Definition of Inductance, Flux linkages of current carrying conductor, Inductance of a single phase two wire line

1. Presentation
on
Inductance ofa TransmissionLines
(Definition of Inductance, Flux linkages of current carrying
conductor, Inductance ofa single phase two wire line)

2. CONTENTS
 Defination Of Inductance
 Flux Linkages of Conductors
1. Flux linkages inside the conductor
2. Flux linkages outside the conductor
 Inductance of a single phase two wire line
 General value of Inductance & Capacitance in Transmission
Lines

3. WHAT IS INDUCTANCE ??

4. 4

5. Consider a conductor of
radius r carrying a current I.
At a distance x from the
center of this conductor, the
magnetic field intensity Hx
can be found from Ampere’s
law:
x xH dl I× =∫Ñ
1. Flux linkages inside the conductor

6.  Where Hx is the magnetic field intensity at each point along a closed path, dl is a
unit vector along that path and Ix is the net current enclosed in the path.
 For the homogeneous materials and a circular path of radius x, the magnitude of Hx
is constant, and dl is always parallel to Hx. Therefore:
2
2
x
x x x
I
xH I H
x
π
π
= ⇒ =
 We had already assumed uniform current
density ;
2
2x
x
I I
r
π
π
=
 Thus, the magnetic intensity at radius x
inside the conductor is
[ ]2
2
x H m
x
H I
rπ
=

7.  The flux density at a distance x from the center of the conductor is
2
[ ]
2
x x T
xI
B H
r
µ
µ
π
= =
 The differential magnetic flux contained in a circular tube of thickness dx
and at a distance x from the center of the conductor is
2
[ ]
2
xI
d dx Wb m
r
µ
φ
π
=
The flux linkages per meter of length due to flux in the tube is the product of
the differential flux and the fraction of current linked:
2 3
2 4
[ ]
2
Wb turns m
x x I
d d dx
r r
π µ
λ φ
π π
−= =
dφ = Bx * (dx * 1)

8. The total internal flux linkages per meter can be found via integration…
3
int 4
0
[ ]
2 8
r
Wb turns m
x I I
d dx
r
µ µ
λ λ
π π
−= = =∫ ∫
Therefore, the internal inductance per meter is
[ ]int
int
8
H ml
I
λ µ
π
= =
[ ]
7
70
int
4 10
8 8
H ml
µ π
π π
−
−×
= = = 0.5×10
If the relative permeability of the conductor is 1 (non-ferromagnetic
materials, such as copper and aluminum), the inductance per meter
reduces to

9. To find the flux linkages external to a conductor, we
will consider the portion of flux between two points
P1 and P2 that lie at distances D1 and D2 from the
center of the conductor.
In the external to the conductor region, the magnetic
intensity at a distance x from the center of conductor
is
2 2
x
x
I I
H
x xπ π
= =
The flux density at a distance x from the center of
conductor is
2
x x
I
B H
x
µ
µ
π
= =

10. The differential magnetic flux contained in a circular tube of thickness
dx and at a distance x from the center of the conductor is :
[ ]
2
I
d dx Wb m
x
µ
φ
π
=
The total external flux linkages per meter can be found via integration
2 2
1 1
1
2
[ ]ln
2 2
D D
ext
D D
Wb turns m
DI I
d dx
x D
µ µ
λ λ
π π
−= = =∫ ∫
The external inductance per meter is :
[ ]t 2
t
1
ln
2
ex
ex H m
D
l
I D
λ µ
π
= =

11. Inductance of a single-phase 2-wire
lineThe inductance of a single-phase line
consisting of two conductors of radii r
spaced by a distance D and both
carrying currents of magnitude I
flowing into the page in “A” conductor
and out of the page in the “B“
conductor.
x xH dl I× =∫Ñ
Since the path of radius x2 encloses both
conductors and the currents are equal
and opposite, the net current enclosed is
0 and, therefore, there are no
contributions to the total inductance
from the magnetic fields at distances
greater than D.
A B

12. Inductance of a single-phase 2-wire transmission
line
The total inductance of a line per unit length in this transmission line is a sum of the
internal inductance and the external inductance between the conductor surface (r)
and the separation distance (D):
[ ]int
1
ln
2 4
ext H m
D
l l l
r
µ
π
 
= + = + ÷
 
By symmetry, the total inductance of the other line is the same, therefore, the total
inductance of a two-wire transmission line is
[ ]
1
ln
4
H m
D
l
r
µ
π
 
= + ÷
 
 Where r’ =is GMR (Geometric mean radius)
 For a solid conductor G.M.R = 0.7788 times the radius of conductor.
 D is the distance between conductors
‘
‘
‘

13. Example based on Inductance of Single phase
2 wire tranmission line

15. Questions ..??
1) When the flux increases , inductance will
-------- ??
2) GMR is abbreviated as _________ and it is
equals to ______ times radius of conductor ??
3) If current is 2A and inductance is 4H , what
will be the value of flux ??

16. Answers
A1) Increase
A2) GEOMETRIC MEAN RADIUS &
GMR = 0.7788 times radius of
conductor
A3) 8 Wb

17. SPECIAL
THANKS TO
YOGESH SIR