Page 234 - Basic Electrical Engineering

2V − 4V + V + 12 = 0 (ii)
B
D
C
Again

or, V − V + 10 = V D
D
C
or, V − 2V + 10 = 0 (iii)
D
C
Solving Eqs. (i), (ii) and (iii), we get

(direction of I is opposite to that shown)
2

Again I = I + I = 0.5 + 0.02 = 0.52 A
5
7
6
again I = I + I = 1.0 + 0.04 = 1.04 A
3
2
1

I + I = I 3
4
5
I = I − I 5
4
3
= 0.04 − 0.02
= 0.02 A