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authorpriyanka2015-06-24 15:03:17 +0530
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+clc
+//Example 6.19
+//Calculate total current through load
+//On applying source transformation
+//From figure 6.65
+i=%i
+V1=10;V2mag=5;V2ph=90;V3mag=14.4;V3ph=225;
+x=V2mag * cos (( V2ph * %pi ) /180) ;
+y=V2mag * sin (( V2ph * %pi ) /180) ;
+V2= complex (x,y)
+a=V3mag * cos (( V3ph * %pi ) /180) ;
+b=V3mag * sin (( V3ph * %pi ) /180) ;
+V3= complex (a,b)
+G1=1/2;G2=1/(2+i*3);G3=1/(2-i*2);
+//By applying Millman Theorem
+disp('V=((V1*G1)+(V2*G2)+(V3*G3))/(G1+G2+G3)')
+V=((V1*G1)+(V2*G2)+(V3*G3))/(G1+G2+G3)
+[Vmag Vang]=polar(V)
+R=1/(G1+G2+G3)
+printf("V=%3.2f(%3.2f deg)V",Vmag,(Vang*180)/%pi)
+disp(R,'R=')
+//Consider the resultant circuit from figure 6.66
+disp('Let the total current through 3+i4 be I')
+//Applying KVL to the circuit
+I=V/(3+i*4+R)
+[Imag Iang]=polar(I)
+printf("I=%3.2f(%3.2f deg)V",Imag,(Iang*180)/%pi)
+