Chapter 10, Problem 44P

Use the superposition principle to obtain v_x in the circuit of Fig. 10.89. Let v_s = 50 sin 2t V and i_s = 12 cos(6t + 10°) A.

Figure 10.89

To determine

Calculate the voltage $v_x$ in the circuit of Figure 10.89 using superposition theorem.

Answer to Problem 44P

The value of voltage $v_x$ in the given circuit is,

$\left[17.678\sin \left(2t-45°\right)+37.95\cos \left(6t-61.57°\right)\right]\text{V}$.

Explanation of Solution

Given data:

Refer to Figure 10.89 in the textbook.

Formula used:

Write the expression to calculate impedance of the capacitor.

$Z_C=\frac{1}{j\omega C}$ (1)

Here,

$C$ is the value of capacitor.

Write the general representation of sinusoidal sine function.

$v_s=v_m\sin \left(\omega t+\varphi \right)$ (2)

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.

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

Here,

$v_m\sin \left(\omega t+\varphi \right)$ 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.

$v_s=v_m\angle \varphi$ (3)

Write the general representation of sinusoidal cosine function.

$i_s=i_m\cos \left(\omega t+\varphi \right)$ (4)

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.

$i_s=i_m\angle \varphi$ (5)

Calculation:

Let $v_x{}^{\prime }$ and $v_x{}^{\prime \text{}\prime }$ are the voltages caused by the voltage and current sources respectively.

Comparing $\left[50\sin 2t\text{V}\right]$ source voltage with equation (2), the magnitude, angular frequency, and phase angle of source voltage are $50\text{V}$, $2\frac{\text{rad}}{\text{s}}$ and $0°$ respectively.

Substitute $50\text{V}$ for $v_m$ and $0°$ for $\varphi$ in equation (3).

$v_s=50\angle 0°\text{V}$

Substitute $2\frac{\text{rad}}{\text{s}}$ for $\omega$ and $50\text{mF}$ for $C$ in equation (1) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(2\frac{\text{rad}}{\text{s}}\right)\left(50\text{mF}\right)}\\ =\frac{1}{j\left(2\frac{\text{rad}}{\text{s}}\right)\left(0.05\frac{\text{s}}{\Omega }\right)}\left\{\begin{array}{l}\because 1\text{mF}=1\times {10}^{-3}\text{F}\\ 1\text{F}=1\frac{\text{s}}{\Omega }\end{array}\right\}\\ =-j10\Omega \end{array}$

To find $v_x{}^{\prime }$, consider that $50\angle 0°\text{V}$ voltage source is in active condition and replace $\left[12\cos \left(6t+10°\right)\text{A}\right]$ current source with its internal impedance (open circuit). The frequency domain representation of the circuit used to find $v_x{}^{\prime }$ is shown in Figure 1.

In Figure 1, apply Kirchhoff’s current law at node $v_x{}^{\prime }$.

$\begin{array}{l}\frac{{v^{\prime }}_x-0}{-j10}+\frac{{v^{\prime }}_x-0}{20}+\frac{{v^{\prime }}_x-\left(50\angle 0°\right)}{20}=0\\ j0.1{v^{\prime }}_x+0.05{v^{\prime }}_x+0.05{v^{\prime }}_x-2.5=0\\ \left(0.1+j0.1\right){v^{\prime }}_x=2.5\end{array}$

Simplify the above equation as follows.

$\begin{array}{c}{v^{\prime }}_x=\frac{2.5}{\left(0.1+j0.1\right)}\\ =12.5-j12.5\text{V}\\ =\text{17}\text{.687}\angle \text{-45}°\text{V}\end{array}$

Represent the voltage $v_x{}^{\prime }$ in time domain.

$v_x{}^{\prime }=17.678\sin \left(2t-45°\right)\text{V}$

Comparing $\left[12\cos \left(6t+10°\right)\text{A}\right]$ source current with equation (4), the magnitude, angular frequency, and phase angle of source current are $12\text{A}$, $6\frac{\text{rad}}{\text{s}}$ and $10°$ respectively.

Substitute $12\text{A}$ for $i_m$ and $10°$ for $\varphi$ in equation (5).

$i_s=12\angle 10°\text{A}$

Substitute $6\frac{\text{rad}}{\text{s}}$ for $\omega$ and $50\text{mF}$ for $C$ in equation (1) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(6\frac{\text{rad}}{\text{s}}\right)\left(50\text{mF}\right)}\\ =\frac{1}{j\left(6\frac{\text{rad}}{\text{s}}\right)\left(0.05\frac{\text{s}}{\Omega }\right)}\left\{\begin{array}{l}\because 1\text{mF}=1\times {10}^{-3}\text{F}\\ 1\text{F}=1\frac{\text{s}}{\Omega }\end{array}\right\}\\ =-j3.333\Omega \end{array}$

To find $v_x{}^{\prime \text{}\prime }$, consider that $12\angle 10°\text{A}$ current source is in active condition and replace $50\angle 0°\text{V}$ voltage source with its internal impedance (short circuit). The frequency domain representation of the circuit used to find $v_x{}^{\prime \text{}\prime }$ is shown in Figure 2.

In Figure 2, reduce the parallel combination of $\left(20\Omega \right)$ and $\left(20\Omega \right)$ resistance.

$\begin{array}{c}{Z}_{\left(20\Omega \right)\parallel \left(20\Omega \right)}=\frac{\left(20\Omega \right)\left(20\Omega \right)}{\left(20\Omega \right)+\left(20\Omega \right)}\\ =10\Omega \end{array}$

The reduced circuit is as shown in Figure 3.

From Figure 3, write the expression for voltage $v_x{}^{\prime \text{}\prime }$.

$\begin{array}{c}{v}_x{}^{\prime \text{}\prime }=\left(-j3.333\Omega \right)\left[\left(12\angle 10°\text{A}\right)\frac{10\Omega }{\left(10-j3.333\right)\Omega }\right]\\ =\left(-j3.333\Omega \right)\left(10.011+j5.42\right)\text{A}\\ =18.066-j33.367\text{V}\\ =\text{37}\text{.95}\angle -\text{61}\text{.57}°\text{V}\end{array}$

Represent the voltage $v_x{}^{\prime \text{}\prime }$ in time domain.

$v_x{}^{\prime \text{}\prime }=37.95\cos \left(6t-61.57°\right)\text{V}$

Write the expression for current $v_x$.

$v_x={v^{\prime }}_x+{v^{″}}_x$

Substitute $17.678\sin \left(2t-45°\right)\text{V}$ for ${v^{\prime }}_x$ and $37.95\cos \left(6t-61.57°\right)\text{V}$ for $v_x{}^{\prime \text{}\prime }$.

$v_x=\left[17.678\sin \left(2t-45°\right)+37.95\cos \left(6t-61.57°\right)\right]\text{V}$

Conclusion:

Thus, the value of voltage $v_x$ in the given circuit is,

$\left[17.678\sin \left(2t-45°\right)+37.95\cos \left(6t-61.57°\right)\right]\text{V}$.