PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

250 Practical MATLAB® Applications for Engineers

The calculations for the currents I 1 , I 2 , and I 3 are verifi ed using MATLAB as follows:

>>Z = [8 – 2j – 3 0; –3 8 + 5j – 5; 0 – 5 7 – 2j]

Z =

8.0000 – 2.0000i –3.0000 0

–3.0000 8.0000 + 5.0000i –5.0000

0 –5.0000 7.0000 – 2.0000i

>>V = [10;0;0]

V =

10

0

0

>>I = inv(z)*V

I =

1.4158 + 0.2159i

0.5861 – 0.3682i

0.4565 – 0.1326i

>>Current _ magnitude = abs(I)

Current _ magnitude =

1.4322

0.6922

0.4754

>>Phase _ angle = angle(I)*180/pi

Phase _ angle =

8.6689

–32.1402

–16.1948

Then the instantaneous currents are

i 1 (t) = 1.43 sin(10t + 8.7°) A

i 2 (t) = 0.693 sin(10t − 32.2°) A

i 3 (t) = 0.476 sin(10t − 16.2°) A

R.3.58 The node equations technique presented in Chapter 2 for the case of the resistive
DC networks can be extended to include AC elements (in phasor form), following
the same steps outlined for the AC loop equations.
The following example illustrates the general approach for the circuit diagram
shown in Figure 3.23. Note that this circuit has three nodes where one is grounded
(reference). Then only two node equations are required, labeled E 1 and E 2.

ANALYTICAL Solution

Observe that the source frequency is ω = 10 rad/s. Then replacing all the elements by

their respective admittances the following is obtained

R 1 = 1/10 and Y 1 = 10

R 2 = 1/3 and Y 2 = 3

XjLj j Y

j

LL

 
 
 j



 () 


()

10

1

50

1

5

1

15

5