Introduction to Electric Circuits

7.4 Mesh current analysis 163

To find A 1 we replace column 1 of A by the column vector on the right-hand side
of the matrix equation. Thus

A 1 = 200 -40 0

0 93 -50

-210 -50 55

= [(200 X 93 X 55) + (-40 X -50 X -210) + (0 X 0 X -50)]

-[(0 X 93 X -210) + (-40 X 0 X 55) + (200 X -50 • -50)]

= (603 000) - (500 000)

= 103 000

11 = 103 000/27 060 - 3.81 A.

To find A 3 we replace the third column of A by the column vector giving

A 3 -- 44 -40 200

-40 93 0

0 -50 -210

= [(44 X 93 X -210) + (-40 x -0 x 0) + (200 x -40 X -50)]

-[(200 X 93 • 0) + (-40 X -40 X -210) + (44 X 0 X -50)]

= (-859 320 + 400 000) - (-336 000)

= - 123 320

13 - A3//A - -123 320//27 060 - -4.56 A

The minus sign indicates that the current is flowing in the opposite direction to

that shown in the circuit diagram (i.e. it is flowing out of the positive terminal of

the voltage source).

Circuits containing voltage and current sources

If the current source is located in one mesh only, then that mesh current is the

source current and the analysis is quite straightforward, as shown by the

following example.

Example 7.15

Determine the current through the resistor R2 in the circuit of Fig. 7.12.

Figure 7.12

,s: 1A l t

R1 = 3~ R3 = 1

l , t I t----

R2 2fl

t,3

O Vs

J

6V