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