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authorSiddharth Agarwal2019-09-03 18:27:40 +0530
committerSiddharth Agarwal2019-09-03 18:27:40 +0530
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+clc
+//Example 11.2
+//Calculate the average power
+printf("Given")
+disp('v=4*cos(%pi/6*t), V=4(0 deg), Z=2(60 deg)')
+Vamp=4;Vang=0;Zamp=2;Zang=60;
+//Let I be the phasor current
+Iamp=Vamp/Zamp
+Iang=Vang-Zang
+P=0.5*Vamp*Zamp*cos((Zang*%pi)/180)
+printf("P=%d W \n",P);
+t=-1:1:15
+t1=-3:1:12
+v=Vamp*cos(%pi/6*t)
+//i=2*cos((%pi/6)*t-(%pi/3))
+i=Iamp*cos(%pi/6*t+((Iang*%pi)/180))
+figure
+a= gca ();
+plot (t,v,t,i)
+xtitle ('v,i vs t' ,'t' ,'v,i');
+a. thickness = 2;
+//Instantaneous power p=v*i
+//On solving
+p=2+4*cos(%pi/3*t+((Iang*%pi)/180))
+figure
+a= gca ();
+plot (t,p)
+xtitle ('p vs t' ,'t' ,'p');
+a. thickness = 2;