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

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 i_{2}   enters the 4-Ω resistor from the left-hand side, i_{2}  must leave the resistor from the right-hand side.) The reference node is selected, and the node voltages v_{1}   and  v_{2}   are now to be determined.

At node 1, applying KCL and Ohm’s law gives

i_{1} = i_{2} + i_{3}  ⇒   5= \frac{v_{1} – v_{2}}{4} + \frac{v_{1} – 0 }{2}

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

20 = v_{1} – v_{2} + 2v_{1}

or

3v_{1} –  v_{2} = 20                      (3.1.1)

At node 2, we do the same thing and get

i_{2} + i_{4} = i_{1} + i_{5}    ⇒   \frac{v_{1} -v_{2}}{4} + 10 = 5 + \frac{v_{2} – 0 }{6}

Multiplying each term by 12 results in

3v_{1} – 3v_{2} + 120 = 60 + 2v_{2}

or

-3v_{1} + 5v_{2} = 60                       (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 v_{1}  and  v_{2}  

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

4 v_{2} = 80        ⇒    v_{2} = 20 V

Substituting v_{2} =20  in Eq. (3.1.1) gives

3v_{1} – 20 = 20        ⇒     v_{1} = \frac{40}{3}= 13.333 V

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

\begin{bmatrix} 3 & -1 \\ -3 & 5\end{bmatrix} \begin{bmatrix}  v_{1} \\ v_{2}\end{bmatrix}=\begin{bmatrix} 20 \\ 60\end{bmatrix}                        (3.1.3)

The determinant of the matrix is

Δ = \begin{vmatrix} 3 & -1 \\ -3 & 5 \end{vmatrix} =15 -3 = 12

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

v_{1} = \frac{Δ_{1}}{Δ}=\frac{\begin{vmatrix} 20 & -1 \\ 60  & 5 \end{vmatrix} }{Δ}=\frac{100 + 60}{12} = 13.333 V

v_{2} = \frac{Δ_{2}}{Δ}=\frac{\begin{vmatrix} 3 & 20 \\ -3 & 60 \end{vmatrix} }{Δ}=\frac{180 + 60}{12} = 20 V

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.

i_{1} = 5 A      i_{2} = \frac{v_{1} – v_{2}}{4}= -1.6668 A ,      i_{3} = \frac{v_{1}}{2} = 6.666 A

i_{4} = 10 A ,          i_{5} = \frac{v_{2}}{6}=3.333 A

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