PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Alternating Current Analysis 267

Then,

I 12

120 0

30 90


 490








∠

∠

∠

I 23

120 120

30 90


 4 150








∠

∠

∠

I 31

120 120

30 90


 430








∠

∠

∠

Observe that the solution for part b is based in transforming the problem into one that

is similar to part a, and the currents just evaluated (part b) are one-third of the currents

of part a, because the loads become three times larger.

The line currents using KCL can be evaluated, and are indicated as follows:

I 1 = I 12 − I 31 + 4 √

__

3 ∠−120° A

I 2 = 4 √

__

3 ∠120° A

I 3 = 4 √

__

3 ∠0° A

The corresponding voltage drops are indicated as follows:

V 1 N = I 1 * Z = 4 √

__

3 ∠−120° * 10 ∠90° V

V 1 N = 40 √

__

3 ∠−30° V

V 2 N = 40 √

__

3 ∠−150° V

V 3 N = 40 √

__

3 ∠90° V

3.4 Examples

Example 3.1

Create the script fi le XL_XC that returns the following plots:

1. mag[XL(ω)] versus ω


2. mag[XC(ω)] versus ω
   for L = 2 H and C = 1 μF, over the frequency range 200 ≤ ω ≤ 2000 rad/s.

MATLAB Solution

% Script file: XL _ XC

L = 2;