Equivalent Inductance

[rail1996]

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

 

[cuky2000]

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:
X_eq = 2j.Pi.f.L_eq and X_sb = 2j.Pi.f.L_sb

The impedance in each parallel branch is:

Z₁=R_eq + j.X_eq and Z₂ = R_sb + j.X_sb

The total equivalent value of two parallel impedance’s could be calculate using the classical Ohm relation.

Z_total = Z₁.Z₂/(Z₁+Z₂) = [(AC+BD) +j. (AD-BC)]/ (C²-D²).

|Z_total| = |(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= (R_eq.R_sb- X_eq.X_sb)
B= R_eq.X_sb+R_sb.X_eq
C = (R_eq+R_sb)
D = (X_eq+X_sb)}

 

[jbartos]

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?

 

[cuky2000]

Thanks jbartos, the sign should be corrected as follow:

R_total= (AC+BD)/(C² + D²)

X_total= j.(BC-AD)/(C² + D²) .

The total equivalent inductance will be:

L_total= [(BC-AD)/(C² + D²)/2.Pi.f