Chapter 3, Problem 1P

Use nodal analysis to find $V_1$ in the circuit in Fig. P3.l.

To determine

The value of voltage $V_1$ by use of nodal analysis.

Answer to Problem 1P

The value of voltage $V_1$ is $35\text{V}$.

Explanation of Solution

Calculation:

The circuit diagram is redrawn as shown in Figure 1.

  

Write the expression for KCL at node $A$ in Figure 1.

  $-i_1+\frac{V_1}{R_2}+\frac{V_1-v_B}{R_3}=0$   ..... (1)

Here, $i_1$ is the current in the circuit, $V_1$ is the voltage at node $A$, $v_B$ is the voltage at node $B$ and $R_2$, $R_3$ are the resistances in the circuit.

Substitute $12\text{mA}$ for $i_1$, $5\text{k}\Omega$ for $R_2$ and $5\text{k}\Omega$ for $R_3$ in equation (1).

  $\begin{array}{c}-12\text{mA}+\frac{V_1}{5\text{k}\Omega }+\frac{V_1-v_B}{5\text{k}\Omega }=0\\ \frac{\left(5\text{k}\Omega \right)\left(-12\text{mA}\right)+V_1+V_1-v_B}{5\text{k}\Omega }=0\\ \frac{\left(5\times {10}^3\Omega \right)\left(-12\times {10}^{-3}\text{A}\right)+2V_1-v_B}{5\text{k}\Omega }=0\\ -60\text{V}+2V_1-v_B=0\end{array}$

Rearrange further.

  $2V_1-v_B=60\text{V}$   ..... (2)

Write the expression for KCL at node $B$ in Figure 1.

  $\frac{v_B-V_1}{R_3}+\frac{v_B}{R_4}+\frac{v_B}{R_5}=0$   ..... (3)

Here, $R_4$ and $R_5$ are the resistances in the circuit.

Substitute $5\text{k}\Omega$ for $R_3$, $4\text{k}\Omega$ for $R_4$ and $4\text{k}\Omega$ for $R_5$ in equation (3).

  $\begin{array}{c}\frac{v_B-V_1}{5\text{k}\Omega }+\frac{v_B}{4\text{k}\Omega }+\frac{v_B}{4\text{k}\Omega }=0\\ \frac{4\left(v_B-V_1\right)+5v_B+5v_B}{20\text{k}\Omega }=0\\ 4v_B-4V_1+10v_B=0\\ -4V_1+14v_B=0\end{array}$

Rearrange for $v_B$.

  $v_B=\frac{4V_1}{14}$

Simplify further.

  $v_B=0.286V_1$

Substitute $0.286V_1$ for $v_B$ in equation (2).

  $\begin{array}{c}2V_1-0.286V_1=60\text{V}\\ \text{1}\text{.714}{V}_1=60\text{V}\\ V_1=\frac{60\text{V}}{\text{1}\text{.714}}\\ =35\text{V}\end{array}$

Conclusion:

Thus, the value of voltage $V_1$ is $35\text{V}$.