matlab A single phase circuit has the following arrangement shown in Fig 1 has t

Question

matlab

A single phase circuit has the following arrangement shown in Fig 1 has the load which is swappable with the following: Inductive Load Z = 1 25 angle 60 degree Ohm: Capacitive Load Z = 2.0 angle - 30 degree Ohm Resistive Load Z = 2.5 angle 0 degree Ohm The power supplied to the circuit can be described as v(t) = V_m cos (omega t + theta_v) where V_m = 100V, theta_ Write Matlab script in a m-file and produce plots for i(t),v(t),p(t), p_ (t)and P_ (t). a) From p_ (t) and p_ (t) plots, estimate the real and reactive power for each load. Draw a conclusion regarding the sign of reactive power for inductive and capacitive loads. b) Using phasor values of current and voltage, calculate the real and reactive power for each load and compare with the results obtained from the curves. c) If the above loads are all connected across the same power supply, determine the total real and reactive power taken from the supply.

Explanation / Answer

For inductive load

clc
Fs = 10e4;
t =0 : 1/Fs : 0.033-1/Fs ;
v = 100/1.414
Z1= 0.625+1.0825i;
i1= v/Z1;
thetaV= angle(i1);
peaki=abs(i1)*1.414;
I = peaki*cos(2*pi*60*t);
V = 100*cos((2*pi*60*t)+thetaV);
p = v*i1;
pmax = abs(p)*1.414;
theatp = angle(p);
P = pmax*cos((2*pi*60*t)-theatp);
prmax = real(p)*1.414;
pxmax = imag(p)*1.414;
Pr = prmax*cos(2*pi*60*t);
Px = pxmax*cos((2*pi*60*t)-theatp);
plot(t,V,t,I);
figure
plot(t,P,t,Pr,t,Px);

For capacitve load

clc
Fs = 10e4;
t =0 : 1/Fs : 0.033-1/Fs ;
v = 100/1.414
Z1= 1.732-1i;
i1= v/Z1;
thetaV= angle(i1);
peaki=abs(i1)*1.414;
I = peaki*cos(2*pi*60*t);
V = 100*cos((2*pi*60*t)+thetaV);
p = v*i1;
pmax = abs(p)*1.414;
theatp = angle(p);
P = pmax*cos((2*pi*60*t)-theatp);
prmax = real(p)*1.414;
pxmax = imag(p)*1.414;
Pr = prmax*cos(2*pi*60*t);
Px = pxmax*cos((2*pi*60*t)-theatp);
plot(t,V,t,I);
figure
plot(t,P,t,Pr,t,Px);

For resistive load

clc
Fs = 10e4;
t =0 : 1/Fs : 0.033-1/Fs ;
v = 100/1.414
Z1= 2.5;
i1= v/Z1;
thetaV= angle(i1);
peaki=abs(i1)*1.414;
I = peaki*cos(2*pi*60*t);
V = 100*cos((2*pi*60*t)+thetaV);
p = v*i1;
pmax = abs(p)*1.414;
theatp = angle(p);
P = pmax*cos((2*pi*60*t)-theatp);
prmax = real(p)*1.414;
pxmax = imag(p)*1.414;
Pr = prmax*cos(2*pi*60*t);
Px = pxmax*cos((2*pi*60*t)-theatp);
plot(t,V,t,I);
figure
plot(t,P,t,Pr,t,Px);

for all three loads kept across supply

clc
Fs = 10e4;
t =0 : 1/Fs : 0.033-1/Fs ;
v = 100/1.414
Y1 =1/(0.625+1.0825i);
Y2= 1/(1.732-1i);
Y3 = 1/2.5
Y= Y1+Y2+Y3;
i1= v*Y;
thetaV= angle(i1);
peaki=abs(i1)*1.414;
I = peaki*cos(2*pi*60*t);
V = 100*cos((2*pi*60*t)+thetaV);
p = v*i1;
pmax = abs(p)*1.414;
theatp = angle(p);
P = pmax*cos((2*pi*60*t)-theatp);
prmax = real(p)*1.414;
pxmax = imag(p)*1.414;
Pr = prmax*cos(2*pi*60*t);
Px = pxmax*cos((2*pi*60*t)-theatp);
plot(t,V,t,I);
figure
plot(t,P,t,Pr,t,Px);

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