Introduction to Electric Circuits

164 Nodal and mesh analysis

Solution

There are two meshes whose currents are I~ and 12, respectively, and we note
that the mesh current I~ is equal to the current source Is. Applying KVL to
mesh 2 and taking the anticlockwise direction to be positive, we have

V s + R3I 2 + R2I 3 -0

Since 13 = 12 - 11 then

V s +R3I 2 + R2I 2-R2I 1 =0

• R I,-
  = Rzls- Vs
  12 : (R2/s- Vs)/(R2 + R3)

Putting in the values we have

12 - [(2 • 1) - 6)]/(2 + 1) - (-4/3) A

The current through the resistor R2 is

13 - I2 - 11- I2 - Is- (-4/3) - 1 - -7//3 - -2.33 A

The negative sign indicates that the current is flowing in the opposite direction

to that shown in the circuit diagram (i.e. it flows downwards through the

resistor).

Supermeshes

We saw that in nodal voltage analysis two nodes having a voltage source

connected between them constitute a supernode. In mesh current analysis two

meshes which have a current source common to both of them form what is

known as a supermesh. As an example, consider the circuit of Fig. 7.13 which

has three meshes identified by the currents I1, 12 and 13. Meshes 2 and 3 have a

common current source Is and together they constitute a supermesh.

Figure 7.13