# Answer to Question #230362 in Electrical Engineering for tonny419

Question #230362

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)

1
2021-08-30T01:54:55-0400 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');

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