Chapter 10, Problem 61E

Determine the individual contribution each current source makes to the two nodal voltages V₁ and V₂ as represented in Fig. 10.67.

■ FIGURE 10.67

To determine

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

Answer to Problem 61E

The expressions for nodal voltages $V_1\text{and}{V}_2$.are $\underset{\_}{6.83\angle -88.8°\text{V}\text{and}8.33\angle -113.73°\text{V}}$, respectively.

Explanation of Solution

Calculation:

Refer to Figure in the respective question.

Case 1:

Redraw Figure, as shown in Figure 1 by making $3\angle -41°\text{mA}$ acting alone and other source $5\angle 13°\text{mA}$ as open.

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

$\begin{array}{l}3\angle -41°=\frac{V_{1a}}{3}-\frac{V_{1a}}{j3}+\frac{V_{1a}-V_{2a}}{j8}-\frac{V_{1a}-V_{2a}}{j5}\\ 3\angle -41°=\frac{V_{1a}}{3}-\frac{V_{1a}}{j3}+\frac{V_{1a}}{j8}-\frac{V_{2a}}{j8}-\frac{V_{1a}}{j5}+\frac{V_{2a}}{j5}\\ 3\angle -41°=V_{1a}\left(\frac{1}{3}-\frac{1}{j3}+\frac{1}{j8}-\frac{1}{j5}\right)+V_{2a}\left(-\frac{1}{j8}+\frac{1}{j5}\right)\\ 3\angle -41°=V_{1a}\left(0.333+j0.333-j0.125+j0.2\right)+V_{2a}\left(j0.125-j0.2\right)\end{array}$

$V_{1a}\left(0.333+j0.408\right)+V_{2a}\left(-j0.075\right)=3\angle -41°$        (1)

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

$\begin{array}{l}\frac{V_{2a}}{3}+\frac{V_{2a}}{j2}+\frac{V_{2a}-V_{1a}}{j8}-\frac{V_{2a}-V_{1a}}{j5}=0\\ \frac{V_{2a}}{3}+\frac{V_{2a}}{j2}+\frac{V_{2a}}{j8}-\frac{V_{1a}}{j8}-\frac{V_{2a}}{j5}+\frac{V_{1a}}{j5}=0\\ V_{1a}\left(-\frac{1}{j8}+\frac{1}{j5}\right)+V_{2a}\left(\frac{1}{3}+\frac{1}{j2}+\frac{1}{j8}-\frac{1}{j5}\right)=0\\ V_{1a}\left(j0.125-j0.2\right)+V_{2a}\left(0.333-j0.5-j0.125+j0.2\right)=0\end{array}$

$V_{1a}\left(-j0.075\right)+V_{2a}\left(0.333-j0.425\right)=0$        (2)

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

$\left[\begin{array}{cc}0.333+j0.408& -j0.075\\ -j0.075& 0.333-j0.425\end{array}\right]\left[\begin{array}{c}{V}_{1a}\\ V_{2a}\end{array}\right]=\left[\begin{array}{c}3\angle -41°\\ 0\end{array}\right]$

$\left[\begin{array}{cc}0.333+j0.408& -j0.075\\ -j0.075& 0.333-j0.425\end{array}\right]\left[\begin{array}{c}{V}_{1a}\\ V_{2a}\end{array}\right]=\left[\begin{array}{c}2.264-j1.968\\ 0\end{array}\right]$        (3)

Write the Matlab code to solve equation (3)

A = [(0.333+i*0.408) (-i*0.075); (-i*0.075) (0.333-i*0.425)];

B = [(2.264-i*1.968);(0)];

C = inv(A)*B

Matlab Output:

C =

  -0.17551 - 5.58282i

   0.49749 + 0.59541i

Polar equations of node voltages are as follows.

$\begin{array}{l}{V}_{1a}=5.585\angle -91.8°\text{V}\\ V_{2a}=0.775\angle 50.11°\text{V}\end{array}$

Case 2:

Redraw Figure, as shown in Figure 2 by making $5\angle 13°\text{mA}$ acting alone and other source $3\angle -41°\text{mA}$ as open.

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

$\begin{array}{l}\frac{V_{1b}}{3}-\frac{V_{1b}}{j3}+\frac{V_{1b}-V_{2b}}{j8}-\frac{V_{1b}-V_{2b}}{j5}=0\\ \frac{V_{1b}}{3}-\frac{V_{1b}}{j3}+\frac{V_{1b}}{j8}-\frac{V_{2b}}{j8}-\frac{V_{1b}}{j5}+\frac{V_{2b}}{j5}=0\\ V_{1b}\left(\frac{1}{3}-\frac{1}{j3}+\frac{1}{j8}-\frac{1}{j5}\right)+V_{2b}\left(-\frac{1}{j8}+\frac{1}{j5}\right)=0\\ V_{1b}\left(0.333+j0.333-j0.125+j0.2\right)+V_{2b}\left(j0.125-j0.2\right)=0\end{array}$

$V_{1b}\left(0.333+j0.408\right)+V_{2b}\left(-j0.075\right)=0$        (4)

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

$\begin{array}{l}\frac{V_{2b}}{3}+\frac{V_{2b}}{j2}+\frac{V_{2b}-V_{1b}}{j8}-\frac{V_{2b}-V_{1b}}{j5}=-5\angle 13°\\ \frac{V_{2b}}{3}+\frac{V_{2b}}{j2}+\frac{V_{2b}}{j8}-\frac{V_{1b}}{j8}-\frac{V_{2b}}{j5}+\frac{V_{1b}}{j5}=-5\angle 13°\\ V_{1b}\left(-\frac{1}{j8}+\frac{1}{j5}\right)+V_{2b}\left(\frac{1}{3}+\frac{1}{j2}+\frac{1}{j8}-\frac{1}{j5}\right)=-5\angle 13°\\ V_{1b}\left(j0.125-j0.2\right)+V_{2b}\left(0.333-j0.5-j0.125+j0.2\right)=-5\angle 13°\end{array}$

$V_{1b}\left(-j0.075\right)+V_{2b}\left(0.333-j0.425\right)=-5\angle 13°$        (5)

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

$\left[\begin{array}{cc}0.333+j0.408& -j0.075\\ -j0.075& 0.333-j0.425\end{array}\right]\left[\begin{array}{c}{V}_{1b}\\ V_{2b}\end{array}\right]=\left[\begin{array}{c}0\\ -5\angle 13°\end{array}\right]$

$\left[\begin{array}{cc}0.333+j0.408& -j0.075\\ -j0.075& 0.333-j0.425\end{array}\right]\left[\begin{array}{c}{V}_{1b}\\ V_{2b}\end{array}\right]=\left[\begin{array}{c}0\\ -4.87-j1.124\end{array}\right]$        (6)

Write the Matlab code to solve equation (6)

A = [(0.333+i*0.408) (-i*0.075); (-i*0.075) (0.333-i*0.425)];

B = [(0);(-4.87-i*1.124)];

C = inv(A)*B

Matlab Output:

C =

   0.3153 - 1.2537i

  -3.8514 - 8.2199i

Polar equations of node voltages are as follows.

$\begin{array}{l}{V}_{1b}=1.292\angle -75.8°\text{V}\\ V_{2b}=9.077\angle -115.10°\text{V}\end{array}$

Find nodal voltage $V_1$.

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

Substitute $5.585\angle -91.8°\text{V}$ for $V_{1a}$ and $1.292\angle -75.8°\text{V}$ for $V_{1b}$ as follows.

$\begin{array}{c}{V}_1=5.585\angle -91.8°+1.292\angle -75.8°\\ =0.1415-j6.83\\ =6.83\angle -88.8°\text{V}\end{array}$

Find nodal voltage $V_2$.

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

Substitute $0.775\angle 50.11°\text{V}$ for $V_{2a}$ and $9.077\angle -115.10°\text{V}$ for $V_{2b}$ as follows.

$\begin{array}{c}{V}_2=0.775\angle 50.11°+9.077\angle -115.10°\\ =-3.353-j7.625\\ =8.33\angle -113.73°\text{V}\end{array}$

Conclusion:

Thus, the expressions for nodal voltages $V_1\text{and}{V}_2$.are $\underset{\_}{6.83\angle -88.8°\text{V}\text{and}8.33\angle -113.73°\text{V}}$, respectively.