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+//Chapter 3 Example 2//
+clc
+clear
+//supply voltage=v,from the figure we can get the required values//
+v=220;// in volts//
+z1=2+%i*8;// in ohms//
+x1=8;
+z2=-%i*6;
+r=5;
+//let the equivalent impedence =zeq//
+zeq=(z1)*(z2)/(z1+z2);
+printf("\n Equivalent impedence = %.2f%.2fi ohms\n",zeq,imag(zeq));
+//current in the circuit=i//
+i=v/(zeq+r);
+printf("\n Load current I = %.2f% +.2fi A\n",i,imag(i));
+//power in the 5 ohm resistor=p//
+p=abs(i)^2*r;
+printf("\n Power in the 5ohm resistor = %.2f W\n",p);
+// current in branch ab =i1,current in branch cd=i2//
+i1=i*(z2)/(z1+z2);
+printf("\n Current in branch ab = %.2f%+.2fi A\n",i1,imag(i1));
+printf("\n = %.2f A\n",abs(i1));
+i2=i*(z1)/(z1+z2);
+printf("\n Current in branch cd= %.2f%.2fi A\n",i2,imag(i2));
+printf("\n = %.2f A\n",abs(i2));
+
+