PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Alternating Current Analysis 273

C =

1.2500e-004

>> XC = -j/(W*C)

XC =

0 -16.0000I

>> Z = R+XC

Z =

20.0000 -16.0000I

>> Zmag = abs(Z);

>> phase = -angle(Z)*180/pi

phase =

38.6598

>> I = V/Zmag;

>> % Print results

>> disp(‘The peak value of the current (in amps) is =’), disp(I);

>> disp(‘with a phase angle(in degrees) of =’), disp(phase)

The peak value of the current (in amps) is =

7.8087

with a phase angle (in degrees) of =

38.6598

The meaning of the aforementioned results is that the current is given by

i(t) = 7. 8 0 8 7 c o s (5 0 0 t + 38.6598°) A

Example 3.6

Create the script fi le curr_volt that returns the plots of v(t) versus t and i(t) versus t, over

the range 0 ≤ t ≤ 0.06 s, for the circuit shown in Figure 3.54, where i(t) = 5 sin(300t) A.

FIGURE 3.54

Network of Example 3.6.

L = 60 mH

i(t)

v(t)

R = 10 Ω