Chapter 10, Problem 63E

To determine

Find the expressions for voltages $V_1\text{and}{V}_2$.

Answer to Problem 63E

The expressions for nodal voltages $V_1\text{and}{V}_2$.are $\underset{\_}{0.018\angle -66.6°\text{V}\text{and}0.0163\angle -70.19°\text{V}}$, respectively.

Explanation of Solution

Given data:

$\begin{array}{l}{i}_1\left(t\right)=4\cos 40t\text{mA}\\ i_2\left(t\right)=4\sin 30t\text{mA}\end{array}$

Formula used:

Consider the general expression for inductive impedance.

$Z_L=j\omega L$        (1)

Here,

$\omega$ is the frequency.

$L$ is the inductance.

Consider the general expression for capacitive impedance.

$Z_C=-j\frac{1}{\omega C}$        (2)

Here,

$\omega$ is the frequency.

$C$ is the capacitance.

Calculation:

Refer to Figure in the respective question.

From Figure , at $\omega =2.5\text{rad}/\text{s}$, Find the value of $L$.

Substitute $2.5\text{rad}/\text{s}$ for $\omega$ and $j3\Omega$ for $Z_L$ in equation (1) as follows.

$\begin{array}{l}j3=j\left(2.5\right)L\\ L=\frac{j3}{j2.5}\\ L=1.2\text{H}\end{array}$

Substitute $2.5\text{rad}/\text{s}$ for $\omega$ and $-j5\Omega$ for $Z_C$ in equation (2) as follows.

$\begin{array}{l}-j5=-j\frac{1}{\left(2.5\right)C}\\ 5=\frac{1}{\left(2.5\right)C}\\ C=\frac{1}{2.5\left(5\right)}\\ C=\frac{1}{12.5}\end{array}$

$C=0.08\text{F}$

Case 1:

Consider $I_1$ acting alone and other source $I_2$ as open, When $I_1$ source is acting alone $\omega =40\text{rad}/\text{s}$.

Substitute $0.08\text{F}$ for $C$ and $40\text{rad}/\text{s}$ for $\omega$ in equation (2) as follows.

$\begin{array}{c}{Z}_C=-j\frac{1}{\left(40\text{rad}/\text{s}\right)\left(0.08\text{F}\right)}\\ =-j\frac{1}{\left(40\right)\left(0.08\right)}\\ =-j\frac{1}{3.2}\\ =-j0.3125\Omega \end{array}$

Substitute $1.2\text{H}$ for $L$ and $40\text{rad}/\text{s}$ for $\omega$ in equation (1) as follows.

$\begin{array}{c}{Z}_L=j\left(40\text{rad}/\text{s}\right)\left(1.2\text{H}\right)\\ =j\left(40\right)\left(1.2\right)\\ =j48\Omega \end{array}$

Redraw Figure, as shown in Figure 1.

Apply KCL at node $V_{1a}$ of Figure 1.

$\begin{array}{l}0.004\angle 0°=\frac{V_{1a}}{j48}+\frac{V_{1a}-V_{2a}}{-j0.3125}\\ 0.004\angle 0°=\frac{V_{1a}}{j48}-\frac{V_{1a}-V_{2a}}{j0.3125}\\ 0.004\angle 0°=\frac{V_{1a}}{j48}-\frac{V_{1a}}{j0.3125}-\frac{V_{2a}}{j0.3125}\\ 0.004\angle 0°=V_{1a}\left(\frac{1}{j48}-\frac{1}{j0.3125}\right)-V_{2a}\left(\frac{1}{j0.3125}\right)\end{array}$

Simplify the equation as follows.

$V_{1a}\left(-j0.0208+j3.2\right)-V_{2a}\left(-j3.2\right)=0.004\angle 0°$

$V_{1a}\left(j3.1792\right)+V_{2a}\left(j3.2\right)=0.004\angle 0°$        (1)

Apply KCL at node $V_{2a}$ of Figure 1.

$\begin{array}{l}\frac{V_{2a}}{2}+\frac{V_{2a}-V_{1a}}{-j0.3125}=0\\ \frac{V_{2a}}{2}-\frac{V_{2a}}{j0.3125}+\frac{V_{1a}}{j0.3125}=0\\ V_{1a}\left(\frac{1}{j0.3125}\right)+V_{2a}\left(\frac{1}{2}-\frac{1}{j0.3125}\right)=0\end{array}$

$V_{1a}\left(-j3.2\right)+V_{2a}\left(0.5+j3.2\right)=0$        (2)

Write the equations (1) and (2) in matrix form.

$\left[\begin{array}{cc}j3.1792& -j3.2\\ -j3.2& 0.5+j3.2\end{array}\right]\left[\begin{array}{c}{V}_{1a}\\ V_{2a}\end{array}\right]=\left[\begin{array}{c}0.004\angle 0°\\ 0\end{array}\right]$

$\left[\begin{array}{cc}j3.1792& -j3.2\\ -j3.2& 0.5+j3.2\end{array}\right]\left[\begin{array}{c}{V}_{1a}\\ V_{2a}\end{array}\right]=\left[\begin{array}{c}0.004+j0\\ 0\end{array}\right]$        (3)

Write the Matlab code to solve equation (3)

A = [(i*3.1792) (-i*3.2); (-i*3.2) (0.5+i*3.2)];

B = [(0.004+i*0);(0)];

C = inv(A)*B

Matlab Output:

C =

   0.0080908 - 0.0009194i

   0.0080382 + 0.0003366i

Polar equations of node voltages are as follows.

$\begin{array}{l}{V}_{1a}=8.142\times {10}^{-3}\angle -6.48°\text{V}\\ V_{2a}=8.045\times {10}^{-3}\angle 2.39°\text{V}\end{array}$

Case 2:

Consider $I_2$ acting alone and other source $I_1$ as open, When $I_2$ source is acting alone $\omega =30\text{rad}/\text{s}$.

Substitute $0.08\text{F}$ for $C$ and $30\text{rad}/\text{s}$ for $\omega$ in equation (2) as follows.

$\begin{array}{c}{Z}_C=-j\frac{1}{\left(30\text{rad}/\text{s}\right)\left(0.08\text{F}\right)}\\ =-j\frac{1}{\left(30\right)\left(0.08\right)}\\ =-j\frac{1}{2.4}\\ =-j0.4166\Omega \end{array}$

Substitute $1.2\text{H}$ for $L$ and $30\text{rad}/\text{s}$ for $\omega$ in equation (1) as follows.

$\begin{array}{c}{Z}_L=j\left(30\text{rad}/\text{s}\right)\left(1.2\text{H}\right)\\ =j\left(30\right)\left(1.2\right)\\ =j36\Omega \end{array}$

Redraw Figure, as shown in Figure 2.

Apply KCL at node $V_{1b}$ of Figure 1.

$\begin{array}{l}-0.004\angle -90°=\frac{V_{1b}}{j36}+\frac{V_{1b}-V_{2b}}{-j0.4166}\\ -0.004\angle -90°=\frac{V_{1b}}{j36}-\frac{V_{1b}}{j0.4166}+\frac{V_{2b}}{j0.4166}\\ -0.004\angle -90°=V_{1b}\left(\frac{1}{j36}-\frac{1}{j0.4166}\right)+V_{2b}\left(\frac{1}{j0.4166}\right)\\ -0.004\angle -90°=V_{1b}\left(-j0.027+j2.4\right)+V_{2b}\left(-j2.4\right)\end{array}$

Simplify the equation as follows.

$V_{1b}\left(j2.373\right)+V_{2b}\left(-j2.4\right)=-0.004\angle -90°$        (4)

Apply KCL at node $V_{2a}$ of Figure 1.

$\begin{array}{l}0.004\angle -90°=\frac{V_{2b}}{2}+\frac{V_{2b}-V_{1b}}{-j0.4166}\\ 0.004\angle -90°=\frac{V_{2b}}{2}-\frac{V_{2b}}{j0.4166}+\frac{V_{1b}}{j0.4166}\\ 0.004\angle -90°=V_{2b}\left(\frac{1}{2}-\frac{1}{j0.4166}\right)+V_{1b}\left(\frac{1}{j0.4166}\right)\end{array}$

$V_{2b}\left(0.5+j2.4\right)+V_{1b}\left(-j2.4\right)=0.004\angle -90°$        (5)

Write the equations (4) and (5) in matrix form.

$\left[\begin{array}{cc}j2.373& -j2.4\\ 0.5+j2.4& -j2.4\end{array}\right]\left[\begin{array}{c}{V}_{1b}\\ V_{2b}\end{array}\right]=\left[\begin{array}{c}-0.004\angle -90°\\ 0.004\angle -90°\end{array}\right]$

$\left[\begin{array}{cc}j2.373& -j2.4\\ 0.5+j2.4& -j2.4\end{array}\right]\left[\begin{array}{c}{V}_{1b}\\ V_{2b}\end{array}\right]=\left[\begin{array}{c}j0.004\\ -j0.004\end{array}\right]$        (6)

Write the Matlab code to solve equation (3)

A = [(i*2.373) (-i*2.4); (0.5+i*2.4) (-i*2.4)];

B = [(i*0.004);(-i*0.004)];

C = inv(A)*B

Matlab Output:

C =

  -0.000861 - 0.015953i

  -0.002518 - 0.015774i

Polar equations of node voltages are as follows.

$\begin{array}{l}{V}_{1b}=0.0159\angle -93.08°\text{V}\\ V_{2b}=0.0159\angle -99.06°\text{V}\end{array}$

Find nodal voltage $V_1$.

$V_1=V_{1a}+V_{1b}$

Substitute $8.142\times {10}^{-3}\angle -6.48°\text{V}$ for $V_{1a}$ and $0.0159\angle -93.08°\text{V}$ for $V_{1b}$ as follows.

$\begin{array}{c}{V}_1=8.142\times {10}^{-3}\angle -6.48°+0.0159\angle -93.08°\\ =7.235\times {10}^{-3}-j0.016\\ =0.018\angle -66.6°\text{V}\end{array}$

Convert $V_1$ in terms of time domain $v_1\left(t\right)$.

$v_1\left(t\right)=0.0$

Find nodal voltage $V_2$.

$V_2=V_{2a}+V_{2b}$

Substitute $.045\times {10}^{-3}\angle 2.39°\text{V}$ for $V_{2a}$ and $0.0159\angle -99.06°\text{V}$ for $V_{2b}$ as follows.

$\begin{array}{c}{V}_2=8.045\times {10}^{-3}\angle 2.39°+0.0159\angle -99.06°\\ =5.53\times {10}^{-3}-j0.0153\\ =0.0163\angle -70.19°\text{V}\end{array}$

Conclusion:

Thus, the expressions for nodal voltages $V_1\text{and}{V}_2$.are $\underset{\_}{0.018\angle -66.6°\text{V}\text{and}0.0163\angle -70.19°\text{V}}$, respectively.