The Parallel Equivalent Of Series Impedance:

For equivalence of the two circuits (a) and (b), Y of Fig.(a) must equal Y of Fig (b)

Therefore Y=1/ (RS+ jXS) = 1/RP +1/jXP.

i.e. RS/ (RS2+XS2) – jXS /(RS2+XS2)=1/RP – j/XP

Fig(A) « EquivalentCircuits » Fig(B)