Network Analysis and Synthesis, 1st Edition by Pearson

Kirchhoff’s Laws, Mesh and Nodal Analysis 35

By applying KVL in mesh II, the following equation is obtained:

−2I

2

− 1I

2

− 1I

2

− 2(I

2

− I

3

) + 5 = 0

or 6I

2

− 2I

3

= 5 (2.11)

By Applying KVL in mesh III, we obtain the following:

−1(I

3

− I

1

) − 2(I

3

− I

2

) − 2I

3

− 2I

3

−5 = 0

or −I

1

− 2I

2

+ 7I

3

= −5 (2.12)

Let us fi nd I

2

from these mesh equations by Cramer’s rule.

∆=

−

−

−−

=−+− +=

50 1

06 2

127

5424 0106 184() ()

Therefore,

∆=

−

−

−−

=−−−

−+

=+

22

55 1

05 2

157

53510502 10 5

125 10

()

()()

()

to find I

−−=

=

∆

∆

==

5 130

130

184

0 706

2

2

I .A

Therefore, the current through branch BC = 0.706 A.

Figure 2.11

2

Ω

1

Ω

5V

5V

5V

2 Ω

2 Ω

2 Ω

2 Ω2 Ω

1 Ω

1 Ω

A

HG F

E

B

CD

−+

Figure 2.12

Mesh III

Mesh I

Mesh II

2

Ω

1

Ω

5V

5V

5V