Chapter 3, Problem 3.46HP

To determine

The value of the node voltages $V_a$ and $V_b$ .

Answer to Problem 3.46HP

The value of node voltage $V_a$ is $-2.708\text{V}$ and $V_b$ is $3.415\text{V}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

Replace all the voltage sources except source $V_1$ with their internal resistance and redraw the circuit.

The required diagram is shown in Figure 2

  

Apply nodal at the node $V_{b1}$ .

  $\begin{array}{l}\frac{V_{b1}-2}{10+4}+\frac{V_{b1}}{6}+\frac{V_{b1}}{6}=0\\ V_{b1}\left[\frac{1}{14}+\frac{1}{6}+\frac{1}{6}\right]=\frac{2}{14}\\ V_{b1}\left[\frac{17}{42}\right]=\frac{2}{14}\\ V_{b1}=\frac{6}{17}\text{V}\end{array}$

The expression to calculate the current through the resistance $R_1$ is given by,

  $I_{R_1}=\frac{2\text{V}-V_{b1}}{10\Omega +4\Omega }$

Substitute $\frac{6}{17}\text{V}$ for $V_{b1}$ in the above equation.

  $\begin{array}{c}{I}_{R_1}=\frac{2\text{V}-\frac{6}{17}\text{V}}{10\Omega +4\Omega }\\ =0.1176\text{A}\end{array}$

The expression to calculate the value of the voltage $V_{a1}$ is given by,

  $V_{a1}=-I_{R_1}\left(10\Omega \right)+2\text{V}$

Substitute $I_{R_1}$ for $I_{R_1}$ in the above equation.

  $\begin{array}{c}{V}_{a1}=-\left(0.1176\text{A}\right)\left(10\Omega \right)+2\text{V}\\ =\text{0}\text{.824}\text{V}\end{array}$

Short circuit the source $V_1$ and consider only voltage source $V_2$ .

The required diagram is shown in Figure 3

  

Apply nodal at the node $V_{b2}$ .

  $\begin{array}{l}\frac{V_{b2}}{10+4}+\frac{V_{b2}}{6}+\frac{V_{b2}-4}{6}=0\\ V_{b2}\left[\frac{1}{14}+\frac{1}{6}+\frac{1}{6}\right]=\frac{4}{6}\\ V_{b2}\left[\frac{17}{42}\right]=\frac{4}{6}\\ V_{b2}=1.65\text{V}\end{array}$

The value of the voltage $V_{a2}$ is given by,

  $V_{a2}=\left(V_{b2}\right)\left(\frac{10\Omega }{10\Omega +4\Omega }\right)$

Substitute $1.65\text{V}$ for $V_{b2}$ in the above equation.

  $\begin{array}{c}{V}_{a2}=\left(1.65\text{V}\right)\left(\frac{10\Omega }{10\Omega +4\Omega }\right)\\ =1.178\text{V}\end{array}$

Replace all the sources shown in Figure 1 and consider only the current source, then redraw the circuit.

The required diagram is shown in Figure 4

  

Apply nodal at the node $V_{a3}$ .

  $\begin{array}{l}\frac{V_{a3}}{10}+2+\frac{V_{a3}-V_{b3}}{4}=0\\ V_{a3}\left[\frac{1}{10}+\frac{1}{4}\right]-V_{b3}\left[\frac{1}{4}\right]=-2\\ V_{a3}\left[\frac{7}{20}\right]-V_{b3}\left[\frac{1}{4}\right]=-2\end{array}$   ....... (1)

Apply nodal at the node $V_{b3}$ .

  $\begin{array}{l}\frac{V_{b3}}{6}-I_1+\frac{V_{b3}}{4}+\frac{V_{b3}-V_{a3}}{4}=0\\ V_{a3}\left[-\frac{1}{4}\right]+V_{b3}\left[\frac{1}{6}+\frac{1}{6}+\frac{1}{4}\right]=2\\ V_{a3}\left[-\frac{1}{4}\right]+V_{b3}\left[\frac{7}{12}\right]=2\end{array}$   ....... (2)

From equation (1) and equation (2) the node voltages are,

  $\begin{array}{l}{V}_{a3}=-4.71\text{V}\\ V_{b3}=1.412\text{V}\end{array}$

The expression to calculate the node voltage $V_a$ is given by,

  $V_a=V_{a1}+V_{a2}+V_{a3}$

Substitute $0.824\text{V}$ for $V_{a1}$, $1.178\text{V}$ for $V_{a2}$ and $-4.71$ for $V_{a_3}$ in the above equation.

  $\begin{array}{c}{V}_a=0.824\text{V}+1.178\text{V}-4.71\\ =-2.708\text{V}\end{array}$

The expression to calculate the node voltage $V_b$ is given by,

  $V_b=V_{b1}+V_{b2}+V_{b3}$

Substitute $\frac{6}{17}\text{V}$ for $V_{b1}$, $1.65\text{V}$ for $V_{b2}$ and $1.412$ for $V_{b3}$ in the above equation.

  $\begin{array}{c}{V}_b=\frac{6}{17}\text{V}+1.65\text{V}+1.412\\ =3.415\text{V}\end{array}$

Conclusion:

Therefore, the value of node voltage $V_a$ is $-2.708\text{V}$ and $V_b$ is $3.415\text{V}$ .