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) |