258 Practical MATLAB® Applications for Engineers
Now consider t he AC source by set t i ng t he DC source to zero (VA = 0 ), and transforming
the source VB into a phasor, the equivalent circuit is redrawn in Figure 3.36.
Solving for the loop currents IB1 and IB3 of Figure 3.36, using loop equations the
following relations are obtained
IB 3 = 4 ______∠0°
i + j
= ________^4 ∠0°
√
__
2 ∠45°
= 2 √
__
2 ∠–45° A
IB 1 = IB 2 =
IB 3
___
2
Then
IB 1 = √
__
2 ∠−45°
IB 2 = IB 1 = √
__
2 ∠−45°
Transforming the aforementioned phasor equations into the instantaneous currents
results in
iB 1 (t) = √
__
2 cos(10 t − 45°) A
iB 2 (t) = √
__
2 cos(10 t − 45°) A
iB 3 (t) = 2 √
__
2 cos(10 t − 45°) A
VA = 10 V
IA3
IA2
L = 100 mH
2 Ω
2 Ω
IA1
+
−
FIGURE 3.35
Network of Figure 3.34 with VB = 0.
IB3
IB2
j Ω
2 Ω
2 Ω
IB3 +
VB = 4
−
0 ° V
FIGURE 3.36
Network of Figure 3.34 with VA = 0.