Introduction to Electric Circuits

7.4 Mesh current analysis 165

Applying KVL to mesh 1 and taking the clockwise direction to be positive, we
have

V- R1/4- R4/5 = 0

The branch current/4 is Ix - I2 and the branch current 15 is 11 -- 13. Therefore

V- R~(I 1 - I2) - R4(I1- 13) = 0

V- RlI1 + RlI2 - R411 + R413 = 0

Rearranging,

(R1 + R4)I1- RlI2- R413 = V (7.27)

Note again that:

1 the mesh current is multiplied by the sum of the resistances around the

mesh;

2 there are two adjacent meshes and for each of these we subtract the product
of the mesh current and the resistance common to it and mesh 1;

3 the right-hand side of the equation is the voltage source in the mesh and is
positive because it acts in the same direction as the mesh current.
Applying KVL to the supermesh and taking the clockwise direction to be
positive, we have

Rl14 - R212 - R313 + R415 = 0
R~(I~ - 12) - R2I 2 - R3I 3 + R4(I1 - ~) = 0
R~I~ - RlI2 - R212 - R313 + R4I~ - R413 = 0
(R1 + R4)I~ - (R1 + R2)I2 - (R3 + R4)13 = 0

Multiplying throughout by -1 and rearranging,

(R1 -Jr- R2)I2 + (R3 + R4)h - (R1 -t- R4)I1 -- 0 (7.28)

Note that in this case:

1 the supermesh current I2 is multiplied by the total resistance through which
it flows and the supermesh current 13 is multiplied by the total resistance
through which it flows and these products are added;

2 there is one adjacent mesh and its current is multiplied by the total
resistance through which it flows and this product is subtracted;

3 there is no voltage source in the supermesh, so that the right-hand side of
the equation is zero.
The 'three-step' method could thus have been used to write down Equations
(7.27) and (7.28) immediately without recourse to Kirchhoff's voltage law. We
see that the current in the branch common to meshes 2 and 3 is Is (=I3 - 12),
so