Chapter 4, Problem 64E

Verify numerical values for each nodal voltage in Exercise 14 by employing LTspice or a similar CAD tool. Submit a printout of an appropriately labeled schematic with the nodal voltages highlighted, along with your hand calculations.

To determine

Verify the solution obtained in Exercise 14 using LTspice.

Explanation of Solution

Calculation:

Refer to FIGURE 4.44 in the textbook.

Apply nodal analysis at node $v_1$.

$\begin{array}{l}\frac{v_1}{2}+\frac{v_1-v_3}{6}=2\\ \frac{6v_1+2\left(v_1-v_3\right)}{12}=2\\ 6v_1+2\left(v_1-v_3\right)=24\\ 8v_1-2v_3=24\end{array}$

$4v_1-v_3=12$ (1)

Apply nodal analysis at node $v_2$.

$\begin{array}{l}-2=\frac{v_2}{5}+\frac{v_2-v_3}{2}+\frac{v_2-v_4}{10}\\ 2v_2+5\left(v_2-v_3\right)+v_2-v_4=-20\end{array}$

$8v_2-5v_3-v_4=-20$ (2)

Apply nodal analysis at node $v_3$.

$\begin{array}{l}1=\frac{v_3-v_4}{5}+\frac{v_3-v_1}{6}+\frac{v_3-v_2}{2}\\ 1=\frac{6\left(v_3-v_4\right)+5\left(v_3-v_1\right)+15\left(v_3-v_2\right)}{30}\end{array}$

$-5v_1-15v_2+26v_3-6v_4=30$ (3)

Apply nodal analysis at node $v_4$.

$\begin{array}{l}\frac{v_4-v_2}{10}+\frac{v_4-v_3}{5}+\frac{v_4}{5}=0\\ v_4-v_2+2\left(v_4-v_3\right)+2v_4=0\end{array}$

$-v_2-2v_3+5v_4=0$ (4)

Solve the equations by Cramer’s rule.

$\left[\begin{array}{c}12\\ -20\\ 30\\ 0\end{array}\right]=\left[\begin{array}{cccc}4& 0& -1& 0\\ 0& 8& -5& -1\\ -5& -15& 26& -6\\ 0& -1& -2& 5\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\\ v_3\\ v_4\end{array}\right]$

Find $\Delta$.

$\begin{array}{c}\Delta =\left|\begin{array}{cccc}4& 0& -1& 0\\ 0& 8& -5& -1\\ -5& -15& 26& -6\\ 0& -1& -2& 5\end{array}\right|\\ =1737\end{array}$

Find $\Delta v_1$.

$\begin{array}{c}\Delta v_1=\left|\begin{array}{cccc}12& 0& -1& 0\\ -20& 8& -5& -1\\ 30& -15& 26& -6\\ 0& -1& -2& 5\end{array}\right|\\ =5346\end{array}$

The value of $v_1$ is,

$\begin{array}{c}{v}_1=\frac{\Delta v_1}{\Delta }\\ =\frac{5346}{1737}\\ =3.078\text{V}\end{array}$

Find $\Delta v_2$.

$\begin{array}{c}\Delta v_2=\left|\begin{array}{cccc}4& 12& -1& 0\\ 0& -20& -5& -1\\ -5& 30& 26& -6\\ 0& 0& -2& 5\end{array}\right|\\ =-4080\end{array}$

The value of $v_2$ is,

$\begin{array}{c}{v}_2=\frac{\Delta v_2}{\Delta }\\ =\frac{-4080}{1737}\\ =-2.349\text{V}\end{array}$

Find $\Delta v_3$.

$\begin{array}{c}\Delta v_3=\left|\begin{array}{cccc}4& 0& 12& 0\\ 0& 8& -20& -1\\ -5& -15& 30& -6\\ 0& -1& 0& 5\end{array}\right|\\ =540\end{array}$

The value of $v_3$ is,

$\begin{array}{c}{v}_3=\frac{\Delta v_3}{\Delta }\\ =\frac{540}{1737}\\ =0.3109\text{V}\end{array}$

Find $\Delta v_4$.

$\begin{array}{c}\Delta v_4=\left|\begin{array}{cccc}4& 0& -1& 12\\ 0& 8& -5& -20\\ -5& -15& 26& 30\\ 0& -1& -2& 0\end{array}\right|\\ =-600\end{array}$

The value of $v_4$ is,

$\begin{array}{c}{v}_4=\frac{\Delta v_4}{\Delta }\\ =\frac{-600}{1737}\\ =-0.3454\text{V}\end{array}$

Apply nodal analysis at node $v_5$.

$\begin{array}{l}\frac{v_5}{1}+\frac{v_5-v_7}{4}=-2\\ \frac{4v_5+v_5-v_7}{4}=-2\end{array}$

$5v_5-v_7=-8$ (5)

Apply nodal analysis at node $v_6$.

$\begin{array}{l}\frac{v_6-v_7}{2}+\frac{v_6}{4}+\frac{v_6-v_8}{4}=2\\ \frac{2\left(v_6-v_7\right)}{4}+\frac{v_6}{4}+\frac{v_6-v_8}{4}=2\\ \frac{2v_6-2v_7+v_6+v_6-v_8}{4}=2\end{array}$

$4v_6-2v_7-v_8=8$ (6)

Apply nodal analysis at node $v_7$.

$\begin{array}{l}\frac{v_7-v_8}{10}+\frac{v_7-v_5}{4}+\frac{v_7-v_6}{2}=6\\ \frac{2\left(v_7-v_8\right)}{20}+\frac{5\left(v_7-v_5\right)}{20}+\frac{10\left(v_7-v_6\right)}{20}=6\\ \frac{2v_7-2v_8+5v_7-5v_5+10v_7-10v_6}{20}=6\end{array}$

$-5v_5-10v_6+17v_7-2v_8=120$ (7)

Apply nodal analysis at node $v_8$.

$\begin{array}{l}\frac{v_8-v_6}{4}+\frac{v_8}{1}+\frac{v_8-v_7}{10}=0\\ \frac{10\left(v_8-v_6\right)}{40}+\frac{40v_8}{40}+\frac{4\left(v_8-v_7\right)}{40}=0\\ \frac{10v_8-10v_6+40v_8+4v_8-4v_7}{40}=0\end{array}$

$-10v_6-4v_7+54v_8=0$ (8)

Solve the equations by Cramer’s rule.

$\left[\begin{array}{cccc}5& 0& -1& 0\\ 0& 4& -2& -1\\ -5& -10& 17& -2\\ 0& -10& -4& 54\end{array}\right]\left[\begin{array}{c}{v}_5\\ v_6\\ v_7\\ v_8\end{array}\right]=\left[\begin{array}{c}-8\\ 8\\ 120\\ 0\end{array}\right]$

Find $\Delta$.

$\begin{array}{c}\Delta =\left|\begin{array}{cccc}5& 0& -1& 0\\ 0& 4& -2& -1\\ -5& -10& 17& -2\\ 0& -10& -4& 54\end{array}\right|\\ =10,520\end{array}$

Find $\Delta v_5$.

$\begin{array}{c}\Delta v_5=\left|\begin{array}{cccc}-8& 0& -1& 0\\ 8& 8& -5& -1\\ 120& -15& 26& -6\\ 0& -1& -2& 5\end{array}\right|\\ =10,720\end{array}$

The value of $v_5$ is,

$\begin{array}{c}{v}_5=\frac{\Delta v_5}{\Delta }\\ =\frac{10,720}{10,520}\\ =1.02\text{V}\end{array}$

Find $\Delta v_6$.

$\begin{array}{c}\Delta v_6=\left|\begin{array}{cccc}5& -8& -1& 0\\ 0& 8& -2& -1\\ -5& 120& 17& -2\\ 0& 0& -4& 54\end{array}\right|\\ =96,960\end{array}$

The value of $v_6$ is,

$\begin{array}{c}{v}_6=\frac{\Delta v_6}{\Delta }\\ =\frac{96,960}{10,520}\\ =9.217\text{V}\end{array}$

Find $\Delta v_7$.

$\begin{array}{c}\Delta v_7=\left|\begin{array}{cccc}5& 0& -8& 0\\ 0& 4& 8& -1\\ -5& -10& 120& -2\\ 0& -10& 0& 54\end{array}\right|\\ =137,760\end{array}$

The value of $v_7$ is,

$\begin{array}{c}{v}_7=\frac{\Delta v_7}{\Delta }\\ =\frac{137,760}{10,520}\\ =13.095\text{V}\end{array}$

Find $\Delta v_8$.

$\begin{array}{c}\Delta v_8=\left|\begin{array}{cccc}5& 0& -1& -8\\ 0& 4& -2& 8\\ -5& -10& 17& 120\\ 0& -10& -4& 0\end{array}\right|\\ =28,160\end{array}$

The value of $v_8$ is,

$\begin{array}{c}{v}_8=\frac{\Delta v_8}{\Delta }\\ =\frac{28,160}{10,520}\\ =2.6768\text{V}\end{array}$

Thus, the nodal voltages are,

S. No	Node	Nodal voltage
1	$v_1$	3.078 V
2	$v_2$	–2.349 V
3	$v_3$	0.3109 V
4	$v_4$	–0.3454 V
5	$v_5$	1.02 V
6	$v_6$	9.217 V
7	$v_7$	13.095 V
8	$v_8$	2.6768 V

Draw the circuit diagram in LTspice as shown in Figure 1.

The output after simulating the LTspice circuit is,

      --- Operating Point ---

V(v3):     0.310881     voltage

V(v1):     3.07772     voltage

V(v2):     -2.34888     voltage

V(v4):     -0.345423     voltage

V(v7):     13.0951     voltage

V(v5):     1.01901     voltage

V(v6):     9.21673     voltage

V(v8):     2.67681     voltage

I(I3):     2     device_current

I(I1):     2     device_current

I(I4):     6     device_current

I(I2):     1     device_current

I(R14):     -1.63498     device_current

I(R13):     0.200345     device_current

I(R12):     2.67681     device_current

I(R11):     1.04183     device_current

I(R10):     2.30418     device_current

I(R9):     1.93916     device_current

I(R8):     1.01901     device_current

I(R7):     3.01901     device_current

I(R6):     -0.0690846     device_current

I(R5):     0.131261     device_current

I(R4):     -0.469775     device_current

I(R3):     1.32988     device_current

I(R2):     1.53886     device_current

I(R1):     -0.46114     device_current

The both calculated and simulated values are approximately equal.

Conclusion:

Therefore, the solution is verified with the LTspice simulation.