Page 147 - Basic Electrical Engineering

Figure 2.40

Solution:

Now applying Kirchhoff’s current law to nodes A and B we have

I = I + I S (i)
2
1

I + I =
2
3
and (ii)
also voltage of node B = V =16 V
0

Voltage across AC + voltage across AB = voltage at node B.

V + 4I = 16 V (iii)
2
1

(iv)

Solving eq (i), (ii), (iii), and (iv) we have

V = 12 V I = 2 A I = 1 A I = I − I = 2 − 1 = 1 A
2
S
1
1
1
2
Therefore, I = 1 A I = − I = 2 A.
3
3
2