Chapter 3, Problem 3.35HP

To determine

(a)

The number of non reference nodes.

Answer to Problem 3.35HP

The number of non reference nodes is $7$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The non reference node is anode that is connected between a voltage node and any other element of the circuit.

From the above figure it is clear that there are in total $7$ non-reference node in the circuit which are $v_1$, ${v^{\prime }}_1,{v^{\prime }}_n,v_2,{v^{\prime }}_2,{v^{\prime }}_3$ and $v_3$.

Conclusion:

Therefore, the number of non-reference nodes is $7$ .

To determine

(b)

The number of unknown node voltage.

Answer to Problem 3.35HP

The number of unknown node voltages is $4$ .

Explanation of Solution

Calculation:

From the figure shown in Figure 1, the number of unknown node voltages are ${v^{\prime }}_1$, ${v^{\prime }}_n$, ${v^{\prime }}_3$ and ${v^{\prime }}_2$ .

Conclusion:

Therefore, the number of unknown node voltages is $4$ .

To determine

(c)

The value of ${v^{\prime }}_1,{v^{\prime }}_2,{v^{\prime }}_3$ and ${v^{\prime }}_n$ .

Answer to Problem 3.35HP

The value of voltage ${v^{\prime }}_n$ is $-7.234\text{V}$, $v_{R_{w1}}$ is $122.28\text{V}$. ${v^{\prime }}_2$ is $-\text{132}\text{.02}\text{V}$, $v_{R_{w3}}$ for $9.738\text{V}$ and ${v^{\prime }}_3$ is $-\text{160}\text{.26}\text{V}$ .

Explanation of Solution

Calculation:

Mark the values and redraw the circuit.

The required diagram is shown in Figure 2

  

Apply nodal at ${v^{\prime }}_n$ .

  $\begin{array}{l}\frac{{v^{\prime }}_n-170}{1.9+0.7}+\frac{{v^{\prime }}_n+170}{2.3+0.7}+\frac{{v^{\prime }}_n+170}{11+0.7}=0\\ {v^{\prime }}_n\left[\frac{1}{2.6}+\frac{1}{3}+\frac{1}{11.7}\right]=\frac{170}{2.6}-\frac{170}{3}-\frac{170}{11.7}\\ {v^{\prime }}_n=-7.234\text{V}\end{array}$

The expression for the voltage across the resistance $R_{w1}$ is given by,

  $v_{R_{w1}}=\left({v^{\prime }}_n-170\right)\left(\frac{0.7}{0.7+1.9}\right)$

Substitute $-7.234\text{V}$ for ${v^{\prime }}_n$ in the above equation.

  $\begin{array}{c}{v}_{R_{w1}}=\left(-7.234\text{V}-170\right)\left(\frac{0.7}{0.7+1.9}\right)\\ =-47.72\text{V}\end{array}$

The expression to calculate the voltage ${v^{\prime }}_1$ is given by,

  ${v^{\prime }}_1=v_{R_{w1}}+170\text{V}$

Substitute $-47.72\text{V}$ for $v_{R_{w1}}$ in the above equation.

  $\begin{array}{c}{v^{\prime }}_1=-47.72\text{V}+170\text{V}\\ =122.28\text{V}\end{array}$

The voltage across the resistance $v_{R_{w2}}$ is given by,

  $v_{R_{w2}}=\left({v^{\prime }}_n+170\right)\left(\frac{0.7}{0.7+2.3}\right)$

Substitute $-7.234\text{V}$ for ${v^{\prime }}_n$ in the above equation.

  $\begin{array}{c}{v}_{R_{w2}}=\left(-7.234\text{V}+170\Omega \right)\left(\frac{0.7\Omega }{0.7\Omega +2.3\Omega }\right)\\ =\left(162.77\text{V}\right)\left(\frac{7}{30}\right)\\ =37.98\text{V}\end{array}$

The expression for the voltage ${v^{\prime }}_2$ is given by,

  ${v^{\prime }}_2=v_{R_{w2}}-170\text{V}$

Substitute $37.98\text{V}$ for $v_{R_{w2}}$ in the above equation.

  $\begin{array}{c}{v^{\prime }}_2=37.98\text{V}-170\text{V}\\ =-\text{132}\text{.02}\text{V}\end{array}$

The voltage across the resistance $v_{R_{w3}}$ is given by,

  $v_{R_{w3}}=\left({v^{\prime }}_n+170\text{V}\right)\left(\frac{0.7\Omega }{0.7\Omega +2.3\Omega }\right)$

Substitute $-7.234\text{V}$ for ${v^{\prime }}_n$ in the above equation.

  $\begin{array}{c}{v}_{R_{w3}}=\left(-7.234\text{V}+170\text{V}\right)\left(\frac{0.7\Omega }{0.7\Omega +2.3\Omega }\right)\\ =9.738\text{V}\end{array}$

The expression to calculate the voltage ${v^{\prime }}_3$ is given by,

  ${v^{\prime }}_3=v_{R_{w3}}-170\text{V}$

Substitute $9.738\text{V}$ for $v_{R_{w3}}$ in the above equation.

  $\begin{array}{c}{v^{\prime }}_3=9.738\text{V}-170\text{V}\\ =-\text{160}\text{.26}\text{V}\end{array}$

Conclusion:

Therefore, the value of voltage ${v^{\prime }}_n$ is $-7.234\text{V}$, $v_{R_{w1}}$ is $122.28\text{V}$. ${v^{\prime }}_2$ is $-\text{132}\text{.02}\text{V}$, $v_{R_{w3}}$ for $9.738\text{V}$ and ${v^{\prime }}_3$ is $-\text{160}\text{.26}\text{V}$ .