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+// Example 7.1
+// AC Power Calculations
+// From Example 6.8 we already found that,
+Z=complex(4.8,6.4);
+V_m=80;
+V_c_m=40;
+I_m=10*10^-3;
+// The total average power supplied by the source is,
+R_omega=4.8*10^3;
+R1=40*10^3;
+R2=5*10^3;
+P=0.5*R_omega*I_m^2; // Average Power
+// This power is actually dissipated by 40kohm and 5kohm resistor
+P_R1= V_m^2/(2*R1);
+P_R2=V_c_m^2/(2*R2);
+disp(P,"Total Average Power Dissipation(in Watt)=")
+disp(P_R1,"Power dissipated across 40kohm(in Watt)=")
+disp(P_R2,"Power dissipated across 5kohm(in Watt)=")
+if P==(P_R1+P_R2) then
+disp("This shows average power dissipation in the due to all resistors")
+end