CHAPTER04 : CIRCUIT DIFFERENTIAL EQUATIONS FORMS AND SOLUTIONS

Example E01 : Pg 125

In [1]:
# ad
Zab = complex(1,-0.5); # impedance appearing across terminals ab
Zbg = complex(1); # impedance appearing across terminals bg
Zbcd = complex(2+1,2); # impedance appearing across terminals bcd
Zad = Zab + (Zbg*Zbcd/(Zbg + Zbcd)); # impedance appearing across terminals ad
print "impedance appearing across terminals ad = ",Zad

# dg 
Zdg = Zbg + (Zab*Zbcd/(Zab+Zbcd)); # impedance appearing across termainals dg
print "impedance appearing across terminals dg = ",Zdg

impedance appearing across terminals ad =  (1.8-0.4j)
impedance appearing across terminals dg =  (1.91780821918-0.219178082192j)