Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 14.2-1
The applied voltage is v
S
(t) 
= 
100
√
2 cos10
3
t V in the circuit 
shown in Fig. 14.2-4. (a) Find the primary (the side with 
source connection) and secondary (the side with no source 
connection) currents and the average power delivered to 
the resistors for (i) k 
= 
0, (ii) k 
= 
0.1, (iii) k 
= 
0.5 and (iv) 
k 
= 
1. (b) Repeat calculations with k 
= 
1 for L
1 
= 
25 H and 
L
2 
= 
100 H.
Solution
(a) The conductive equivalent circuit for coupled coils is used to obtain the phasor equivalent circuits 
for the four values of coupling coefficients. The phasor equivalent circuits are shown in Fig. 14.2-5. 
The solution process is illustrated for k 
= 
0.1.
V s 
= 
100
∠
0
°
V rms. The two mesh equations are expressed in matrix form as
10
25
5
5
40
100
100 0
0
1
2
+
−
−
+











 =
∠






j
j
j
j
I
I
Solving these equations, we get I
1
= 
3.737
∠-
67.86
°
A rms and I
2
= 
0.174
∠-
46
°
A rms.
Average power dissipated in 10 
W
resistor is 3.737
2
×
10 
= 
139.65 W
Average power dissipated in 40 
W
resistor is 0.174
2
× 
40 
= 
1.21 W
Average power delivered by source 
= 
100
× 
3.737
× 
cos67.9
°
= 
140.83 W.
Impedance seen by source 
= 
Z
in
 
= 
100
∠
0
°
÷
3.737
∠
67.86
°
= 
10.09
+ 
j24.78 
W
10 
Ω
10 
Ω
j
25 
Ω
j
100 
Ω
40 
Ω
10 
Ω
j
20 
Ω
j
95 
Ω
40 
Ω
10 
Ω
j
0 
Ω
j
75 
Ω
40 
Ω
10 
Ω
–
j
25 
Ω
j
50 
Ω
j
50 
Ω
j
25 
Ω
k
= 0
k
= 0.1
k
= 0.5
k
= 1
j
5 
Ω
40 
Ω
j
0 
Ω
40 
Ω
25 mH 100 mH
10 
Ω
40 
Ω
20 mH 95 mH
10 
Ω
10 
Ω
40 
Ω
40 
Ω
0 mH
–25 mH
75 mH
50 mH
0 mH
+
–
+
–
+
–
+
–
+
–
+
–
+
–
+
–
i
1
i
1
i
1
i
1
i
2
I
1
I
2
I
1
I
2
I
1
I
2
I
1
I
2
5 mH
25 mH
50 mH
i
2
i
2
i
2
V
S
(
t
)
V
S
(
t
)
V
S
V
S
V
S
V
S
V
S
(
t
)
V
S
(
t
)
Fig. 14.2-5 
Phasor equivalent circuits of the circuit in for different values of coupling 
coefficient
The table that follows summarises the results of calculations for all four values of k.
Fig. 14.2-4 
Circuit for 
Example: 14.2-1 
40 
Ω
10 
Ω
100 mH
25 mH
k
+
–
v
S 
(
t
)

---

14.10
Magnetically Coupled Circuits
K
I
1
(A rms)
I
2
(A rms)
P
10
W
(W)
P
40
W
(W)
PF
Z
in
(
W
)
0
3.714
∠-
68.2
°
0
137.94
0
0.371
10
+ 
j25
0.1
3.74
∠-
67.86
°
0.174
∠-
46
°
139.65
1.21
0.376
10.09
+ 
j24.78
0.5
4.334
∠-
58.2
°
1.006
∠-
36.4
°
187.83
40.48
0.527
12.16 
+ 
j19.61
1
5.281
∠-
10.49
°
2.451
∠
11.31
°
278.89
240.29
0.983
18.62 
+ 
j 3.45i
(b) We observe from the table of results that as the coupling coefficient increases (i) the primary 
and secondary currents increase in magnitude and become more active in content than reactive, 
(ii) power delivered to the resistor in the secondary side increases, (iii) the power factor of circuit 
increases and (iv) the source experiences a decreasing input impedance that becomes more and more
resistive.
Now, we raise the self-inductance level of the coils by 1000 fold. But we keep the ratio between 
them at 4 as before. The mesh equations for k 
= 
1 are
10
25 10
50 10
50 10
40
10
100 0
3
3
3
5
1
2
+
×
−
×
−
×
+













 =
∠ °
j
j
j
j
I
I
00






We observe that the reactive terms in the matrix entries overpower the resistive terms. The solution 
is obtained as I
1
= 
5
∠-
0.01
°
A rms and I
2
= 
2.5
∠
0.01
°
A rms.
Average power dissipated in 10 
W
resistor is 5
2
× 
10 
= 
250 W
Average power dissipated in 40 
W
resistor is 2.5
2
× 
40 
= 
250 W
Average power delivered by source 
= 
100
× 
5
× 
cos0.01
°
= 
500 W.
Impedance seen by source 
= 
Z
in
 
= 
100
∠
0
°
÷
5
∠-
0.01
°
= 
20
+ 
j0.004 
W
The input impedance as seen by the source is almost purely resistive at 20 
W
. The resistor that is in 
series with the source in the primary side must be contributing 10 
W
to the input impedance. Therefore, 
the transformer and its secondary load of 40 
W
resistor contributes 10 
W
to input impedance. Thus, the 
equivalent impedance of the transformer plus 40 
W
load seems to be an almost pure resistance of 10 
W
. The ratio between 10 
W
and 40 
W
happens to coincide with the ratio L
1
: L
2
– that is a coincidence 
that calls for further analysis.
Let us calculate the voltage appearing across primary and secondary windings of the transformer.
Voltage across primary winding 
= 
100
∠
0
°
-
10
× 
5
∠-
0.01
°
≈
50
∠
0
°
V rms
Voltage across primary winding 
= 
40
× 
2.5
∠
0.01
°
≈
100
∠
0
°
V rms
Thus, primary voltage: secondary voltage 
≈
1:2 
= 
L
L
1
2
:
and using the solution for currents, 
primary current: secondary current 
≈
2:1 
= 
L
L
2
1
:
. That makes it too many coincidences to pass 
over lightly. Therefore, we analyse the transformer with k 
= 
1 in greater detail now.
14.3 
the Perfectly couPled tranSforMer and the Ideal tranSforMer
Figure 14.3-1 shows a two-winding transformer with primary self-inductance of L
1 
and secondary 
self-inductance of L
2
with coupling coefficient of unity. We neglect the winding resistance 
for the time being. The two circuits shown in this figure differ only in the dot polarity in the 
transformer.

---

The Perfectly Coupled Transformer and The Ideal Transformer 
14.11
+
–
+
k
= 1
(a)
–
+
–
I
S
I
p
v
p
v
s
Z
L
L
1
L
2
+
–
(b)
k 
= 1
I
S
I
p
+
–
v
p
L
1
+
–
v
s
Z
L
L
2
Fig. 14.3-1 
A passively terminated two-winding transformer with 
k
= 
1
The primary is driven by a source 
= 
2V
t
cos
w
V and secondary is passively terminated in an 
impedance Z
L
= 
R
L
+ 
j X
L
W
. Passive termination implies that there are no sources of any kind in the 
load circuit.
The mesh currents are I
p
and I
s
themselves and the mesh equations for circuit in Fig. 14.3-1 (a) in 
matrix form are
j L
j M
j M
Z
j L
I
I
V
p
s
w
w
w
w
1
2
0
0
−
−
+











 =
∠ °






L
Since k 
= 
1, the value of M 
= 
L L
1 2
. Solving for primary and secondary current phasors using this 
value for M, we get
I
V
j L
L
L
V
Z
V
j L
L
L Z
I
V
L
L
p
p
s
=
∠
+




∠
=
+






= ∠
0
0
1
1
0
1
2
1
1
2
1
2
1
w
w
L
L






=






=
=






∴
=
1
1
2
1
2
1
2
1
Z
V
L
L
Z
V
I Z
V
L
L
V
V
L
L
p
s
s
p
s
p
L
L
L
We have used the fact that the source voltage phasor V
∠
0
°
itself is the primary voltage phasor V
p
to make substitutions in the equation above.
We note that the secondary voltage and primary voltage are in phase and related by a real scaling 
factor of
L
L
2
1
. However, this number is nothing but 
n
n
2
1
, where n
1
and n
2 
are the number of turns 
employed in primary and secondary windings. This is so since self-inductance of a coil is proportional 
to the square of the number of turns in the coil. We represent this ratio as turns ratio ‘n’. 
Therefore, secondary voltage magnitude in a transformer with unity coupling coefficient 
is turns ratio times the primary voltage magnitude.
The equation for I
p
is the sum of two terms. Each term involves a division of primary voltage phasor 
by impedance. Therefore, the equation for I
p
suggests that the input impedance of the circuit is a parallel 
combination of two impedances – an inductor of L
1
and another impedance of value 
L
L
Z
Z
n
L
1
2
2
L
=
.

---

14.12
Magnetically Coupled Circuits
Therefore, a transformer with unity coupling coefficient 
reflects
the secondary load 
impedance 
Z
L
onto the primary side as a 
turns-ratio transformed impedance 
of value 
Z
L
/
n
2
where 
n
= 
secondary turns/primary turns.
Thus, the entire transformer and load circuit can be replaced by the equivalent circuit shown in Fig. 
14.3-2 as far as the input side v–i behaviour under sinusoidal steady-state condition is concerned. We 
will see later that this is true not only for sinusoidal steady-
state analysis but also for time-domain analysis too.
The primary winding resistance is included as a series 
resistance in the primary side, whereas the secondary 
winding resistance is included as a part of R
L
in the load 
for analysis purposes.
Now, we consider the circuit in Fig. 14.3-1 (b). Reader 
may easily verify that the mesh equations for this circuit 
will be,
j L
j M
j M
Z
j L
I
I
V
p
s
w
w
w
w
1
2
0
0
L
+











 =
∠ °






Both the off-diagonal terms in the mesh impedance matrix have changed sign. The solution for 
primary and secondary current phasors with k 
= 
1 will be, 
I
V
j L
L
L
V
Z
V
j L
L
L Z
V
j L
n
p
p
p
=
∠
+




∠
=
+





 =
+
0
0
1
1
1
1
2
1
1
2
1
1
w
w
w
L
L
22
2
1
2
1
0
1
1
Z
I
V
L
L
Z
nV
Z
V
I Z
V
L
L
s
p
s
s
p
L
L
L
L






= − ∠






= −
=
= −





 = −
nV
p
There is no difference in solution except for 180
°
phase shift in secondary voltage. That is anyway 
expected due to the change in relative polarity of windings. The voltage magnitude ratio remains at 
turns ratio n. 
But the most important aspect to be noted is that the same equivalent circuit shown in Fig. 14.3-2 
will describe the transformer in this case too. See the equation for primary current phasor. In fact, this 
is true even if k 
≠
1.
The input impedance of a passively terminated transformer is independent of its 
dot
point assignment.
There are many important applications of transformers with k 
≈
1 in which we make use of the 
inductor L
1
that appears in the input impedance to advantage. A tuned amplifier widely employed in 
communication circuits is such an application example. 
There are many applications of transformers with k 
≈
1 in electrical power engineering and 
electronics engineering where that inductance in the input impedance of transformer is not desirable. 
Fig. 14.3-2 
Input equivalent 
circuit of transformer 
with 
k
= 
1