A load consisting of a 50-Ω resistor in parallel with a 8-µF capacitor is connected across a current source delivering i_{\mathrm{S}}(t) = 50 cos (2500t) mA. Find the average power delivered to the load.
The load receives the entire current from the source, so we know the magnitude of the load current equals the magnitude of the source current. We can combine the parallel impedances to find the equivalent impedance and then use the real part to compute the average power delivered to the load.
clear all
IL = 50e-3;
w = 2500;
R = 50;
C = 8e-6;
ZC = 1/j/w/C;
ZL = 1/(1/R+1/ZC);
PLavg = real(ZL)/2*IL^2
PLavg =
31.2500e-003
P = 31.25 mW.