For the circuit of Fig. 4.46, determine all four nodal voltages.FIGURE 4.46

The given circuit is labeled and redrawn as

From the above circuit,

$$ v_{2}=5 \mathrm{~V} $$

Here we consider the nodes 1 and 3 super node. One current source and two resistors are connected.

Applying KCL at super node

$$ \begin{aligned} &\frac{v_{1}-v_{2}}{10}+2+\frac{v_{3}}{1}=0 \\ &\frac{v_{1}-v_{2}+20+10 v_{3}}{10}=0 \\ &v_{1}-v_{2}+10 v_{3}=-20 \end{aligned} $$

Substitute value of \(v_{2}\) in the above equation, we get

$$ \begin{aligned} &v_{1}-5+10 v_{3}=-20 \\ &v_{1}+10 v_{3}=-15 \ldots \ldots \ldots(1) \end{aligned} $$

Applying \(\mathrm{KCL}\) at node \(v_{4}\), we get

$$ \begin{aligned} &-2+\frac{v_{4}-v_{2}}{4}+\frac{v_{4}}{2}=0 \\ &\frac{v_{4}-v_{2}+2 v_{4}}{4}=2 \\ &3 v_{4}-v_{2}=8 \end{aligned} $$

Substitute \(v_{2}\) in the above equation,

$$ \begin{aligned} &3 v_{4}-5=8 \\ &3 v_{4}=13 \\ &\therefore v_{4}=4.333 \mathrm{~V} \end{aligned} $$

\(6 \mathrm{~V}\) voltage source is super node in between nodes 1 and 3 and its equation is,

$$ \begin{aligned} &v_{1}-v_{3}=6 \\ &v_{1}=6+v_{3} \ldots \ldots \ldots(2) \end{aligned} $$

Substitute \(v_{1}\) in equation (1) then

$$ \begin{aligned} &6+v_{3}+10 v_{3}=-15 \\ &6+11 v_{3}=-15 \\ &11 v_{3}=-21 \\ &\therefore v_{3}=-1.909 \mathrm{~V} \end{aligned} $$

Substitute obtained value of \(v_{3}\) in equation (2) then we get

$$ \begin{aligned} &v_{1}-(-1.909)=6 \\ &\therefore v_{1}=4.091 \mathrm{~V} \end{aligned} $$