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:
- mag[XL(ω)] versus ω
- 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;