Introduction to Electric Circuits

162 Nodal and mesh analysis

Solution

There are three meshes identified by the currents I~, I2 and 13. Applying KVL to
mesh 1 and taking the clockwise direction to be positive, we have

Vsl - R lI1- R214 = 0

The branch current/4 is I~ - I2, so

Vsl - RII1 - R2(I1 - I2) = 0

Vsl - R lll- R211 + R212 = 0

(R1 + R2)I~ - R212 = Vsl (7.23)

Note that we could have written down this equation immediately using the
three steps outlined above.
For mesh 2, which has two adjacent meshes, steps (1) and (2) give
(R 2 + R 3 4- R4)I 2 - R21~ - R413 for the left-hand side of the equation. There is
no voltage source in this mesh so the right-hand side is simply zero. The mesh
equation is therefore
-R211 + (R2 + R3 + R4)I2- R413 = 0 (7.24)

For mesh 3, which has only one adjacent mesh, the left-hand side of the
equation is
-R412 + (R4 + R5)/3

The voltage source is acting in the opposite direction to the mesh current so the
right-hand side of the equation is -Vs2. The mesh equation is therefore
-R,I2 + (R4 + R5)/3 = -Vs2 (7.25)
Putting in the values for the resistances and voltages. Equations (7.23), (7.24)
and (7.25) may be written in matrix form as

i<4+40 -40 (40 + 40 3 + 50) -50 0 i i j: 1 i 0001 =

0 -50 (50 + 5) 13 - 10

Using Cramer's rule I~ - A~/A,, Now

A - 44 -40 0

-40 93 -50

0 -50 55

= [(44 x 93 x 55) + (-40 x -50 x 0) + (0 x -40 x -50)]

-[(0 x 93 x 0) + (-40 x -40 x 55) + (44 x -50 x -50)]

= 225 060 - (88 000 + 110 000)

= 27 060

(7.26)