Circuit Theory/Example70/code

Node vs1:=(1+I)/6; vs2:=0.5 + I*sqrt(3)/2; numeric::solve([(vs1-va)/5 - va - (va + vs2)/(sqrt(3)*I) = 0],[va])

Mesh vs1:=(1+I)/6; vs2:=0.5 + I*sqrt(3)/2; numeric::solve([i1-i2-vs1+5*i1,i2*sqrt(3)*I - vs2 + i2 - i1,i3 = i1-i2],[i1,i2,i3])

Thevenin voltage vs1:=(1+I)/6; vs2:=0.5 + I*sqrt(3)/2; numeric::solve([vth=(vs1+vs2)*sqrt(3)*I/(5 + sqrt(3)*I)-vs2],[vth])

Norton current vs1:=(1+I)/6; vs2:=0.5 + I*sqrt(3)/2; numeric::solve([xn=vs1/5 - vs2/(sqrt(3)*I)],[xn])

calculating rth 1/(1/5 + 1/(sqrt(3)*i)) = 0.5357 + 1.5465*i

checking rth vth = 0.7479777026 + 0.5491833001*i; xn = - 0.4666666667 + 0.3220084679*i; rth = vth/xn

current in thevenin equivalent is vth = 0.7479777026 + 0.5491833001*i; rth=0.5357 + 1.5465*i ixx = vth/(rth + 1)

power is ( 0.7479777026^2 + 0.5491833001^2)/(2*  0.5357)

checking with simulation period is 2*pi = 6.2832 seconds

for the 1 ohm resistor va = - 0.4206268848 + 0.06596618565*i = va(t) = 0.4258 * cos(t + 2.9860) (171 degrees) xn = - 0.4666666667 + 0.3220084679*i i(t) = 0.5670 * cos(t+ 2.5376) (145.4 degrees)