Use nodal analysis to find Vo in the circuit in Fig. P3.25.Figure P3.25

The nodes \(V_{o}\) and \(V_{1}\) forms a super node \(S_{2}\) and the nodes \(V_{2}\) and \(V_{3}\) forms another super node \(S_{1}\) as shown in Figure 1

Write the voltage constraint equation for \(6 \mathrm{~V}\) voltage source.

$$ \begin{aligned} &V_{2}-V_{3}=6 \mathrm{~V} \\ &V_{3}=V_{2}-6 \mathrm{~V} \ldots \ldots(1) \end{aligned} $$

Write the voltage constraint equation for \(12 \mathrm{~V}\) voltage source.

$$ \begin{aligned} &V_{o}-V_{1}=12 \mathrm{~V} \\ &V_{1}=V_{o}-12 \mathrm{~V} \ldots . . \end{aligned} $$

Apply Kirchhoff's current law at super node \(S_{2}\).

$$ \begin{aligned} &2 \mathrm{~mA}+\frac{V_{o}}{1 \mathrm{k} \Omega}+\frac{V_{1}-V_{2}}{1 \mathrm{k} \Omega}+\frac{V_{1}}{1 \mathrm{k} \Omega}=0 \\ &\frac{V_{o}+V_{1}-V_{2}+V_{1}}{1 \mathrm{k} \Omega}=-2 \mathrm{~mA} \\ &V_{o}+2 V_{1}-V_{2}=-(2 \mathrm{~mA})(1 \mathrm{k} \Omega) \\ &V_{o}+2 V_{1}-V_{2}=-2 \mathrm{~V} \ldots \ldots(3) \end{aligned} $$

Apply Kirchhoff's current law at super node \(S_{1}\).

$$ \begin{aligned} &\frac{V_{2}-V_{4}}{2 \mathrm{k} \Omega}+\frac{V_{2}-V_{1}}{1 \mathrm{k} \Omega}-2 \mathrm{~mA}+4 \mathrm{~mA}+\frac{V_{3}}{1 \mathrm{k} \Omega}=0 \\ &\frac{V_{2}-V_{4}+2\left(V_{2}-V_{1}\right)+2 V_{3}}{2 \mathrm{k} \Omega}=2 \mathrm{~mA}-4 \mathrm{~mA} \\ &V_{2}-V_{4}+2 V_{2}-2 V_{1}+2 V_{3}=-(2 \mathrm{~mA})(2 \mathrm{k} \Omega) \\ &-2 V_{1}+3 V_{2}+2 V_{3}-V_{4}=-4 \mathrm{~V} \ldots \ldots \text { (4) } \end{aligned} $$

Apply Kirchhoff's current law at node \(V_{4}\).

$$ \begin{aligned} &\frac{V_{4}-V_{2}}{2 \mathrm{k} \Omega}+\frac{V_{4}}{2 \mathrm{k} \Omega}-4 \mathrm{~mA}=0 \\ &\frac{V_{4}-V_{2}+V_{4}}{2 \mathrm{k} \Omega}=4 \mathrm{~mA} \\ &-V_{2}+2 V_{4}=(4 \mathrm{~mA})(2 \mathrm{k} \Omega) \\ &2 V_{4}=V_{2}+8 \mathrm{~V} \\ &V_{4}=\frac{V_{2}}{2}+4 \mathrm{~V} \ldots \ldots \text { (5) } \end{aligned} $$

Recall equation (4).

$$ -2 V_{1}+3 V_{2}+2 V_{3}-V_{4}=-4 \mathrm{~V} $$

Substitute \(V_{o}-12 \mathrm{~V}\) for \(V_{1}, V_{2}-6 \mathrm{~V}\) for \(V_{3}\) and \(\frac{V_{2}}{2}+4 \mathrm{~V}\) for \(V_{4}\)

$$ \begin{aligned} &-2\left(V_{o}-12 \mathrm{~V}\right)+3 V_{2}+2\left(V_{2}-6 \mathrm{~V}\right)-\left(\frac{V_{2}}{2}+4 \mathrm{~V}\right)=-4 \mathrm{~V} \\ &-2 V_{o}+24 \mathrm{~V}+3 V_{2}+2 V_{2}-12 \mathrm{~V}-\frac{V_{2}}{2}-4 \mathrm{~V}=-4 \mathrm{~V} \\ &-2 V_{o}+5 V_{2}-\frac{V_{2}}{2}+8 \mathrm{~V}=-4 \mathrm{~V} \\ &\frac{-4 V_{o}+10 V_{2}-V_{2}}{2}=-12 \mathrm{~V} \\ &-4 V_{o}+9 V_{2}=-24 \mathrm{~V} \ldots \ldots(6) \end{aligned} $$

Recall equation (3).

$$ V_{o}+2 V_{1}-V_{2}=-2 \mathrm{~V} $$

Substitute \(V_{o}-12 \mathrm{~V}\) for \(V_{1}\)

$$ \begin{aligned} &V_{o}+2\left(V_{o}-12 \mathrm{~V}\right)-V_{2}=-2 \mathrm{~V} \\ &V_{o}+2 V_{o}-24 \mathrm{~V}-V_{2}=-2 \mathrm{~V} \\ &3 V_{o}-V_{2}=-2 \mathrm{~V}+24 \mathrm{~V} \\ &3 V_{o}-V_{2}=22 \mathrm{~V} \\ &V_{2}=3 V_{o}-22 \mathrm{~V} \ldots \ldots(7) \end{aligned} $$

Recall equation (6).

$$ -4 V_{o}+9 V_{2}=-24 \mathrm{~V} $$

Substitute \(3 V_{o}-22 \mathrm{~V}\) for \(V_{2}\).

$$ \begin{aligned} &-4 V_{o}+9\left(3 V_{o}-22 \mathrm{~V}\right)=-24 \mathrm{~V} \\ &-4 V_{o}+27 V_{o}-198 \mathrm{~V}=-24 \mathrm{~V} \\ &23 V_{o}=-24 \mathrm{~V}+198 \mathrm{~V} \\ &V_{o}=\frac{174 \mathrm{~V}}{23} \\ &=7.56 \mathrm{~V} \end{aligned} $$

Therefore, the value of the voltage \(V_{o}\) in the circuit is \(7.56 \mathrm{~V}\).