Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

14.7 
flux exPulSIon By a Shorted coIl
One does not make a good transformer and keep its secondary shorted! However, there are situations 
and applications in which an electrical system can be modeled as a transformer with a low resistance 
secondary winding kept shorted. Therefore, we look at the input impedance of a transformer with the 
secondary winding shorted as in Fig.14.7-1. 
Z
p
(s) may be determined by mesh analysis. The result 
will be,
Z s
s L L
k
s L r
L r
r r
r
sL
p
p s
p s
s p
p s
s
s
( )
(
)
(
)
=
−
+
+
+
+
2
2
1
(14.6-10)
The impedance at DC frequency is r
p
W
. 
The impedance at high frequency is 
≈
s(1
-
k
2
)L
p
, 
indicating that the transformer behaves as an inductance of 
(1
-
k
2
)L
p
H at high frequencies when its secondary is kept 
shorted.
Assume that the shorted coil has negligibly small resistance. Taking r
s
= 
0, we get,
Z s
r
s
k L
p
p
p
( )
(
)
= +
−
1
2
.
Now, if the transformer is perfectly coupled (i.e., k 
= 
1), then the input impedance becomes 
equal to
primary winding resistance at all frequencies. Hence, the short-circuit impedance of a good 
transformer with k 
= 
1, r
p
= 
r
s
= 
0 is zero at all frequencies.
The expressions for I
p
(s) and I
s
(s) with r
2
= 
0 may be derived as
I s
V s
s
k L
r
I s
V s
s
k L
r
k
n
I
p
p
p
p
s
M
L
p
p
p
p
s
( )
( )
(
)
( )
( )
(
)
=
−
+
=
−
+
=
1
1
2
2
and 
(( )
s
Currents entering the dots generate positive flux linkages in coils. Let 
y
p
be the flux linkage in 
primary coil and 
y
s
be the flux linkage in the secondary coil. Then,
Fig. 14.7-1 
A transformer with 
secondary shorted
k
I
p
(
s
)
Z
p
(
s
)
I
s
(
s
)
r
p
L
s
L
p
r
s

---

Flux Expulsion by a Shorted Coil 
14.25
y
p
p p
s
p p
p
p p
p s p
s
L I s
MI s
L I s
k
n
MI s
L I s
k
n
L L I s
( )
( )
( )
( )
( )
( )
( )
=
−
=
−
=
−
2
==
−
= −
+
=
−
−
+
+
L I s
k
s
L I s
MI s
MV s
s
k L
r
M
p p
s s
p
p
p
p
( )[
]
( )
( )
( )
( )
(
)
1
1
2
2
2
y
V
V s
s
k L
r
p
p
p
( )
(
)
1
0
2
−
+
=
Therefore, the secondary coil that started with zero flux linkage continues to hold zero flux linkage 
at all time after that. Thus,
A shorted coil with zero resistance expels flux completely at all frequencies in the 
magnetic path where it is located.
It does this by allowing suitable current to flow in it in order to cancel the mutual flux generated by 
current in the driven coil. Both primary and secondary currents rise exponentially with a time constant 
of 
(
)
1
2
-
k L
r
p
p
s in step response.
But what if the windings are perfectly coupled in addition? Then,
I s
V s
r
I s
k
n
V s
r
p
p
p
s
p
p
( )
( )
( )
( )
=
=
and 
If v
p
(t) is a unit step function, the current rises instantaneously from 0 to 1/r
p
in the primary 
winding and from 0 to k/nr
p
in the secondary winding. Further, the flux linkages in both windings 
continue to be at zero.
There was no impulse voltage in the circuit. How was it possible for the inductor currents to change 
instantaneously without impulse voltages in the circuit?
Currents in individual coils can change instantaneously in a coupled-coil system. Faraday’s law 
indicates that the induced emf across a coil is proportional to the rate of change of flux linkage in 
it. Therefore, flux linkage of a coil cannot change instantaneously unless impulse voltage is applied 
to the coil. This reduces to the statement that current in an inductor cannot change instantaneously 
unless impulse voltage is applied in the case of inductors that are not magnetically coupled. However, 
there is a distinction between instantaneous change of flux linkage and current if the flux linkage in 
a coil can be produced by more than one current as in the case of coupled inductors. Therefore, the 
more general principle of ‘no instantaneous change in flux linkage in a coil unless impulse voltage is 
applied’ should be applied in the case of coupled-coils.
The flux linkage in a shorted coil with zero resistance remains constant in time. This 
statement is sometimes referred to as ‘Constant Flux Linkage Theorem’.
Note that in the step response of a perfectly coupled transformer with shorted secondary (r
s
= 
0), 
the flux linkage of secondary winding remained at zero though the current increased instantaneously. 

---

14.26
Magnetically Coupled Circuits
The magnetic coupling was perfect, and the primary cannot have flux linkage in it in that case unless 
secondary has it. Therefore, the primary winding flux linkage also remained at zero despite step 
change in primary current.
We may derive a relationship between the instantaneous changes in primary and secondary 
windings as follows:
y
y
y
p
p p
s
s
s s
p
p
t
L i t
Mi t
t
L i t
Mi t
t
L
( )
( )
( )
( )
( )
( )
( )
=
−
= −
+
∆
=
Therefore
pp
p
s
s
s
s
p
i t
M i t
t
L i t
M i t
∆
− ∆
∆
= − ∆
− ∆
( )
( )
( )
( )
( )
y
But, there cannot be instantaneous changes in flux linkages in the absence of impulse voltages. 
Hence, 
∆
y
p
(t) and 
∆
y
s
(t) are zero. Therefore,
L
M
M
L
i
i
p
s
p
p
−
−






∆
∆








=






0
0
The determinant of the matrix on the left side is nonzero for all k < 1. Therefore, 
∆
i
p
and 
∆
i
s
can 
only be zero with imperfect coupling. That is, there can be no instantaneous change in coil currents 
in a two-coil system if the coils are imperfectly coupled and there is no impulse voltage applied or 
supported somewhere in the system. 
However, if k 
= 
1, the determinant of the square matrix in the left side of the equation is zero and 
there can be a non-zero solution for 
∆
i
p
and 
∆
i
s
. In fact, any pair of values that satisfy the constraint 
that 
∆
i
s
= ∆
i
p
/n will be permitted. The exact value by which the primary current jumps will be decided 
by the jump in primary voltage and the primary resistance. Since the flux linkage in primary winding 
does not change, all the primary voltage will have to be absorbed by the primary resistance at all t.
Its resistance compromises the effectiveness of flux expulsion from shorted coil. The expression 
for flux linkage in shorted coil when k 
≠
1 and r
s
≠
0 is
y
s
p
s
p s
p s
s p
p s
s
V s
Mr
s
k L L
s L r
L r
r r
( )
( )
(
)
(
)
=
−
+
+
+
2
2
1
DC value of this ratio is M/r
p
. Therefore, DC flux will not be expelled under steady state. Similarly, 
low frequency AC flux will also manage to get into the shorted coil under steady state conditions. 
However, the ratio goes to low values at high frequency. Therefore non-zero resistance in shorted coil 
results in DC and low-frequency fluxes penetrating into the coil. High-frequency flux is expelled more 
or less effectively.
The principle of flux expulsion detailed in this section is employed in shielding sensitive electronic 
equipment from electromagnetic interference.