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+clc

+//Example 6.10

+//Calculate the voltage across 20 ohm capacitor

+//Consider the circuit to be solved by superposition principle

+disp('Consider the current source 2(90 deg)only')

+//From figure 6.32

+//Let I1 be the current through -i*4 capacitive reactance

+Imag=2;Iph=90;

+i=%i

+x=Imag * cos (( Iph * %pi ) /180) ;

+y=Imag * sin (( Iph * %pi ) /180) ;

+I= complex (x,y)

+I1=(I*(i*15))/(i*5+i*15-i*4)

+//Let V20 be the voltage across -i*4 capacitive reactance

+V200=(-i*4)*I1

+printf("V20=%3.2fV \n",V200)

+disp('Consider the 20 V voltage source only')

+V=20;

+//From figure 6.35

+//let V201 be the voltage across -i*5 capacitive reactance

+V201=-V

+printf("V201=%d V \n",V201)

+disp('Consider the current source 1(90 deg)only')

+I1mag=1;I1ang=90;

+//From figure 6.37

+//Let V202 be the voltage across -i*5 capacitive reactance

+V202=(-i*5)*I1mag*i

+printf("V202=%3.2fV \n",V202)

+//Let V20 be the voltage across -i*20 capacitive reactance

+V20=V200+V201+V202

+printf("\n V20=%3.2fV \n",V20)