Introduction to Electric Circuits

166 Nodal and mesh analysis

/3 = Is + 12 (7.29)

Substituting for ~ from Equation (7.29) in Equation (7.27) we have

(R1 + R4)/1- RlI2- R4(Is + I2) = V

(R, + R4)I 1 - RlI 2 - R4I s - R412 = V

(R1 + R4)I 1 -(R 1 4- R4)I 2 = V 4- R4I s (7.30)

Substituting for 13 from Equation (7.29) in Equation (7.28) we have

(R, + R2)I2 + (R3 + R4)(Is + 12) - (R, + R4)I1 = 0

(R1 + R2)I2 + R3Is + R312 + R4Is + R412- (R~ + R4)I~ = 0

(R1 + R2 + R3 + R4)I2- (R1 + R4)I1 = -(R3 + R4)Is (7.31)

We now have two equations (7.30) and (7.31) from which I~ and 12 can be

determined. Once 12 is known 13 follows immediately from Equation (7.29).

Example 7.16

Determine (1) the mesh current I2 and (2) the voltage at node X in the circuit of

Fig. 7.14.

R1 = 6D.

Figure 7.14

@

Vsl = 12V

X R3 = 15~

i

R2 3~ Vs2 = 3V

Solution

There are three meshes in total but since there is a current source common to

meshes 1 and 2, these can be considered to be a supermesh. Applying the three-

step method to the supermesh:

1 we multiply the mesh current 11 by the resistance through which it flows

(R~) and add this to the product of the mesh current I2 and the resistance

through which it flows (R2);

2 there is one adjacent mesh whose current is 13. We multiply this by the

resistance (R2) common to the two meshes and subtract the product;

3 there is a voltage source (Vsl) which acts in the opposite direction to the

mesh current flowing through it so the right-hand side of the supermesh

equation is - Vs~