Use nodal analysis to find Vo in the network in Fig. P3.24.Figure P3.24

Apply Kirchhoff's current law at node \(V_{2}\) in Figure 1 and solve.

$$ \begin{aligned} &\frac{V_{2}-V_{o}}{1 \mathrm{k}}+\frac{V_{2}-V_{1}}{2 \mathrm{k}}+\frac{V_{2}-12}{2 \mathrm{k}}-\left(2 \times 10^{-3}\right)=0 \\ &2 V_{2}-2 V_{o}+V_{2}-V_{1}+V_{2}-12=\left(2 \times 10^{-3}\right)(2 \mathrm{k}) \\ &4 V_{2}-V_{1}-2 V_{o}=4+12 \\ &V_{1}=4 V_{2}-2 V_{o}-16 \ldots \ldots \text { (1) } \end{aligned} $$

Apply Kirchhoff's current law at node \(V_{1}\) in Figure 1 and solve.

$$ \begin{aligned} &\frac{V_{1}-V_{2}}{2 k}+\frac{V_{1}-V_{o}}{1 \mathrm{k}}+\frac{V_{1}-12}{1 \mathrm{k}}=0 \\ &V_{1}-V_{2}+2 V_{1}-2 V_{o}+2 V_{1}-24=0 \\ &5 V_{1}-V_{2}-2 V_{o}=24 \ldots \ldots (2)\end{aligned} $$

Substitute equation (1) in (2) and solve.

$$ \begin{aligned} &5\left(4 V_{2}-2 V_{o}-16\right)-V_{2}-2 V_{o}=24 \\ &20 V_{2}-10 V_{o}-80-V_{2}-2 V_{o}=24 \\ &19 V_{2}-12 V_{o}=104 \\ &V_{2}=\frac{12 V_{o}+104}{19} \ldots \ldots \text { (3) } \end{aligned} $$

Apply Kirchhoff's current law at node \(V_{o}\) in Figure 1 and solve.

$$ \begin{aligned} &\frac{V_{o}-V_{2}}{1 \mathrm{k}}+\frac{V_{o}-V_{1}}{1 \mathrm{k}}+\frac{V_{o}}{1 \mathrm{k}}=0 \\ &V_{o}-V_{2}+V_{o}-V_{1}+V_{o}=0 \\ &3 V_{o}-V_{2}-V_{1}=0 \ldots \ldots (4)\end{aligned} $$

Substitute equation ( 1\()\) and (3) in (4) and solve for \(V_{0}\).

$$ \begin{aligned} &3 V_{0}-\left(\frac{12 V_{o}+104}{19}\right)-\left(4\left(\frac{12 V_{o}+104}{19}\right)-2 V_{o}-16\right)=0 \\ &(19)\left(3 V_{o}\right)-12 V_{o}-104-48 V_{o}-416+(19)\left(2 V_{o}\right)+(16)(19)=0 \\ &57 V_{o}-12 V_{o}-104-48 V_{o}-416+38 V_{o}+304=0 \\ &95 V_{o}-60 V_{o}=520-304 \end{aligned} $$

Simplify the expression further.

$$ \begin{aligned} 35 V_{o} &=216 \\ V_{o} &=\frac{216}{35} \\ &=6.17 \mathrm{~V} \end{aligned} $$