## Inductance of Transmission Line

### Introduction

• From knowledge of Basic Electrical Engineering, we know that a current-carrying conductor produces a magnetic field around it. These magnetic lines are concentric circles with their direction specified by right hand thumb rule.
• In AC transmission the nature of current is varying therefore we have to consider inductance in addition to resistance (R).
• The opposition offered to change in current is known as inductance.
• Inductance is zero for DC current.
• The varying magnetic field induces an EMF in a conductor, and its magnitude is given by-

$e=N\dfrac { d\phi }{ dt }$

If N = 1,

$e=\dfrac { d\phi }{ dt }$

EMF induced in an inductor due to alternating current is given by-

$e=L\dfrac { di }{ dt }$

From above two equation –

$L=\dfrac { d\phi }{ di }$

If we assumed permeability is constant that is flux linkages vary linearly with current.

$L=\dfrac { \phi }{ i }$

If we consider N, then the above equation becomes-
(in the above expression value of N is 1)

$N\phi =Li$

$\lambda =Li$

Where, $\lambda$= flux linkage Unit: Wb-turns.

$\lambda$ ,i are the RMS values.

Note:

$\phi$ = flux Unit: Wb

$\lambda$= flux linkage Unit: Wb-turns i.e. $N\times \phi$

### Flux linkage inside the conductor

Assumption-

• The path of the return current is far away from the conductor under consideration therefore it does not disturb the magnetic field of the conductor.
• Magnetic lines of flux are concentric with the conductor.
• Current is uniformly distributed over the cross section of conductor.
• Conductor and medium is considered non-magnetic that is relative permeability ($\mu_{r}$) is 1.
$L_{int}=0.5\times { 10 }^{ -7 }$

Unit: H/m

If relative permeability ($\mu_{r}$) is given or the conductor is not non-magnetic then –

$L_{int}=0.5\times { 10 }^{ -7 }{ \mu }_{ r }$

Unit: H/m

### Flux linkage outside the conductor

Assumptions – same as above

$L_{ ext }=2\times { 10 }^{ -7 }\ln { \dfrac { { D }_{ 2 } }{ { D }_{ 1 } } }$

Unit: H/m

Here relative permeability is 1 as it is outside the conductor and medium is air.

### Inductance of single phase two wire line

Assumptions –

• The conductor carries an equal current but in the opposite direction.
• A line of flux set up by the current in conductor 1, at a distance equal to or more than D + $r_2$ from the center of conductor 1, links a net zero current.
• The fraction of current linked by the above line of flux at a distance equal to or less than D – $r_2$ is 1.
• The fraction of current linked by the above line of flux between distance D- $r_2$ and D + $r_2$ varies from 1 to 0.
• All the external flux set up by current in conductor 1 links all the current up to the center of conductor 2 and that the flux beyond the center of conductor 2 does not link any current.
• Inductance of conductor 1 is given by –
$L_1= L_{ ext } + L_{ int }=2\times { 10 }^{ -7 }\ln { \dfrac { { D } }{ r_1{ e }^{ -\frac { 1 }{ 4 } } } }$
$L_1=2\times { 10 }^{ -7 }\ln { \dfrac { { D } }{ { r’_1 } } }$

Unit: H/m

Where, $r’_1=r_1{ e }^{ -\frac { 1 }{ 4 } } = 0.7788r_1$

or

$L_1=0.4605\times \log\dfrac{D}{r’_1}$

Unit: mH/km

Similarly, Inductance of conductor 2 is given by –

$L_2=2\times { 10 }^{ -7 }\ln { \dfrac { { D } }{ { r’_2 } } }$

Unit: H/m

The total inductance of a single phase two wire line is –

$L=L_1+L_2=4\times { 10 }^{ -7 }\ln { \dfrac { { D } }{ { r’ } } }$

Unit: H/m

Where, $r’_1 = r’_2 = r’$

Note: r’ can be assumed to be that of a fictitious conductor that has no internal flux but with the same inductance as that of a conductor with radius r.

$L = 0.921\log\dfrac{D}{r’}$

Unit: mH/km

Note: The inductance per conductor is one half of the loop inductance.

### The inductance of composite conductor lines

${ L }_{ m }=2\times { 10 }^{ -7 }\ln { [\frac { { [({ D }_{ a{ a }^{ ‘ } }{ D }_{ a{ b }^{ ‘ } }{ D }_{ a{ n } })({ D }_{ b{ a }^{ ‘ } }{ D }_{ b{ b }^{ ‘ } }{ D }_{ b{ n } })\ldots({ D }_{ m{ a }^{ ‘ } }{ D }_{ m{ b }^{ ‘ } }{ D }_{ m{ n } })] }^{ \frac { 1 }{ mn } } }{ { [({ D }_{ a{ a } }{ D }_{ a{ b } }{ D }_{ a{ m } })({ D }_{ b{ a } }{ D }_{ b{ b } }{ D }_{ b{ m } })\ldots({ D }_{ m{ a } }{ D }_{ m{ b } }{ D }_{ m{ m } })] }^{ \frac { 1 }{ { m }^{ 2 } } } } ] }$
${ L }_{ m }=2\times { 10 }^{ -7 }\ln { \dfrac { { D }_{ m } }{ { D }_{ s } } }$

Unit: H/m

${ L }_{ m }=0.4605\log { \dfrac { { D }_{ m } }{ { D }_{ s } } }$

Unit: mH/km

The above formula is very important please note the following points-

1. For numerator-
1. Numerator is (mn)th root of product m and n terms.
2. In the numerator, It is the distance from all the ‘m’ sub conductors of ‘M’ conductor to all ‘n’ sub conductors of conductor ‘N’.
3. The numerator is known as GMD (Geometric mean distance) it is written as Dm.
4. Note distance is calculated only once that is from ‘M’ to ‘N’ and not ‘N’ to ‘M’ (If you want to calculate inductance of Conductor ‘M’) remember this while solving numerical.
2. For Denominator-
1. The denominator is ($m^2$)th root. (Where, ‘m’ is the number of conductors in conductor ‘M’)
2. It is the distance between sub conductors of ‘M’.
3. The denominator is known as self GMD or GMR  (Geometric mean radius, $D_s$)
• Note in denominator distance Daa, Dbb, Dcc, ….Dmm will be replaced by r’ (GMR) Where, $r’= 0.7788r$
• Recall the expression for inductance of a single phase two wire line.
• $L=4\times { 10 }^{ -7 }\ln { \dfrac { { D } }{ { r’ } } }$ Where D = distance between conductors which is nothing but GMD; r’ = GMR.
• All formulae of inductance have term $\ln { \frac { GMD }{ GMR } }$
• Inductance of transmission line  is directly proportional to GMD.
• Inductance of transmission line is inversely proportional to GMR.
• Please remember this relation, we will recall it again in the bundled conductor.

### Inductance of three phase line with equilateral spacing

${ L }=2\times { 10 }^{ -7 }\ln { \dfrac { { D }_{ m } }{ r’ } }$

Unit: H/m

${ L }=0.4605\log { \dfrac { { D } }{ r’ } }$

Unit: mH/km

For stranded conductors r’ will be replaced by $D_s$.

${ L }=2\times { 10 }^{ -7 }\ln { \dfrac { { D }_{ m } }{ { D }_{ s } } }$

Unit: H/m

Note: In practice the conductors do not have equilateral spacing and hence flux linkage and inductance of conductor in each phase is not same.

### Inductance of three phase line with unsymmetrical spacing

$L=2\times { 10 }^{ -7 }\ln { \dfrac { \sqrt[3]{d_{12}d_{23}d_{31}} }{ r’ } }$

Unit: H/m

$L=0.4605\log { \dfrac { \sqrt[3]{d_{12}d_{23}d_{31}} }{ r’ } }$

Unit: mH/km

## Special case of double circuit transmission line

### Inductance of double circuit three phase line

$L=2\times 10^{-7}\ln\left[2^{\frac{1}{6}}\left(\dfrac{D}{r’}\right)^{\frac{1}{2}}\left(\dfrac{m}{n}\right)^{\frac{1}{3}}\right]$

Unit: H/phase/m

### Conductors situated at the corners of a regular hexagon

$L=10^{-7} \ln\dfrac{\sqrt{3}D}{2r’}$

## Transposition of transmission line

• Different inductance in each phase results in different voltage drop in each phase which causes an unbalanced circuit therefore the position of conductors is interchanged at a special transposition tower. Transposition of transmission line is the exchange of positions of the conductors, in order to reduce crosstalk and to make average inductance of all phases same.
• Modern transmission lines are normally not transposed However, intermediate switching stations, where the transposition takes place, are implemented whenever it is required.

### Advantages of transposition of transmission lines

• Transposition of transmission line makes it possible for each phase conductor to have the same average inductance.
• It reduces the disturbances in the adjacent communication lines.

For better understanding, you can view these images of the transposition tower.

Note: If you are preparing for a university examination, then for derivations of the above formulae you can refer NPTEL. For GATE exam above information is enough.

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