Network Analysis and Synthesis, 1st Edition by Pearson

Kirchhoff’s Laws, Mesh and Nodal Analysis 65

From mesh I, it is clear that i

1

= 1 A; and

from mesh III, i

3

= −3 A.

By applying KVL in mesh II, that is, by

equating voltage drops to voltage rise, we get

21

4

21 23

()

()

ii ii

−+ −=

or

−−=

2+3ii

123

4

i

By substituting the value of i

1

and i

3

in the

equation, we get

−+−− =

21

334

2

() ()i

or

−+ +=

23

34

2

i

or 33 1

22

ii==

,. ,ie

.A

Therefore, the current through 1 Ω resistor

=−ii

23

=−=

134

−

() A

Example 2.31 Find the current through 8 Ω resistor

in the network shown in Figure 2.59 using nodal

analysis.

Solution: Let node voltages be V

1

and V

2

as shown

in Figure 2.60.

By applying KCL at node 1, we write

I

1

= I

2

+ I

3

and I

3