text{s}}\cdot \text{F}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =-\frac{j}{\left(50\times {10}^{-3}\right)\frac{\text{1}}{\text{s}}\cdot \frac{\text{s}}{\Omega }}\left\{\because 1\text{F}=\frac{\text{1}\text{s}}{1\Omega }\right\}\\ =-j20\Omega \end{array}$

Substitute $10\Omega$ for $Z_{R_1}$, $10\Omega$ for $Z_{R_2}$, $j2\Omega$ for $Z_L$ and $-j20\Omega$ for $Z_C$ in equation (6) to find $Z_{\text{eq}}$.

$\begin{array}{c}{Z}_{\text{eq}}=\left(10\Omega \right)+\left(\left(-j20\Omega \right)\parallel \left(10\Omega +j2\Omega \right)\right)\\ =10\Omega +\frac{\left(-j20\Omega \right)\left(10\Omega +j2\Omega \right)}{-j20\Omega +10\Omega +j2\Omega }\\ =10\Omega +\frac{\left(203.96\angle -78.69°\right)\Omega }{\left(20.59\angle -60.94°\right)\Omega }\\ =\left(19.667\angle -8.83°\right)\Omega \end{array}$

The reduced circuit of Figure 3 is drawn as Figure 4.

Refer to Figure 4, the current $I_o$ is calculated as follows.

$I_o=\frac{V_s}{Z_{\text{eq}}}$

Substitute $\left(19.667\angle -8.83°\right)\Omega$ for $Z_{\text{eq}}$ and $\left(50\angle 0°\right)\text{V}$ for $V_s$ in above equation to find $I_o$.

$\begin{array}{c}{I}_o=\frac{\left(50\angle 0°\right)\text{V}}{\left(19.667\angle -8.83°\right)\Omega }\\ =\left(2.542\angle 8.83°\right)\text{A}\end{array}$

Convert the above equation from phasor form to time domain form.

$i_o\left(t\right)=2.542\sin \left(\omega t+8.83°\right)\text{A}$

Substitute $1$ for $\omega$ in above equation to find $i_o\left(t\right)$.

$\begin{array}{c}{i}_o\left(t\right)=2.542\sin \left(\left(1\right)t+8.83°\right)\text{A}\\ =2.542\sin \left(t+8.83°\right)\text{A}\end{array}$

Conclusion:

Thus, the value of the current $i_o\left(t\right)$ in the circuit of Figure 9.47 when $\omega =1\frac{\text{rad}}{\text{s}}$ is $2.542\sin \left(\left(1\right)t+8.83°\right)\text{A}$.

To determine

Find the value of the current $i_o\left(t\right)$ in the circuit of Figure 9.47 when $\omega =5\frac{\text{rad}}{\text{s}}$.

Answer to Problem 40P

The value of the current $i_o\left(t\right)$ in the circuit of Figure 9.47 when $\omega =5\frac{\text{rad}}{\text{s}}$ is $4.123\sin \left(5t+22.834°\right)\text{A}$.

Explanation of Solution

Given data:

The value of the angular frequency $\left(\omega \right)$ is $5\frac{\text{rad}}{\text{s}}$.

Calculation:

Substitute $5\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (4) to find $Z_L$.

$\begin{array}{c}{Z}_L=j\left(5\frac{\text{rad}}{\text{s}}\right)\left(2\text{H}\right)\\ =j\left(5\frac{\text{1}}{\text{s}}\right)\left(2\Omega \cdot \text{s}\right)\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =j10\Omega \end{array}$

Substitute $5\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (5) to find $Z_C$.

$\begin{array}{c}{Z}_C=-\frac{j}{\left(5\frac{\text{rad}}{\text{s}}\right)\left(50\text{mF}\right)}\\ =-\frac{j}{\left(5\right)\left(50\times {10}^{-3}\right)\frac{\text{rad}}{\text{s}}\cdot \text{F}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =-\frac{j}{\left(250\times {10}^{-3}\right)\frac{\text{1}}{\text{s}}\cdot \frac{\text{s}}{\Omega }}\left\{\because 1\text{F}=\frac{\text{1}\text{s}}{1\Omega }\right\}\\ =-j4\Omega \end{array}$

Substitute $10\Omega$ for $Z_{R_1}$, $10\Omega$ for $Z_{R_2}$, $j10\Omega$ for $Z_L$ and $-j4\Omega$ for $Z_C$ in equation (6) to find $Z_{\text{eq}}$.

$\begin{array}{c}{Z}_{\text{eq}}=\left(10\Omega \right)+\left(\left(-j4\Omega \right)\parallel \left(10\Omega +j10\Omega \right)\right)\\ =10\Omega +\frac{\left(-j4\Omega \right)\left(10\Omega +j10\Omega \right)}{-j4\Omega +10\Omega +j10\Omega }\\ =10\Omega +\frac{\left(56.568\angle -45°\right)\Omega }{\left(11.662\angle 30.963°\right)\Omega }\\ =\left(12.12683\angle -22.834°\right)\Omega \end{array}$

Refer to Figure 4, the current $I_o$ is calculated as follows.

$I_o=\frac{V_s}{Z_{\text{eq}}}$

Substitute $\left(12.12683\angle -22.834°\right)\Omega$ for $Z_{\text{eq}}$ and $\left(50\angle 0°\right)\text{V}$ for $V_s$ in above equation to find $I_o$.

$\begin{array}{c}{I}_o=\frac{\left(50\angle 0°\right)\text{V}}{\left(12.12683\angle -22.834°\right)\Omega }\\ =\left(4.123\angle 22.834°\right)\text{A}\end{array}$

Convert the above equation from phasor form to time domain form.

$i_o\left(t\right)=4.123\sin \left(\omega t+22.834°\right)\text{A}$

Substitute $5$ for $\omega$ in above equation to find $i_o\left(t\right)$.

$\begin{array}{c}{i}_o\left(t\right)=4.123\sin \left(\left(5\right)t+22.834°\right)\text{A}\\ =4.123\sin \left(5t+22.834°\right)\text{A}\end{array}$

Conclusion:

Thus, the value of the current $i_o\left(t\right)$ in the circuit of Figure 9.47 when $\omega =5\frac{\text{rad}}{\text{s}}$ is $4.123\sin \left(5t+22.834°\right)\text{A}$.

To determine

Find the value of the current $i_o\left(t\right)$ in the circuit of Figure 9.47 when $\omega =10\frac{\text{rad}}{\text{s}}$.

Answer to Problem 40P

The value of the current $i_o\left(t\right)$ in the circuit of Figure 9.47 when $\omega =10\frac{\text{rad}}{\text{s}}$ is $4.843\sin \left(10t+12.13°\right)\text{A}$.

Explanation of Solution

Given data:

The value of the angular frequency $\left(\omega \right)$ is $10\frac{\text{rad}}{\text{s}}$.

Calculation:

Substitute $10\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (4) to find $Z_L$.

$\begin{array}{c}{Z}_L=j\left(10\frac{\text{rad}}{\text{s}}\right)\left(2\text{H}\right)\\ =j\left(10\frac{\text{1}}{\text{s}}\right)\left(2\Omega \cdot \text{s}\right)\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =j20\Omega \end{array}$

Substitute $10\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (5) to find $Z_C$.

$\begin{array}{c}{Z}_C=-\frac{j}{\left(10\frac{\text{rad}}{\text{s}}\right)\left(50\text{mF}\right)}\\ =-\frac{j}{\left(10\right)\left(50\times {10}^{-3}\right)\frac{\text{rad}}{\text{s}}\cdot \text{F}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =-\frac{j}{\left(500\times {10}^{-3}\right)\frac{\text{1}}{\text{s}}\cdot \frac{\text{s}}{\Omega }}\left\{\because 1\text{F}=\frac{\text{1}\text{s}}{1\Omega }\right\}\\ =-j2\Omega \end{array}$

Substitute $10\Omega$ for $Z_{R_1}$, $10\Omega$ for $Z_{R_2}$, $j20\Omega$ for $Z_L$ and $-j2\Omega$ for $Z_C$ in equation (6) to find $Z_{\text{eq}}$.

$\begin{array}{c}{Z}_{\text{eq}}=\left(10\Omega \right)+\left(\left(-j2\Omega \right)\parallel \left(10\Omega +j20\Omega \right)\right)\\ =10\Omega +\frac{\left(-j2\Omega \right)\left(10\Omega +j20\Omega \right)}{-j2\Omega +10\Omega +j20\Omega }\\ =10\Omega +\frac{\left(44.721\angle -26.565°\right)\Omega }{\left(20.591\angle 60.945°\right)\Omega }\\ =\left(10.325\angle -12.13°\right)\Omega \end{array}$

Refer to Figure 4, the current $I_o$ is calculated as follows.

$I_o=\frac{V_s}{Z_{\text{eq}}}$

Substitute $\left(10.325\angle -12.13°\right)\Omega$ for $Z_{\text{eq}}$ and $\left(50\angle 0°\right)\text{V}$ for $V_s$ in above equation to find $I_o$.

$\begin{array}{c}{I}_o=\frac{\left(50\angle 0°\right)\text{V}}{\left(10.325\angle -12.13°\right)\Omega }\\ =\left(4.843\angle 12.13°\right)\text{A}\end{array}$

Convert the above equation from phasor form to time domain form.

$i_o\left(t\right)=4.843\sin \left(\omega t+12.13°\right)\text{A}$

Substitute $10$ for $\omega$ in above equation to find $i_o\left(t\right)$.

$\begin{array}{c}{i}_o\left(t\right)=4.843\sin \left(\left(10\right)t+12.13°\right)\text{A}\\ =4.843\sin \left(10t+12.13°\right)\text{A}\end{array}$

Conclusion:

Thus, the value of the current $i_o\left(t\right)$ in the circuit of Figure 9.47 when $\omega =10\frac{\text{rad}}{\text{s}}$ is $4.843\sin \left(10t+12.13°\right)\text{A}$.