Electrical Circuits II

(ECE233b)

Magnetically Coupled Networks

Anestis Dounavis

The University of Western Ontario

Faculty of Engineering Science
Mutual Inductance
i Ampere’s law predicts that the flow of
electric current will create a magnetic filed.
+
V N  Faraday’s law predicts that time varying
- magnetic field will create a voltage potential.
d
vt 
dt
where vt is the voltage per turn induced by a time-varying
magnetic flux , through a coil.

A coil having N turns induces a total voltage of

d dλ
v  Nvt  N 
dt dt
where  is called the flux linkage and =N (weber-turn).
Mutual Inductance
i d dλ
v  Nvt  N  ; λ  N
+ dt dt
V N 
- The flux linkage  induced by the current i
has the following relationship in a constant
permeable medium
λ  Li
where L is a constant, referred to as the inductance (henry).

The voltage induced can know be written as

dλ di
i v L
+ dt dt
V L or if time variation is sinusoidal
-
V  jLI (phasor representation)
Mutual Inductance
i1 i2
If the magnetic field that induces the
+ + current is produced by some external
v1 v2 source (i.e. adjacent circuit) then this
- - produces a mutual inductance
11 22
12 21
 There are no electrical connections between adjacent circuits
 A portion of the magnetic flux is shared between the coils
The total fluxes are
λ1  L11 i1  L12 i 2
λ2  L21i1  L22 i 2
L11  L1
For linear circuits: L12  L21  M & substituting
L22  L2
Voltages of coupled network become
dλ1 di1 di 2 i1 M i2
v1   L1 M + +
dt dt dt Circuit
dλ2 di1 di 2 diagram v1 L1 L2 v2
v2  M  L2 - -
dt dt dt
 Polarity (dot convention)
i1 M i2
Dot convention indicates if 11 and 22
+ +
are added or subtracted
v1 L1 L2 v2
- -
Circuit equations for voltages and currents adhere to passive
element sign convention
i i
Example + +
v R v = iR v R v = -iR
- -

 If reference current enters the dot, the induced voltage is

positive at the dot in the other coil.
 If reference current leaves the dot, the induced voltage is
negative at the dot in the other coil.
 Polarity (dot convention)
 If reference current enters the dot, the induced voltage is
positive at the dot in the other coil.
 If reference current leaves the dot, the induced voltage is
negative at the dot in the other coil.
Case 1
i1 M i2 i1 L1 L2 i2
di1 di 2
+ + + + v 1  L1 M
di2 + + di1 dt dt
v1 L1 L2 v2 v1 M dt - -
M v2 di1 di 2
dt
- - - - v2  M  L2
dt dt
Case 2
i1 M i2 i1 L1 L2 i2
di1 di 2
+ + + v 1  L1 M
di2 - - di1 + dt dt
v1 L1 L2 v2 v1 M dt + +
M v2 di 1 di 2
dt
- - - - v 2  M  L2
dt dt
Example 1
Determine the equations for v1 and v2 for i1 M i2
the circuit shown - +
v1 L1 L2 v2
+ -

Example 2
Determine the equations for v1 and v2 for i1 M i2
the circuit shown - +
v1 L 1 L2 v2
+ -
Example 3
Two coupled coils can be interconnected in 4 possible ways (see
Example 2 pp. 320). L L
1 2

For the following interconnection determine the equivalent

inductance
L1 L2

+ V -
Example 4
Find the current I1 and I2 and the output voltage Vo, for the circuit
shown

4 I1 j1 2 I2

+
-
24|00 j4 j8
+ -j4 Vo
-
Example 5

Write the KVL equations in the standard form for the network
shown

L1 I1 I2 L2

R2
R1 R3
+
V1 -
Energy Analysis
The energy stored in a circuit is defined as t1 t1

E   p(t)dt   v(t)i(t)dt
0 0

For magnetically coupled inductors, the total energy in a network

can be expressed as (derivation pp 323-325)
i1 M i2
1 2 1 + +
E(t)  L1 i1  L2 i 2  Mi1 i2
2

2 2 v1 L1 L2 v2
- -

Or if one of the dot is at the bottom i1 i2

M
1 2 1 + +
E(t)  L1 i1  L2 i 22  Mi1 i2 L1 L2
2 2 v1 v2
- -
Energy Analysis
Therefore, the general form of the energy stored in the
magnetically coupled inductors at any instant of time is
1 2 1
E(t)  L1 i1  L2 i 22  Mi1 i2
2 2
The energy expression can be rewritten in matrix form as:
1  L1  M   i1  
E(t)  i1 i2     
2  M L2  i2  
The two coupled inductors represent a passive network, therefore
the energy stored within this network must be nonnegative

This means that L1>0, L2>0 and the matrix  L1 M

 M L2 

is nonnegative definite.
Energy Analysis
Since the matrix  L1  M  is nonnegative definite, then
 M L2 

the determinant of the matrix must also be nonnegative definite
  L1  M 
det    
  L L  M 2
0
  M
1 2
L2  
Hence for a passive network | M | L1 L2
M
The coupling between the 2 inductors is defined as k 
L1 L2
where k is defined as the coupling coefficient and ranges from
No flux common to 0  k 1 All flux common to 2 coils
the 2 coils (i.e. M=0) (no “leakage” flux)
In practice k  1 in transformers
Example 6
The network shown operates at 60 Hz. Compute the energy
stored in the mutually coupled inductors at time t=10ms.

2 I1 j1 2 I2

+
12|300 j j2
- -j2
Ideal Transformers
Assume lossless magnetic core (no hysteresis or eddy currents
and perfect coupling (k=1))

i(t)1  i(t)2
+ +
v(t)1 N1 N2 v(t)2
-  -

If the same flux  goes through each winding then

Faraday’s law: d d
v 1  N1 v 2  N2
dt dt

The ratio of these two equations yield v1 N1


v 2 N2
 Ideal Transformers
Using Amperes Law (assuming ideal magnetic core with infinite
permeability), it can be shown that
i1 - N2
N1i1  N2 i 2  0 or 
i2 N1
Equivalent Circuit Diagram
i1 N1:N2 i2
+ + v1 N1 i1 - N2
v1 L1 L2 v2 where  
v 2 N2 i2 N1
- -
Usually the direction of i2 is often reversed:
i1 1:n i2
+ + v1 N1 1 i1 N2
v1 v where    n
L1 L2 2 v 2 N2 n i 2 N1
- -
v1 source primary source n turn ratio
v2 source secondary source
Impedance Transformation
Consider the following circuit: i1 1:n i2
+ +
What’s the input impedance? v1 v2 ZL
Z1 L1 L2
v1 v2 n 1 v2 1 - -
Z1    2  2 ZL
i1 ni 2 n i2 n v2
ZL 
Summary i2
v2 ZL N2
i1  ni 2 v1  Z1  2 where n
n n N1
Note that the signs of these equations depend on the polarity
w.r.t. the dot. For example:
i1 1:n i2 i1  ni 2
ZL
+ + v2 Z1  2
v1 v2 ZL
v1   n
Z1 L1 L2 n
- - Voltages and currents are
reversed (not the impedance)
 Impedance Transformation
Consider the following circuit:
i1 1:n i2
Z1 Z2
+ +
VS1 +- v1 v2
+
- VS2
- -
To simplify analysis the transformer and load can be replaced
with equivalent circuit
Equivalent Circuit Reflected to Secondary
i1 i2 i2 = 0, since far end
1:n
Z1 + terminals are open circuited
+ +
VS1 +- v1 v2 vTH i1 = ni2 = 0
- - - vTH = v2 = nv1 = nVS1
i1 1:n i2
Z1 v2 = nv1 = 0
+ +
+
VS1 - v1 v2 iN i 1 VS1
iN  i2  
- - n nZ1
Impedance Transformation
vTH
ZTH   n 2 Z1
iN
The original circuit
i1 1:n i2
Z1 Z2
+ +
+ +
VS1 - v1 v2 - VS2
- -

can be modeled with the following equivalent circuit

i2
n2Z12 Z2
+ +
nVS1 - - VS2
Impedance Transformation
Equivalent Circuit Reflected to Primary
i1 1:n i2
Z1 Z2
+ +
+ +
VS1 - v1 v2 - VS2
- -

In a similar manner we can replace the ideal transformer and its

secondary circuit with an equivalent circuit as shown

i1
Z1 Z2 /n 2
+ +
VS1 - - (VS2)/n
Example 7
Compute the current I1 and the output voltage Vo for the circuit
shown

2 -j2 I1 2 -j2
1:2
+ + +
+
12|00 - V1 V2 2 Vo
- - -
Example 8
For the circuit shown, find Vo

2 -j2
4|00
1:2

+
-
+
+
12|00 - 2 Vo
-
Example 9
For the circuit shown, find I1

2 -j2
12|00
I1
1:2

+
-
+
36|00 - 2