Page 390 - Handbook of Electrical Engineering

378 HANDBOOK OF ELECTRICAL ENGINEERING

This has the form:-
(13.11)
y 2 = a 21 I c + a 22 I s
Where

y 2 =−jωM 3s I 3
a 21 =+jωM sc
a 22 =−(R ss + jωL s )
R ss =+R s + R e

The solution of the simultaneous equations (13.10) and (13.11) for the two currents I s and I c is:-

y 1 a 21 − y 2 a 11
I s = amps (13.12)
a 12 a 21 − a 11 a 22
and
y 2 a 12 − y 1 a 22
I c = amps (13.13)
a 12 a 21 − a 11 a 22
Some simplifications can be made after comparing the various mutual and self-inductances.
The following assumptions are valid:-

because the majority of the flux between the screen and the core couples
M sc = L s
the screen and the core.

Let M = M 3s ≈ M 3c

And M sc M 3s or M 3c because of the relative dimensions and separation distances.
The denominator of (13.12) and (13.13) becomes:-

2
a 12 a 21 − a 11 a 22 = R ss R cc + jω(R cc L s + R ss L c ) + ω (L s (L s − L c ))

In which the extreme right-hand term is very small in the range of frequencies of interest, and can
be ignored. Therefore the denominator becomes:-

a 12 a 21 − a 11 a 22 = R ss R cc + jω(R cc L s + R ss L c )

The I s numerator of (13.12) becomes:-

2
y 1 a 21 − y 2 a 11 = (+ω M(L s − L c ) + jωMR cc )I 3

The I c numerator of (13.13) becomes:-
2
y 2 a 12 − y 1 a 22 = (−ω M(L s − M) + jωMR ss )I 3