Chapter 10, Problem 46P

Solve for v_o(t) in the circuit of Fig. 10.91 using the superposition principle.

Figure 10.91

To determine

Calculate the voltage $v_o\left(t\right)$ in the circuit of Figure 10.91 using superposition theorem.

Answer to Problem 46P

The value of voltage $v_o\left(t\right)$ in the given circuit is $\left[15+32.18\sin \left(2t+26.57°\right)+16.1\cos \left(3t-26.57°\right)\right]\text{V}$.

Explanation of Solution

Given data:

Refer to Figure 10.91 in the textbook.

Formula used:

Write the expression to calculate impedance of the inductor.

$Z_L=j\omega L$        (1)

Here,

$\omega$ is the angular frequency, and

$L$ is the value of inductor.

Write the expression to calculate impedance of the capacitor.

$Z_C=\frac{1}{j\omega C}$        (2)

Here,

$C$ is the value of capacitor.

Write the general representation of sinusoidal cosine function.

$v_s=v_m\cos \left(\omega t+\varphi \right)$        (3)

Here,

$\varphi$ is the phase angle, and

$v_m$ is the magnitude of source voltage.

Write the general expression to phasor transform of sinusoidal function from time domain to frequency domain.

$v_s=P\left\{v_m\cos \left(\omega t+\varphi \right)\right\}=v_m{e}^{i\varphi }$

Here,

$v_m\cos \left(\omega t+\varphi \right)$ is the time domain representation of source voltage, and

$v_m{e}^{i\varphi }$ is the frequency domain representation of source voltage.

Write the polar form representation of frequency domain.

$v_s=v_m\angle \varphi$        (4)

Write the general representation of sinusoidal sine function.

$i_s=i_m\cos \left(\omega t+\varphi \right)$        (5)

Here,

$\varphi$ is the phase angle, and

$i_m$ is the magnitude of source current.

Write the general expression to phasor transform of sinusoidal function from time domain to frequency domain.

$i_s=P\left\{i_m\cos \left(\omega t+\varphi \right)\right\}=i_m{e}^{i\varphi }$

Here,

$i_m\cos \left(\omega t+\varphi \right)$ is the time domain representation of source current, and

$i_m{e}^{i\varphi }$ is the frequency domain representation of source current.

Write the polar form representation of frequency domain.

$i_s=i_m\angle \varphi$        (6)

Calculation:

Let ${v^{\prime }}_o$, ${v^{″}}_o$, and ${v^{‴}}_o$ are the voltages due to $15\text{V}$, $\left[6\sin \left(2t\right)\text{A}\right]$ current source, and $\left[18\cos \left(3t\right)\text{V}\right]$ voltage source respectively.

Consider that $15\text{V}$ voltage source only is in active condition to find ${v^{\prime }}_o$. Replace other voltage source with short circuit, and current source with open circuit. The reduced figure is shown in Figure 1.

The capacitor acts like an open circuit for DC source. Therefore, the entire source voltage will appear across the capacitor.

${v^{\prime }}_o=15\text{V}$

Consider that $\left[6\sin \left(2t\right)\text{A}\right]$ current source only is in active condition to find ${v^{″}}_o$. Replace all other voltage source with short circuit. The reduced figure is shown in Figure 2.

Comparing $\left[6\sin \left(2t\right)\text{A}\right]$ source current with equation (5), the magnitude, angular frequency, and phase angle of source current are $6\text{A}$, $2\frac{\text{rad}}{\text{s}}$ and $0°$ respectively.

Substitute $6\text{A}$ for $i_m$ and $0°$ for $\varphi$ in equation (6).

$i_s=6\angle 0°\text{A}$

Substitute $2\frac{\text{rad}}{\text{s}}$ for $\omega$ and $2\text{H}$ for $L$ in equation (1) to find $Z_L$.

$\begin{array}{c}{Z}_L=j\left(2\frac{\text{rad}}{\text{s}}\right)\left(2\text{H}\right)\left\{\because 1\text{H}=1\Omega \cdot \text{s}\right\}\\ =j4\Omega \end{array}$

Substitute $2\frac{\text{rad}}{\text{s}}$ for $\omega$ and $\frac{1}{12}\text{F}$ for $C$ in equation (2) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(2\frac{\text{rad}}{\text{s}}\right)\left(\frac{1}{12}\text{F}\right)}\\ =\frac{1}{j\left(2\frac{\text{rad}}{\text{s}}\right)\left(\frac{1}{12}\frac{\text{s}}{\Omega }\right)}\left\{\because 1\text{F}=1\frac{\text{s}}{\Omega }\right\}\\ =-j6\Omega \end{array}$

The frequency domain representation of figure 2 is shown in Figure 3.

In Figure 3, the reduce the parallel combination of $\left(6\Omega \right)$ and $\left(j4\Omega \right)$.

$\begin{array}{c}{Z}_{\left(6\Omega \right)\parallel \left(j4\Omega \right)}=\frac{\left(6\Omega \right)\left(j4\Omega \right)}{\left(6+j4\right)\Omega }\\ =3.328\angle 56.31°\Omega \end{array}$

The reduced circuit is shown in Figure 4.

From Figure 4, calculate the current flows through the capacitor using current division rule.

$\begin{array}{c}{i}_{\left(-j6\Omega \right)}=\frac{3.328\angle 56.31°}{\left(3.328\angle 56.31°\right)+\left(-j6\right)}\left(6\angle 0°\right)\text{A}\\ =\frac{\left(1.846+j2.769\right)}{\left(1.846+j2.769-j6\right)}\left(6\angle 0°\right)\text{A}\\ =\frac{\left(1.846+j2.769\right)}{\left(1.846+j2.769-j6\right)}\left(6\angle 0°\right)\text{A}\\ =\text{5}\text{.366}\angle \text{116}\text{.57}°\text{A}\end{array}$

Write the expression to calculate the voltage ${v^{″}}_o$.

${v^{″}}_o=\left(-j6\Omega \right)i_{\left(-j6\Omega \right)}$

Substitute $\text{5}\text{.366}\angle \text{116}\text{.57}°\text{A}$ for $i_{\left(-j6\Omega \right)}$.

$\begin{array}{c}{v^{″}}_o=\left(-j6\Omega \right)\left(\text{5}\text{.366}\angle \text{116}\text{.57}°\text{A}\right)\\ =32.18\angle 26.57°\text{V}\end{array}$

Write the time domain representation of voltage ${v^{″}}_o$.

${v^{″}}_o=32.18\sin \left(2t+26.57°\right)\text{V}$

Consider that $\left[18\cos \left(3t\right)\text{V}\right]$ voltage source only is in active condition to find ${v^{‴}}_o$. Replace all other voltage source with short circuit and current source with open circuit. The reduced figure is shown in Figure 5.

Comparing $\left[18\cos \left(3t\right)\text{V}\right]$ source voltage with equation (3), the magnitude, angular frequency, and phase angle of source voltage are $18\text{V}$, $3\frac{\text{rad}}{\text{s}}$ and $0°$ respectively.

Substitute $18\text{V}$ for $v_m$ and $0°$ for $\varphi$ in equation (4).

$v_s=18\angle 0°\text{V}$

Substitute $3\frac{\text{rad}}{\text{s}}$ for $\omega$ and $2\text{H}$ for $L$ in equation (1) to find $Z_L$.

$\begin{array}{c}{Z}_L=j\left(3\frac{\text{rad}}{\text{s}}\right)\left(2\text{H}\right)\left\{\because 1\text{H}=1\Omega \cdot \text{s}\right\}\\ =j6\Omega \end{array}$

Substitute $3\frac{\text{rad}}{\text{s}}$ for $\omega$ and $\frac{1}{12}\text{F}$ for $C$ in equation (2) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(3\frac{\text{rad}}{\text{s}}\right)\left(\frac{1}{12}\text{F}\right)}\\ =\frac{1}{j\left(3\frac{\text{rad}}{\text{s}}\right)\left(\frac{1}{12}\frac{\text{s}}{\Omega }\right)}\left\{\because 1\text{F}=1\frac{\text{s}}{\Omega }\right\}\\ =-j4\Omega \end{array}$

The frequency domain representation of figure 5 is shown in Figure 6.

Apply Kirchhoff’s current law at node ${v^{‴}}_o$ (non-reference node).

$\begin{array}{l}\frac{{v^{‴}}_o-0}{-j4}+\frac{{v^{‴}}_o-18}{6}+\frac{{v^{‴}}_o-0}{j6}=0\\ \frac{{v^{‴}}_o}{-j4}+\frac{{v^{‴}}_o}{6}-\frac{18}{6}+\frac{{v^{‴}}_o}{j6}=0\\ j0.25{v^{‴}}_o+0.1666{v^{‴}}_o-3-j0.1666{v^{‴}}_o=0\\ \left(0.1666+j0.0834\right){v^{‴}}_o=3\end{array}$

Simplify the equation as follows.

$\begin{array}{c}{v^{‴}}_o=\frac{3}{0.1666+j0.0834}\\ =14.398-j7.21\text{V}\\ {v^{‴}}_o\text{=16}\text{.1}\angle -26.57\text{V}\end{array}$

Write the time domain representation of voltage ${v^{‴}}_o$.

${v^{‴}}_o=16.1\cos \left(3t-26.57°\right)\text{V}$

Write the expression for $v_o$.

$v_o={v^{\prime }}_o+{v^{″}}_o+{v^{‴}}_o$

Substitute $15\text{V}$ for ${v^{\prime }}_o$, $32.18\sin \left(2t+26.57°\right)\text{V}$ for ${v^{″}}_o$ and $16.1\cos \left(3t-26.57°\right)\text{V}$ for ${v^{‴}}_o$.

$v_o=\left[15+32.18\sin \left(2t+26.57°\right)+16.1\cos \left(3t-26.57°\right)\right]\text{V}$

Conclusion:

Thus, the value of voltage $v_o\left(t\right)$ in the given circuit is $\left[15+32.18\sin \left(2t+26.57°\right)+16.1\cos \left(3t-26.57°\right)\right]\text{V}$.