Question 8.56: A load consisting of a 1-kΩ resistor in series with a 12-µF ......
A load consisting of a 1-kΩ resistor in series with a 12-µF capacitor is connected across a voltage source v_{\mathrm{S}}(t) = 150 cos (377t) V. Find the phasor voltage, current, and average power delivered to the load.
In this case, the load voltage equals the source voltage, because the load is connected in parallel with the source. The solution is presented in the following MATLAB code.
Script File
clear all
w = 377;
Vs = 150;
R = 1e3;
C = 12e-6;
ZC = 1/j/w/C;
% Find the load impedance
ZL = R+ZC;
% Find the load current
IL = Vs/ZL
MagIL = abs(IL)
PhaseIL = angle(IL)*180/pi
% Compute the average power
PLavg = real(ZL)/2*MagIL^2
IL =
143.0124e-003 + 31.6119e-003i
MagIL =
146.4645e-003
PhaseIL =
12.4644e+000
PLavg =
10.7259e+000
\begin{aligned}& \mathrm{V}_{\mathrm{L}}=150+j 0 \mathrm{~V}=150 ~\angle ~0^{\circ} \mathrm{V} .\\ \\& \mathrm{I}_{\mathrm{L}}=143.01+j 31.61 \mathrm{~mA}=146.46~ \angle ~12.46^{\circ} \mathrm{mA} . \\\\& P=10.7259 \mathrm{~W} .\end{aligned}