Calculate the node voltages in the circuit shown in Fig. 3.3(a).

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Consider Fig. 3.3 (b), where the circuit in Fig. 3.3 (a) has been prepared for nodal analysis. Notice how the currents are selected for the application of KCL. Except for the branches with current sources, the labeling of the currents is arbitrary but consistent. (By consistent, we mean that if, for example, we assume that [latex]i_{2}[/latex] enters the 4Ω resistor from the left-hand side, [latex]i_{2}[/latex] must leave the resistor from the right-hand side.) The reference node is selected, and the node voltages [latex]v_{1}[/latex] and [latex]v_{2}[/latex] are now to be determined.
At node 1, applying KCL and Ohm’s law gives

[latex]i_{1}=i_{2}+i_{3} \quad \Longrightarrow \quad 5=\frac{v_{1}-v_{2}}{4}+\frac{v_{1}-0}{2}[/latex]

Multiplying each term in the last equation by 4, we obtain

[latex]20=v_{1}-v_{2}+2 v_{1}[/latex]

or

[latex]3 v_{1}-v_{2}=20[/latex] (3.1.1)

At node 2, we do the same thing and get

[latex]i_{2}+i_{4}=i_{1}+i_{5} \quad \Longrightarrow \quad \frac{v_{1}-v_{2}}{4}+10=5+\frac{v_{2}-0}{6}[/latex]

Multiplying each term by 12 results in

[latex]3 v_{1}-3 v_{2}+120=60+2 v_{2}[/latex]

or

[latex]-3 v_{1}+5 v_{2}=60[/latex] (3.1.2)

Now we have two simultaneous Eqs. (3.1.1) and (3.1.2). We can solve the equations using any method and obtain the values of [latex]v_{1}[/latex] and [latex]v_{2}[/latex].

METHOD 1 Using the elimination technique, we add Eqs. (3.1.1) and (3.1.2).

[latex]4 v_{2}=80 \quad \Longrightarrow \quad v_{2}=20 V[/latex]

Substituting [latex]v_{2}=20[/latex] in Eq. (3.1.1) gives

[latex]3 v_{1}-20=20 \quad \Longrightarrow \quad v_{1}=\frac{40}{3}=13.33 V[/latex]

METHOD 2 To use Cramer’s rule, we need to put Eqs. (3.1.1) and (3.1.2) in matrix form as

[latex]\left[\begin{array}{rr}3 & -1 \-3 & 5\end{array} ight]\left[\begin{array}{l}v_{1} \v_{2}\end{array} ight]=\left[\begin{array}{l}20 \60\end{array} ight][/latex] (3.1.3)

The determinant of the matrix is

[latex]\Delta=\left|\begin{array}{rr}3 & -1 \-3 & 5\end{array} ight|=15-3=12[/latex]

We now obtain [latex]v_{1}[/latex] and [latex]v_{2}[/latex] as

[latex]v_{1}=\frac{\Delta_{1}}{\Delta}=\frac{\left|\begin{array}{rr}20 & -1 \60 & 5\end{array} ight|}{\Delta}=\frac{100+60}{12}=13.33 V[/latex]
[latex]v_{2}=\frac{\Delta_{2}}{\Delta}=\frac{\left|\begin{array}{rr}3 & 20 \-3 & 60\end{array} ight|}{\Delta}=\frac{180+60}{12}=20 V[/latex]

giving us the same result as did the elimination method.
If we need the currents, we can easily calculate them from the values of the nodal voltages.

[latex]i_{1}=5 A , \quad i_{2}=\frac{v_{1}-v_{2}}{4}=-1.6667 A , \quad i_{3}=\frac{v_{1}}{2}=6.666[/latex]
[latex]i_{4}=10 A , \quad i_{5}=\frac{v_{2}}{6}=3.333 A[/latex]

The fact that [latex]i_{2}[/latex] is negative shows that the current flows in the direction opposite to the one assumed.

Question 3.1: Calculate the node voltages in the circuit shown in Fig. 3.3...

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