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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)