Alternating Current Analysis
271
>> imp _ Z
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Magnitude of Z (in Ohms) and phase (in degrees) is given by :
Zmag =
3.6761
Phase _ deg =
-36.0274
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Example 3.4
Evaluate by hand and by using MATLAB the voltage v(t) across the series RL circuit
shown in Figure 3.52, if the current is i(t) = 5 cos(100t) A.
ANALYTICAL Solution
I = 5 ∠0°
Z = 40 + j * 100 * 0.3 = 40 + j30
V = Z * I = (40 + j30) * 5 ∠0° = 250 ∠(36.52°)
then
v(t) = 250 cos(100t + 36.87°)
MATLAB Solution
% Script file: vol _ RL
W=100;
Z=40+j*W*.3;
V=5*Z;
Vmax=abs(V);
Phase=angle(V);
Phasedegree=Phase*180/pi;
% Print results
disp(‘^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^’)
disp(‘The peak value (in volts) and the phase (in degrees) of v(t)
are=’);
Vmax,
Phasedegree,
disp(‘^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^’)
FIGURE 3.52
Network of Example 3.4.
R 1 = 40 Ω
L = 0.3 H
v (t )
i(t )
CRC_47760_CH003.indd 271
CRC_47760_CH003.indd 271
7/23/2008 1:27:44 PM
7/23/2008 1:27:44 PM