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1. PresentedBy:
AnisurRahman,
ID: 1402EEE00033
Department of EEE
Manarat InternationalUniversity(MIU)
Ashulia,Khagan,Savar,Dhaka,Bangladesh

2. Definition Of Inductance
Flux Linkages of Conductors
1. Flux linkages inside the conductor
2. Flux linkages outside the conductor
 Flux linkages of one conductor in a group of
conductors
 Inductance of a single phase two wire line
 Inductance of 3-phase overhead line
 Bundled conductors

3. •

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
The internal inductance per meter is
 int
int
8
H ml
I
 

  
 
7
70
int
4 10
8 8
H ml
 
 




    
If the relative permeability of the conductor is 1 (non-ferromagnetic
materials, such as copper and aluminum), the inductance per meter
reduces to

6. The external inductance per meter is :
The total external flux linkages per meter can be found via integration
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.

7. Theoretically , the flux due to a conductor extends from
the center of the conductor to right unto infinity.
Assuming that the flux linkages will extend unto a point P
very far from the group of the conductors , and the
distances are as shown in fig.

8. Consider a group of conductors 1, 2, 3,…… n such that the
sum of currents in all these conductors is zero.
If the currents carried by respective strands are I1, I2, I3, …In
we have I1+I2+I3+…+In = 0
 ψ1p1 = All flux linkages of conductor 1 due to tis own current
I1 , internal and external , unto point P ….
Ψ1p1 = 2 × 10 I1 ln D1p/r1‘ Wb. T/Mt
Ψ1p2 = Flux linkages with conductor 1 due to current in
conductor-2
Ψ1p2 = 2 × 10 × I2 ln D2p/D12
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9. Similarly ψ1p3…….ψ1pn
Ψ1p=
2 × 10 { I1 ln D1p/r'1 + I2 ln D2p/D12 +...+ In ln Dnp/D1n}
Flux linkages with conductor 1 due to I1 , I2…..In
Net flux linkages ψ1p
Ψ1p = 2 × 10 { I1 ln 1/r1'+ I2 ln 1/D12 +…+ In ln 1/D1n }
Wb-turns/m
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10. The 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

11. 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


 
     
 
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


 
   
 
 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
‘
‘
‘

12. In a 3-phase transmission line, the inductance of each
conductor is considered instead of loop inductance.
The conductor of a 3-phase overhead line may be placed
symmetrically or unsymmetrically on the towers.
With Symmetrical Spacing :
A 3-phase line in which the space between any two
conductor is the same as shown in fig. the line is called
symmetrical line.
Fig. shows the conductor of a 3-phase line conductor has
a radius r meters and spacing between the conductors is
D meters.

13. Under balanced three-phase phasor currents, the algebraic
sum of the currents in the conductors is zero.
Hence, Ia + Ib + Ic = 0
The flux linkages of the conductor ‘a’ are
ψa = 2 × 10 [ Ia ln 1/Daa + Ib ln 1/Dab + Ic ln 1/Dac ] Wb-T/m
Inductance of conductor a,
La = ψa/Ia
= 2 × 10 ln D/r‘ H/m
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14. A 3-phase line in which the space between the conductors is
different as shown in fig., the line is called unsymmetrical line.
Consider 3-phase line with conductors a, b and c each of
radius r meters.

15. Let, the spacing between them be Dab, Dbc and Dca and the
currents flowing through them be Ia, Ib and Ic respectively as
shown in fig.
From fig, the flux linkages of the conductor a
Ψa = 2 × 10 [ Ia ln 1/r' + Ib ln 1/D12 + Ic ln 1/D31 ] Wb-T/m
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16. A bundle conductor is a conductor made up of two or
more sub-conductors and is used as one phase
conductor.
Lines of 400kv and higher voltages invariably use
bundled conductors.
Sub-conductors of a bundled conductor are separated
from each other by a constant distance varying from 0.2
m to 0.6 m depending upon designed voltage and
surrounding conditions throughout the length of the line
with the help of spacers.

17. It reduces corona loss.
It reduces radio interference
The bundled conductor lines transmit bulk power with
reduced losses, thereby giving increased transmission
efficiency
Bundle conductor lines have a higher capacitance to
neutral so they have higher charging current, which helps
in improving power factor
By bundling, the GMR is increased, the inductance per
phase is reduced. As a result reactance per phase is
reduced.