Coupled Inductors and Transformers
Self-Inductance
When current i flows through the coil, a
magnetic flux  is produced around it.
d
d di
di
v=N
=N
=L
dt
di dt
dt
Inductance:
d
L=N
di
This inductance is commonly called self-inductance, because it
relates the voltage induced in a coil by the time-varying current in the
same coil.
1
Mutual Inductance
1 = 11 + 21
d1
d1 di1
di1
= N1
= L1
v1 = N1
dt
di1 dt
dt
d21
d21 di1
di1
v2 = N 2
= N2
= M 21
dt
di1 dt
dt
Reverse direction:
2 = 22 + 12
d2
v2 = N 2
v1 = N1
dt
= N2
d2 di2
di
= L2 2
di2 dt
dt
d12
d di
di
= N1 12 2 = M12 2
dt
di2 dt
dt
Mutual inductance is the ability of one inductor to induce a voltage
across a neighboring inductor.
2
For linear magnetic coupling:
M12 = M 21 = M
Determining dot markings:
-
-
-
Arbitrarily mark one terminal of one coil (D)
Assign a current into the dotted terminal
Use the right-hand rule to determine the
direction of magnetic flux
Arbitrarily pick one terminal of the second
coil (A) and assign a current into this
terminal
Use the right-hand to determine the direction
of magnetic flux
Compare the direction of the two fluxes.
If they have the same direction, place a dot
on the selected terminal (A) of the second
coil.
If the fluxes have opposite directions, place
the dot at the other terminal (B) of the
3
second coil.
Dot Convention: If a current enters the dotted terminal of one coil, the
reference polarity of the mutual voltage in the second coil is positive
at the dotted terminal of the second coil.
4
Energy Calculations
1 2 1 2
L1i1 + L2i2  Mi1i2
2
2
1 2 1 2
L1i1 + L2i2 - Mi1i2 ³ 0
2
2
Energy stored in magnetically coupled coils: w =
Since the stored energy cannot be negative
To complete the square, add and subtract i1i2 L1L2
(
1
i1 L1 - i2 L2
2

)
2
+ i1i2
(
)
L1L2 - M ³ 0
L1L2 - M ³ 0
or M £ L1L2
Coefficient of coupling k
M
k=
L1L2
0 £ k £1
5
Example 1
Calculate the phasor currents I1. and I2.
Mesh 1:
-12 + (- j 4 + j5)I1 - j3I 2 = 0
Mesh 2:
- j3I1 + (12 + j 6)I 2 = 0
Substituting:
Substituting:
or
jI1 - j3I 2 = 12
or I1 =
(12 + j 6)I 2
= (2 - j 4)I 2
j3
12
j14.04o
or I 2 =
A
= 2.91e
4- j
j (2 - j 4)I 2 - j3I 2 = 12
- j 63.43o
I1 = (2 - j 4)I 2 = 4.472 e
⋅ 2.91e
j14.04o
- j 49.39o
= 13.01e
A
6
Example 2
Find the steady-state expressions
for the currents ig and iL when
vg = 200 cos 10,000t V.
Find the coupling coefficient.
Phasor circuit:
200V  (5  j10)Ig  j5IL  0
Ig  (10  j15) A
 
(15  j10)IL  j5Ig  0 
IL  5 A
ig (t )  18.03cos(10,000t  56.31) A
iL (t )  5cos(10,000t ) A
k
M
0.0005

 0.5
L1L2
(0.001)(0.001)
7
Example 3
Determine the coupling coefficient and the energy stored in the coupled
inductors at t = 1.5 s.
Coupling coefficient: k =
M
L1L2
=
1
2 ⋅1
= 0.7071
8
(1)
(2)
9
Linear Transformers
Find the input and output
impedances of the transformer.
VS = ( Z S + R1 + j L1 )I1 - j MI 2
0 = - j MI1 + ( R2 + j L2 + Z L )I 2
VS
Z ab = Zin =
- ZS 
I1
Similarly,
j M
 I2 =
I1
R2 + j L2 + Z L
2M 2
Zin = R1 + j L1 +
R2 + j L2 + Z L
2M 2
Zcd = Zout = R2 + j L2 +
R1 + j L1 + Z S
10
Example 4
Find a) the input impedance of the transformer and b) the Thevenin
equivalent wrt terminals at V2.
æç
ö÷
2
 2M 2
1200
÷÷ 
= çç200 + j3600 +
Zin = R1 + j L1 +
R2 + j L2 + Z L ççè
100 + j1600 + 800 - j 2500 ø÷÷÷
æ
12002 (900 + j900) ÷ö÷
çç
÷÷  = (1000 + j 4400) 
= ç200 + j3600 +
çç
2
2
900 + 900
è
ø÷÷
æç
ö÷
2
 2M 2
1200
÷÷ 
= çç100 + j1600 +
Zout = R2 + j L2 +
R1 + j L1 + Z S ççè
200 + j3600 + 500 + j100 ÷÷÷ø
æ
12002 (700 - j3700) ö÷÷
çç
÷÷  = (171.09 + j1224.26)  = ZTh
= ç100 + j1600 +
çç
2
2
÷÷ø
700 + 3700
è
11
Example 4 cont’d
Note: The open-circuit voltage V2 will be j1200 times the value of I1.
I1 =
o
300V
= 79.67e- j 79.29 mA
(700 + j3700)
- j 79.29o
VTh = j1200 ⋅ 79.67e
mA=95.6 e
j10.71o
V
12
Example 5
Find the input impedance and the current from the voltage source.
13
Ideal Transformer
Linear Transformer
Ideal Transformer Properties:
k 1
L1 , L2  
R1 , R2  0
How can we analyze a circuit with an ideal transformer where there is no coil impedance
and no mutual impedance?
V
V2  jMI1;
I1  1
jL1

V2  jM

V1 V2

N1 N 2
V1

jL1
L1 L2
V1 
L1
L2
N
V1  2
L1
N1
14
Ideal Transformer
V1
N1
=
V2
N2
I1 N1 = I 2 N 2
or
V1
N
= 1
V2
N2
or
I1
N
= 2
I2
N1
If the coil voltages V1 and V2 are both positive or negative at the dot-marked
terminal, use a plus sign. Otherwise, use a negative sign.
If the Coil currents I1 and I2 are both directed into or out of the dot-marked
terminal, use a minus sign. Otherwise, use a plus sign.
15
Input Impedance
1:n
V2
V1 =
n
where
N2
n=
N1
I1 = nI 2
V1
1 V2
Zin = = 2
I1 n I 2
,
V2
1
= Z L  Zin = 2 Z L
I2
n
æ N1 ÷ö
Zin = ççç ÷÷ Z L
è N ÷ø
2
2
Transformers can be used for impedance matching.
Complex Power
V2
*
S1 = V I = (nI 2 ) = V2I*2 = S 2
n
In an ideal transformer, the complex power supplied to the
primary coil is delivered to the secondary coil without loss.
*
1 1
16
Example 6
The source voltage in the
phasor domain circuit is
25 kV. Find the amplitude
and phase angle of V2 and
I2.
I1 =
Vs
2
æ N1 ÷ö
ç
Z1 + çç ÷÷ Z 2
çè N 2 ÷ø
=
25 ⋅103
1500 + j 6000 + (25) (4 - j14.4)
2
A= (4+j3) A = 5e
j 36.87o
A
17
Example 7
An ideal transformer is rated as 2400/120 V, 9.6 kVA, and has 50 turns
on the secondary side. Calculate: (a) the turns ratio, (b) the number of
turns on the primary side, (c) the current ratings for the primary and
secondary windings.
(a) n =
V2
120
=
= 0.05
V1 2400
N2
50
(b) n =
 0.05 =
N1
N1
50
 N1 =
= 1000 turns
0.05
(c) S = V1I1 = V2 I 2 = 9.6 kVA
9600 9600
 I1 =
=
= 4A
2400
V1
9600 9600
I2 =
=
= 80 A
V2
120
18
Example 8
Ideal transformers are also used in power transmission from a
generating plant to a residence. Find the turns ratio for each of the ideal
transformers.
20k 220k

1
n1
 n1  11
220k 69k

1
n2
69k 13.8k

1
n3
13.8k 220

1
n4
 n2  0.314  n3  0.2  1 5  n4  1 62.7
19
Example 9
In the ideal transformer circuit, find Vo and the complex power supplied
by the source.
*
*
20
Example 10
The variable resistor is adjusted until
maximum average power is delivered to
it. Find the value of the resistor and the
power delivered to it. How much power
is associated with the ideal transformer?
Calculate Voc:
 840  60I  V1  20I1  0
Voc  V2  20I1
4I1  I2  0;

V1 4   V2 1
V1  840
Voc   840 4  20 ( 0)  210 180  Vrms
21
Calculate Isc and RTh:
 840  60I  V1  20 I1  Isc   0
20 Isc  I1   V2  0
4I1  I2 ;


V1 4   V2 1
Isc  6  6180  A rms
RTh  210 180  6180   35
Power analysis:
840  60I1  V1  20 I1  I2   0
35I2  20 I2  I1   V2  0
4I1  I2
V1 4   V2 1
 I1  0.75 A rms ;
PS   (840)(0.75)  630 W;
P20  20(3.75) 2  281.25 W;
I2  3  3180 A rms
P60  60(0.75) 2  33.75 W
P35  35(3) 2  315 W
PTransformer  PS  P60  P20  P35  0 (!)
22
Equivalent Circuit
The T-equivalent circuit is most commonly used for transformers.
é V1 ù é j L1
ê ú=ê
ê V2 ú ê j M
ë û ë
j M ù é I1 ù
úê ú
j L2 úû êëI 2 úû
By comparison:
é V1 ù é j ( La + Lc )
ê ú=ê
ê V2 ú ê
j Lc
ë û êë
La = L1 - M
ù é I1 ù
j Lc
úê ú
j ( Lb + Lc )úúû êëI 2 úû
Lb = L2 - M
Lc = M
23
Example 11
Determine the T-equivalent circuit of the linear transformer.
La = L1 - M = 10H - 2H = 8H
Lb = L2 - M = 4H - 2H = 2H
Lc = M = 2H
24