0195136047.pdf

78 CIRCUIT ANALYSIS TECHNIQUES

33 I 1 − 20 I 2 − 8 I 3 = 10

− 20 I 1 + 55 I 2 − 10 I 3 = 0

− 8 I 1 − 10 I 2 + 22 I 3 = 20

STEP 4:Simultaneously solve the independent equations for the unknown mesh currents by

Gauss elimination or Cramer’s rule.

In this example the solution yields

I 1 = 1 .132 A; I 2 = 0 .711 A; I 3 = 1 .645 A

The current through the 10-V source isI 1 = 1 .132 A, which is the same as in Example 2.2.1. The

voltage across the 10-resistor isVBC= 10 (I 2 −I 3 )= 10 ( 0. 711 − 1. 645 )=− 9 .34 V, which

is the same as in Example 2.2.1.

Looking at Examples 2.2.1 and 2.2.3, it can be seen that there is no specific advantage

for either method since the number of equations needed for the solution is three in either case.

Such may not be the case in a number of other problems, in which case one should choose

judiciously the more convenient method, usually with the lower number of equations to be

solved.

The mesh-current method deals routinely with voltage sources. When we have current

sources with shunt conductances, the source-transformation technique may be used effectively to

convert the current source to a voltage source. However, in cases where we haveconstrained

meshes, that is, the two mesh currents are constrained by a current source, the concept of

asupermeshbecomes useful for the circuit analysis, as shown in the following illustrative

example.

EXAMPLE 2.2.4

For the network shown in Figure E2.2.4, find the current delivered by the 10-V source and the

voltage across the 3-resistor by means of mesh-current analysis.

1 Ω

3 Ω

2 Ω 4 Ω

+

+−

−

5 A 10 V

Supermesh

I 1 I 2

I 3

Vx =? Figure E2.2.4

Solution

Note that we cannot express the voltage across the current source in terms of the mesh currentsI 1

andI 2. The current source does, however,constrainthe mesh currents by the following equation:

I 2 −I 1 = 5