The document discusses self-induced emf, self-inductance, mutually induced emf, and mutual inductance. It states that self-induced emf is proportional to the rate of change of current through a coil and self-inductance is the proportionality constant. Mutually induced emf occurs when the magnetic flux passing through one coil changes due to a changing current in another nearby coil, and mutual inductance is the proportionality constant between the induced emf and rate of change of current. The coupling coefficient k describes the degree of coupling between two coils and is defined as the ratio of mutual flux to self flux.

1. ELE101/102 Dept of E&E,MIT Manipal 1
Self Induced emf & Self Inductance
The induced emf, e, in a coil is proportional to the rate
of the change of the magnetic flux passing through it
due to its own current. This emf is termed as Self
Induced EMF
The induced emf e is proportional to the rate of change
of current through coil and this proportionality constant
is called the self inductance, L.
The negative sign is used to indicate that EMF is
opposing the cause producing it
dt
di
Le1 −=

2. ELE101/102 Dept of E&E,MIT Manipal 2
Mutually Induced emf & Mutual Inductance
If two coils of wire are placed near
each other, a change of current in
one coil will induce emfs e1 in the first
coil and e2 in the second coil.
The induced emf, e2, in coil 2 is
proportional to the rate of the change
of the magnetic flux passing through
it and hence proportional to rate of
change of current in first coil and is
termed as Mutually induced EMF.
e2
e1

3. ELE101/102 Dept of E&E,MIT Manipal 3
The induced emf e2 is
proportional to the rate of change
of current through coil 1 and this
proportionality constant is called
the mutual inductance, M
The mutually induced emf is
expressed as
This induced emf can also be
expressed as
Self & Mutual Inductance…
M
N 1 N 2
L 1 L 2
φ 1 1 φ 1 2
φ 1 = φ 1 1 + φ 1 2
i 1
V 1 e 2
+
-
e 1
1
2
di
e M
dt
=
12
2 2
d
e N
dt
ϕ
=

4. ELE101/102 Dept of E&E,MIT Manipal 4
Therefore
If is constant, is constant
and
Unit: Henry (H)
Mutual Inductance…
M
N 1 N 2
L 1 L 2
φ 1 1 φ 1 2
φ 1 = φ 1 1 + φ 1 2
i 1
V 1 e 2
+
-
e 1
1
12
2NM
di
dφ
=
rµ
1
12
di
dφ
1
12
2
I
Φ
NM =

5. ELE101/102 Dept of E&E,MIT Manipal 5
Coupling Coefficient
1 2
1 1 2 2
1 2
1 2
2 12 1 21
1 2
12 1 21 2
2
1 2 2
1 2
Self Inductances and are
and
Mutual Inductance
where and
is thecoupling coeficie
L L
M
;
nt
or
N N
L L
I I
N N
M
I I
k k
k
M M
L L k
k L L
Φ Φ
= =
Φ Φ
= =
Φ = Φ Φ = Φ
= =

6. ELE101/102 Dept of E&E,MIT Manipal 6
Example
Coil 1 of a pair of coupled coils has a continuous current
of 5A, and the corresponding fluxes φ1 and φ12 are
0.6mWb and 0.4 mWb respectively. If the turns are
N1=500 and N2=1500, find L1, L2, M and k.
Ans:
 k = Φ12/Φ1 = 0.667
 M = N2 Φ12/I1=0.12H
 L1 = N1 Φ1/I1= 0.06 H
 L2 = 0.539H

7. ELE101/102 Dept of E&E,MIT Manipal 6
Example
Coil 1 of a pair of coupled coils has a continuous current
of 5A, and the corresponding fluxes φ1 and φ12 are
0.6mWb and 0.4 mWb respectively. If the turns are
N1=500 and N2=1500, find L1, L2, M and k.
Ans:
 k = Φ12/Φ1 = 0.667
 M = N2 Φ12/I1=0.12H
 L1 = N1 Φ1/I1= 0.06 H
 L2 = 0.539H