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Question 8.23: A current source delivering i(t) = 240 cos (500t) mA is conn......

A current source delivering i(t) = 240 cos (500t) mA is connected across a parallel combination of a 1-kΩ resistor and a 2-µF capacitor. Find the steady-state current i_R(t) through the resistor and the steady-state current i_C(t) through the capacitor. Draw a phasor diagram showing \rm{I}, \rm{I}_{\mathrm{C}} and \rm{I}_{\mathrm{R}}.

Step-by-Step
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This problem can be solved by converting to the phasor domain, using current division, and then
converting back to the time domain. The following MATLAB code completes the calculations.

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Script File

clear all
w = 500;
R = 1e3;
C = 2e-6;
% Create the source phasor
Is = 240e-3*exp(0*j*pi/180);
% Create the capacitor impedance
ZC = 1/j/w/C;
% Use current division to find the resistor current
IR = Is/R/(1/R+1/ZC);
MagIR = abs(IR)
PhaseIR = angle(IR)*180/pi
% Use current division to find the capacitor current
IC = Is/ZC/(1/R+1/ZC);
MagIC = abs(IC)
PhaseIC = angle(IC)*180/pi
% Verify KCL holds
CheckKCL = Is-IR-IC
MagIR =
169.7056e-003
PhaseIR =
-45.0000e+000
MagIC =
169.7056e-003
PhaseIC =
45.0000e+000
CheckKCL =
13.8778e-018 - 13.8778e-018i
% Create the phasor plot
axis(0.25*[-1 1 -1 1])
line([-100 100],[0 0],...
'Color','k',...
'LineWidth',0.5)
line([0 0],[-100 100],...
'Color','k',...
'LineWidth',0.5)
grid on
hold on
xlabel('Real')
ylabel('Imag')

line([0,real(Is)],[0,imag(Is)],...
'LineWidth',3,...
'Marker','o',...
'MarkerSize',3,...
'Color','g')
line([0,real(IR)],[0,imag(IR)],...
'LineWidth',2,...
'Marker','o',...
'MarkerSize',3,...
'Color','b')
line([0,real(IC)],[0,imag(IC)],...
'LineWidth',2,...
'Marker','o',...
'MarkerSize',3,...
'Color','b')

\begin{aligned}& i_{\mathrm{R}}(t)=169.7 \cos \left(500 t-45^{\circ}\right) \mathrm{mA} .\\ \\& i_{\mathrm{C}}(t)=169.7 \cos \left(500 t+45^{\circ}\right) \mathrm{mA} .\end{aligned}

fig 8.23

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