PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

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.