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Equivalent Inductance

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

RE: Equivalent Inductance

Assuming that Leq and Lsb are the inductance of two parallel impedances. If this is the case, consider calculating first the inductive reactance (X) as follow:
                                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

Suggestion to the previous posting:
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

Thanks jbartos, the sign should be corrected as follow:

    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
     

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