Equivalent Inductance
Equivalent Inductance
(OP)
Hi All,
Still doing my system fault level calculations. I am stuck in trying to work out the equivalent inductances of two RL circuits. One circuit has Req & Leq and the other has Rsb & Lsb.
From previous work, I have the formula for the equivalent inductance Lp = (Rp/Req)^2 * Leq + (Rp/Rsb)^2 * Lsb where Rp is the equivalent resistance which equals Req & Rsb connected in parallel.
Wish to know how this formula come about as my new case now have 3 circuits rather than two.
Many thanks in advance & Merry Xmas
Still doing my system fault level calculations. I am stuck in trying to work out the equivalent inductances of two RL circuits. One circuit has Req & Leq and the other has Rsb & Lsb.
From previous work, I have the formula for the equivalent inductance Lp = (Rp/Req)^2 * Leq + (Rp/Rsb)^2 * Lsb where Rp is the equivalent resistance which equals Req & Rsb connected in parallel.
Wish to know how this formula come about as my new case now have 3 circuits rather than two.
Many thanks in advance & Merry Xmas






RE: Equivalent Inductance
Xeq = 2j.Pi.f.Leq and Xsb = 2j.Pi.f.Lsb
The impedance in each parallel branch is:
Z1=Req + j.Xeq and Z2 = Rsb + j.Xsb
The total equivalent value of two parallel impedance’s could be calculate using the classical Ohm relation.
Ztotal = Z1.Z2/(Z1+Z2) = [(AC+BD) +j. (AD-BC)]/ (C2-D2).
|Ztotal| = |(A+jB)/(C+jD)| = SQRT[(AC+BD)2 +(AD-BC)2 ] / (C2-D2).
Angle = Tan-1(AD-BC)/(AC+BD).
Where:
Pi=3.1416
j = SQRT (-1)
f = system frequency (typical 50 or 60 Hz for most commercial power systems)
A= (Req.Rsb- Xeq.Xsb)
B= Req.Xsb+Rsb.Xeq
C = (Req+Rsb)
D = (Xeq+Xsb)
RE: Equivalent Inductance
Please, would you clarify C^2-D^2 in terms of C^2+D^2, and AD-BC in terms of BC-AD?
RE: Equivalent Inductance
Rtotal= (AC+BD)/(C2 + D2)
Xtotal= j.(BC-AD)/(C2 + D2) .
The total equivalent inductance will be:
Ltotal= [(BC-AD)/(C2 + D2)/2.Pi.f