Introduction to Electric Circuits

154 Nodal and mesh analysis

To determine A~, we replace column 1 by the column vector on the right-hand

side of Equation (7.10). Thus

52 -O.5

m 1 ---

42 0.74

= (52 • 0.74) - (-0.5 • 42) = 38.48 + 21 = 59.48

Therefore

V2- 59.48//0.342- 173.9 V

The potential difference across R2 is

V2- V5 = 173.9- 0- 173.9 V

Again using Cramer's rule, V3 = A2/A. Now to find A2 we replace column 2

in A with the column vector on the right-hand side of Equation (7.10). Thus

0.8 52

A 2 :

-0.5 42

= (0.8 X 42) - (52 X -0.5) = 33.6 + 26 = 59.6

Therefore

V3 = 59.6/0.342 = 174.3 V

The current supplied by the voltage source Vs2 is given by

Is2 = (V4- V3)/R5
-- (Vs2- V3)/R 5
= (220- 174.3)/5
= 9.14 A

Application to reactive a.c. circuits

The examples shown so far have been of purely resistive circuits. For a.c.

circuits containing reactance the method applies equally but we must use

complex impedances.

Example 7.11

Determine the potential differences across the admittance Y2 in the circuit of

Fig. 7.5. Shown overleaf.

Solution

The nodes are identified as 1, 2, 3, and 4, their voltages being V~, V2, V3 and V4.
Let node 4 be the reference node so that V4 = 0. Also