A voltage e=250sinωt+50sin(3ωt+pi/3)+20 sin(5ωt+(5pi/6)) is applied to a series Circuit of resistance 20Ω and an inductance of 0.05H. Take ω=314rad/s
3.1 Derive an expression for the current. (5)
Calculate the following:
3.2 The rms value of the current and voltage. (2)
3.3 The total power supplied. (2)
3.4 The power factor. (1)
The rms, power factor and power delivered are computed using the matlab code given below:
close all,
clear all,
clc,
Fs=1000;
t = 0:(1/Fs):1;
Show = 100;
w = 314;
F = w/(2*pi);
R = 200;
L = 0.05;
Xl = w*L;
Z = R + Xl;
% 250 sinωt+50 sin(3ωt+π/3)+20 sin(5ωt+(5π/6))
e1 = 250 * sin(w*t);
e2 = 50*sin(3*w*t+(pi/3));
e3 = 20*sin(5*w*t + (5*pi)/6);
e = e1+e2+e3;
plot(t(1:Show),e(1:Show)); title('Plot: e = 250 sinωt+50 sin(3ωt+π/3)+20 sin(5ωt+(5π/6)) '); ylabel('--- e --->'); xlabel('--- t --->'); grid on,
Amp = max(e);
fprintf('\nGiven R = %d Ohms\t\tand L = %.3f H',R,L);
fprintf('\nXl is given by Xl = wL = %d*%.2f = %.3f',w,L,Xl);
fprintf('\nTherefore, Impedence Z = R + Xl = %d + %.2f = %.3f',R,Xl,Z);
fprintf('\nFrom graph, the max. Amplitude of the signal: e = 250 sinωt+50 sin(3ωt+π/3)+20 sin(5ωt+(5π/6)) = %f',Amp);
MaxI = Amp/Xl;
fprintf('\nMax. Current is given by: Imax = Max. Amp./Z = %.3f/%.3f = %.3f',Amp,Z,MaxI);
Irms = MaxI/1.414;
Vrms = Amp/1.414;
fprintf('\n\nRMS value of I = Imax/1.414 = %.3f/1.414 = %.3f',MaxI,Irms);
fprintf('\n\nRMS value of V = Vmax/1.414 = %.3f/1.414 = %.3f',Amp,Vrms);
P = Vrms*Irms;
fprintf('\n\nPower supplied = P = Irms * Vrms = %.3f * %.3f = %.3f',Irms,Vrms,P);
PF = cos(w*L/R);
fprintf('\n\nPower Factor = Cos(phi) = wL/R = %d*%.3f/%d = %.3f',w,L,R,PF);
fprintf('\n\n');
Matlab Output:
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