# [SI-LIST] Effective inductance calculation of a 3-wire system

• To: si-list@xxxxxxxxxxxxx
• Date: Mon, 13 Jun 2005 12:59:55 -0400 (EDT)

Hi,

I am facing the following problem while calculating the effective
inductance of an aggressor wire in a 3-wire system.

We know that when there are 2 wires carrying equal current in the same
direction the effective inductance of one of the wires is given by the
relation:

L(total) = L_a + L_ab, where L_a is the partial self inductance
of wire 'a' and L_ab is the partial mutual inductance of wire 'a' with
wire 'b'.

Also, L(total) = L_a - L_ab, when the direction of current propagation is
opposite with respect to each other (or out of phase)

My question is, what happens when I consider a 3-wire system? Suppose I
want to calculate the L(total) of the middle wire 'a' in the figure given
below:

________________________________  ---> 'b'

________________________________  ---> 'a'

________________________________  <--- 'c'

The direction of current is also shown in this case. For the ease of
understanding, current in the aggressor wire 'a' and one of the victim
wire 'b' flows EAST, while the current in the other victim wire 'c' flows
WEST. In this scenario, how do I calculate the L(total) for the aggressor
'a'.

Will it be simply: L(total) = L_a + L_ab - L_ac
or, L(total) = L_a + 0.5*L_ab - 0.5*L_ac, when I consider 'b' and 'c' to
be separated by the same distance from 'a' and the return current being
splitted equally in the two victim wires.

or, will the expression be something else?

Thanks and Regards,
Amit

Research Assistant
527 ERC
University of Cincinnati
OH - 45221
ph(lab) : 513-556-3025

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