Chapter 2. Resistance and Inductance of Transmission Lines

Resistance and Inductance of Transmission Lines 19



The conductor resistance is affected by three factors: frequency, spiraling, and temperature.

The dc resistance of a stranded conductor is greater than the value given by eqn. (1.1)

because spiralling of the strands makes them longer than the conductor itself. The increase in

resistance due to spiralling is around 1% for three strand conductors and about 2% for

concentrically stranded conductors.

When an alternating current flows through a conductor, the current distribution is not

uniform over the conductor cross-sectional area and the degree of non-uniformity increases

with increase in frequency. The current density is greatest at the surface of the conductor. This

causes the ac resistance to be somewhat higher than the dc resistance. This effect is known as

skin effect. The ac resistance is usually referred as the effective resistance of the conductor.

The conductor resistance increases with the increase of temperature. Since the value of r

is given at a specific temperature and the line operates at higher temperature, the actual

resistance is higher than the value found in eqn. (2.1). For small changes in temperature, the

resistance increases linearly as temperature increases and the resistance at a temperature T is

given by

RT = Ro(1 + ao · T)

where



...(2.2)



RT = resistance at T°C

Ro = resistance at 0°C

a0 = temperature coefficient of resistance at 0°C.



By using eqn. (2.2), the resistance R2 at a temperature T2°C can be found if the resistance

R1 at a temperature T1°C is known, i.e.



R2

T + 1 ao

= 2

R1

T1 + 1 a o

For aluminium



...(2.3)



1/ao » 228



2.3 INDUCTANCE—BASIC CONCEPTS

From our basic understanding of electromagnetic field theory, we know that a conductor

carrying current has a magnetic field around it. The magnetic lines of force are concentric

circles having their centres at the centre of the conductor and are arranged in planes

perpendicular to the conductor.

The voltage induced in a conductor is given by



dy

volt

...(2.4)

dt

where y represents the flux linkages of the conductor in Wb-Turns. Eqn. (2.4) can be written in

the form

d y di

di

×

= L×

E=

...(2.5)

di dt

dt

E=



dy

is the inductance in Henrys.

di

In a linear magnetic circuit, flux linkages vary linearly with the current such that the

inductance remains constant and is given by



when L =



20 Electrical Power Systems



L=

or



y

Henry

i



y = Li



...(2.6)



If the current is alternating, eqn. (2.6) can be written as

l = LI



...(2.7)



where l and I are the rms values of flux linkages and current respectively.

Making use of Ampere’s law which relates magnetic field intensity H to the current Ienclosed.



z



H × dl = Ienclosed



...(2.8)



The flux density (wb/mt2) is given by

B=m·H



...(2.9)



–7



where m = mo · mr. mo = 4p × 10 Henry/mt is the permeability of free space and mr is the relative

permeability.

One can obtain the flux linked with the circuit by integrating the flux density over any

surface bounded by the conductors composing the circuit. Then using eqn. (2.6), the inductance

can be computed.



d

in eqn. (1.4) by jw, the steady state ac voltage drop due to alternating flux

dt

linkages can be obtained as

Replacing



V = jwLI = jwl



...(2.10)



Similarly, the mutual inductance between two circuits is defined as the flux linkages of one

circuit due to current in the second circuit, i.e.,

M21 =



l 21

Henry

I1



...(2.11)



The voltage drop in circuit 2 due to current in circuit 1 is given by

V2 = jw M21 · I1 = jwl21 volts



...(2.12)



The concept of mutual inductance is required while considering the coupling between

parallel lines and the effect of power lines on telephone lines.



2.4 INDUCTANCE OF A SINGLE CONDUCTOR

Transmission lines are composed of parallel conductors and can be assumed as infinitely long.

First we will develop expressions for flux linkages of an isolated current carrying cylindrical

conductor with return path lying at infinity. This will form a single turn circuit and magnetic

flux lines are concentric closed circles with direction given by the right-hand rule. To calculate

the inductance of a conductor, it is necessary to consider the flux inside the conductor as well

as the external flux. This division is helpful as the internal flux progressively links a smaller

amount of current as we proceed inwards towards the centre of the conductor and the external

flux always links the total current inside the conductor.



Resistance and Inductance of Transmission Lines 21



2.4.1 Internal Inductance

Figure 2.1 shows the cross-section of a long cylindrical

conductor of radius r carrying a sinusoidal current of rms

value I.

The mmf round a concentric closed circular path of

radius x internal to the conductor as shown in Fig. 2.1 is

where



z



H x × dl = Ix



...(2.13)



Hx = magnetic field intensity

(AT/m) at a distance x

meters from the centre of

Fig. 2.1: Flux linkages of a long

the conductor.

round conductor.

Ix = current enclosed (Amp) upto distance x.



Since the field is symmetrical, Hx is constant for all points equidistant from the centre.

Therefore, from eqn. (2.13), we have

2px Hx = Ix

Neglecting skin effect and assuming uniform current density we have

Ix =



FG IJ FG IJ

H K H K

px 2

pr



2



I =



x2

r2



...(2.14)



...(2.15)



I



From eqns. (1.14) and (1.15) we obtain,

Hx =



Ix



AT/m



2p r 2



...(2.16)



For a nonmagnetic conductor with constant permeability mo, the magnetic flux density Bx

at a distance x from the centre is

mo × I

x

Bx = mo · Hx =

...(2.17)

2p r 2

where mo is the permeability of free space (or air) and is equal to 4p × 10–7 H/m.

The differential flux dfx for a small region of thickness dx and one meter length of the

conductor is

dfx = Bx · dx · 1 =



mo × I



x · dx

...(2.18)

2p r 2

The flux dfx links only the fraction of the conductor. Therefore, on the assumption of

uniform current density, only the fractional turn (px2/pr2) of the total current is linked by the

flux, i.e.



Fx I

dl = G J df

Hr K

2



2



x



x



=



mo × I

2p r 4



x 3 dx



...(2.19)



Integrating from o to r, we get the total internal flux linkages as



z

r



lint =



o



moI

2p r



4



x 3 dx =



moI

Wb–T/m

8p



...(2.20)



22 Electrical Power Systems



or



lint =



or



4 p ´ 10 -7

× I Wb–T/m

8p



l int

1

= ´ 10 -7 H/m

I

2

is independent of the radius of the conductor.

Lint =



Note that Lint



...(2.21)



2.5 INDUCTANCE DUE TO EXTERNAL FLUX LINKAGE

Figure 2.2 shows two points P and Q at distances D1 and

D2 from a conductor which carries a current of I Amperes.

Since the magnetic lines of flux are concentric circles around

the conductor, whole of the flux between points P and Q

lies within the concentric cylindrical surfaces which pass

through these points. The field intensity at a distance x is

Hx =



I

AT/m

2p x



...(2.22)



and Flux density

Bx =



Fig. 2.2: Flux linkages between two

external points 2, 3.



mo × I

Wb/m2

2 px



...(2.23)



The flux outside the conductors links the entire current I and hence the flux linkage dlx is

numerically equal to the flux dfx. The flux dfx for a small region of thickness dx and one meter

length of the conductor is given by

mo × I

dx Wb/m length of the conductor

2 px

...(2.24)

Therefore, the total flux linkages of the conductor due to flux between points P and Q is



dlx = dfx = Bx · dx · 1 =



z



D2



lPQ =



D1



FG IJ Wb–T/m

H K



moI

m I

D

dx = o ln 2

2 px

2p

D1



...(2.25)



The inductance between two points external the conductor is then

Lext =



\



l PQ

I



=



FG IJ H/m

H K

F D IJ H/m

ln G

HD K



D2

mo

ln

2p

D1



Lext = 2 × 10–7



2



...(2.26)



1



2.6 INDUCTANCE OF A SINGLE PHASE TWO WIRE LINE

Figure 2.3 shows a single phase line consisting of two solid round conductors of radius r1 and

r2 spaced distance D apart. The conductor carry equal currents but in the opposite directions.



Resistance and Inductance of Transmission Lines 23



These currents set up magnetic field lines that links between the conductors as shown in

Fig. 2.3.

Inductance of conductor 1 due to internal flux

is given by eqn. (2.21). As a simplifying assumption

we can assume that all the external flux set up by

current in conductor 1 links all the current upto

the centre of conductor 2 and that the flux beyond

the centre of conductor 2 does not link any current.

This assumption gives quite accurate results

especially when D is much greater than r1 and r2.

Fig. 2.3: Single phase two wire lines.

Thus, to obtain the inductance of conductor 1 due

to the external flux linkage, substituting D1 = r1

and D2 = D in eqn. (2.26).

L1(ext) = 2 × 10–7 ln



FG D IJ H/m

Hr K



...(2.27)



1



The total inductance of conductor 1 is then

L1 = Lint + L1(ext)

\



L1 =



D

1

´ 10 -7 + 2 ´ 10 -7 ln

r1

2



FG 1 + ln D IJ

H4 r K

F

DI

= 2 × 10 G ln e + ln J

r K

H

F D IJ H/m

= 2 × 10 ln G

Hr e K

F DI

= 0.4605 log G J mH/km

H r¢ K

= 2 ´ 10 -7



1



1

4



–7



1



–7



1



-1 4



1



where



...(2.28)

...(2.29)



r1¢ = r1 e -1/4 = 0.7788 r1



The radius r1¢ is the radius of a fictious conductor which has no internal inductance but has

the same total inductance as the actual conductor.

Similarly, the inductance of conductor 2 is

L2 = 0.4605 log

The total inductance of the circuit is

L = L1 + L2



FG D IJ mH/km

H r¢ K

2



...(2.30)



24 Electrical Power Systems



= 0.921 log

If the two conductors are identical, i.e.,



IJ mH/km

r r J

K



FG

GH



D



¢

1



...(2.31)



¢

2



r1¢ = r2¢ = r¢, then

L = 0.921 log



FG D IJ mH/km

H r¢ K



...(2.32)



Eqn. (2.32) gives the inductance of a two wire line in which one conductor acts as a return

conductor for the other. This is known as loop inductance.

From eqn. (2.29), the inductance of conductor 1 can be written as:



FG

H



L1 = 0.4605 log



D

1

+ 0.4605 log

r¢

1



Similarly, the inductance of conductor 2,



FG

H



L2 = 0.4605 log



D

1

+ 0.4605 log

r¢

1



IJ mH/km

K

IJ mH/km

K



...(2.33)



...(2.34)



As both the conductors are identical, hence we can write L1 = L2 = L. Therefore, inductance

per phase per km length of the line is given by



FG

H



L = 0.4605 log



D

1

+ 0.4605 log

r¢

1



IJ mH/km

K



...(2.35)



From eqn. (2.35), it is clear that the first term is only a function of the fictious radius of the

conductor. The first term of eqn. (2.35) can be defined as the inductance due to both the internal

flux and that external to conductor 1 to a radius of 1 mt. The second term of eqn. (2.35) is

dependent only upon the conductor spacing and this is known as inductance spacing factor.



2.7 SELF AND MUTUAL INDUCTANCES

The inductance per phase for the single-phase two wire line (Fig. 2.3) can also be expressed in

terms of self inductance of each conductor and their mutual inductances. Let us consider the

single phase circuit represented by two coils characterized by the self inductances L11 and L22

and the mutual inductance M12.

Figure 2.4 shows the single-phase line viewed as two

magnetically coupled coils and the magnetic polarity is

shown by dot symbols.

The flux linkages l1 and l2 can be written as:

l1 = L11I1 + M12I2

l2 = M21I1 + L22I2

Since



...(2.36)

...(2.37)



I2 = –I1, we get

l1 = (L11 – M12)I1



...(2.38)



l2 = (–M21 + L22)I2



...(2.39)



Fig. 2.4: The single phase two wire

lines viewed as two magnetically

coupled coils.



Resistance and Inductance of Transmission Lines 25



Therefore, we can write

L1 = L11 – M12



...(2.40)



L2 = –M21 + L22



...(2.41)



Comparing eqns. (2.40) and (2.41) with eqns. (2.33) and (2.34), we get



FG IJ mH/km

H K

= 0.4605 log F 1 I mH/km

GH D JK



L11 = L22 = 0.4605 log 1

r¢



...(2.42)



M12 = M21



...(2.43)



The above described approach of self and mutual inductances can be extended to a group of

conductors. Let us consider n conductors carrying phasor currents I1, I2, ..., In, such that

I1 + I2 + ... + In = 0



...(2.44)



Generalize formula for the flux linkages of conductor i is given by

li = Lii I i +



or we can write



n



å M ij I j



...(2.45)



j =1

j¹i



FG 1

I

1 J

l = 0.4605 G I log + å I log

mWb–T/km

D J

GH r

JK

n



i



i



¢

i



j =1

j ¹i



j



...(2.46)



ij



2.8 TYPE OF CONDUCTORS

So far transmission lines consisting of single solid round conductors for forward and return

paths have been considered. Transmission line conductors used in practice are always stranded

to provide the necessary flexibility for stringing. Stranded conductors are also known as composite

conductors as they compose of two or more elements or strands electrically in parallel. The

conductors used for transmission lines are stranded copper conductors, hollow copper conductors

and ACSR conductors. In overhead transmission lines, ACSR conductor is most commonly

used. The low tensile strength of aluminium conductors is made up by providing central strands

of high tensile strength steel. Such a conductor is known as aluminium conductor steel reinforced

(ACSR).

ACSR conductors have the few advantages:

1. It is cheaper than copper conductors of equal resistance.

2. Corona losses are reduced because of the larger diameter of the conductor.

3. It has superior mechanical strength and hence span of larger lengths which results in

smaller number of supports for a particular length of transmission line.

The total number of strands (S) in concentrically stranded conductor with total annular

space filled with strands of uniform diameter (d) is given by

S = 3y2 – 3y + 1



...(2.47)



26 Electrical Power Systems



where y is the number of layers where in the single central strand is counted as the first

layer. The overall diameter (D) of a stranded conductor is

D = (2y – 1)d



...(2.48)



Figure 2.5 shows the cross-sectional view of an

ACSR conductor with 24 strands of aluminium and 7

strands of steel.

Expanded ACSR conductors are used in extra high

voltage (EHV) transmission line. By the use of a filler

such as paper or hessian between various layers of

strands so as to increase the overall conductor diameter

to reduce the corona loss and electrical stress at

conductor surface. Bundled conductors are commonly

used in EHV transmission line to reduce the corona

loss and also reduce the radio interference with

communication circuits.



Fig. 2.5: Cross-sectional view of

ACSR conductor (7 steel strands and

24 aluminium strands).



2.9 INDUCTANCE OF COMPOSITE CONDUCTORS

In the previous sections, solid round conductors

were considered for the calculation of

inductance. However, stranded conductors are

used for practical transmission lines. Figure 2.6

shows a single phase line comprising composite

conductors X and Y. The current in X is I

referenced into the page and the return current

in Y is –I. Conductor X having n identical strands

or subconductors, each with radius rx. Conductor

Fig. 2.6: Single phase line consisting of

two composite conductors.

Y having m identical strands or subconductors

with radius ry. It is assumed that the current

is equally divided among the subconductors. Thus, each subconductor of X, carry a current I/n

and each subconductor of Y, carry a current –I/m.

Applying eqn. (2.46) to subconductor a, we get

la = 0.4605



–0.4605



or



R

|

S

|

T



I

1

1

1

1

1

+ log

+ log

+ . . . + log

log + log

¢

n

Dab

Dac

Dad

Dan

rx



R

S

T



I

1

1

1

1

+ log

+ log

+ ... + log

log

m

Daa¢

Dab¢

Dac¢

Dam



R(D × D

|

l = 0.4605 I log S

| (r D D

T

a



aa¢



¢

x



ab ¢



ab



The inductance of subconductor a is



ac



1



× Dac ¢ .... Dam ) m



1



Dad .... Dan ) n



U

|

V

|

W



U

V

W



U

|

V

|

W



...(2.49)



Resistance and Inductance of Transmission Lines 27



R(D × D

|

L = 0.4605 n log S

| (r D D

T

aa¢



a



¢

x



1



× Dac¢ .... Dam ) m



ab ¢



ab



ac



Dad .... Dan )



1

n



U

|

V

|

W



...(2.50)



The average inductance of any one subconductor of composite conductor X is:

Lavg =



La + L b + Lc + ... + Ln

n



...(2.51)



Since conductor X is composed of n subconductors electrically in parallel, its inductance is

Lx =



Lavg

n



=



La + Lb + Lc + ... + Ln



...(2.52)



n2



Substituting the values of La, Lb, Lc, ..., Ln in eqn. (2.52) we get,

Lx = 0.4605 log



l

= l( D



FG D IJ mH/km

HD K

m



...(2.53)



SX



q



Dm = ( Daa¢ Dab¢ .... Dam ) .... ( Dna¢ Dnb¢ .... Dnm )



where



DSX



aa



q



Dab .... Dan ) .... ( Dna Dnb .... Dnn )



Daa = Dbb = .... = Dnn = rx¢



where



1/n 2



1/mn



...(2.54)

...(2.55)



Dm is the mn th root of the mn terms, which are the products of all possible mutual distances

from the n subconductors of conductor X to m subconductors of conductor B. It is called the

mutual geometric mean distance (mutual GMD). DSX is the n2 root of the product of n2 terms

consisting of rX¢ of every strand times the distance from each strand to all other strands within

group X. The DSX is defined as the self geometric mean distance (self GMD) of conductor X.

The inductance of the composite conductor Y can also determined in a similar manner.

In this case, mutual GMD will remain same, i.e., Dm is same but self GMD DSY will be

different.



2.10 INDUCTANCE OF THREE PHASE TRANSMISSION

LINES WITH SYMMETRICAL SPACING

Figure 2.7 shows the conductors of a three phase transmission

line with symmetrical spacing. Radius of conductor in each

phase is r.

Using eqn. (2.46), the total flux linkage of conductor in

phase a is given by



FGH



IJK



1

1

1

+ I b log + I c log

...(2.56)

r¢

D

D

Assuming balanced three phase currents, we have



la = 0.4605 I a log



Ia + Ib + Ic = 0

or



Ib + Ic = –Ia



...(2.57)



Fig. 2.7: Three-phase line with

symmetrical spacing.



28 Electrical Power Systems



Using eqns. (2.56) and (2.57), we get



FG

H



la = 0.4605 I a log



1

1

- I a log

r¢

D



IJ

K



FG D IJ mWb–T/km

H r¢ K

F DI

= 0.4605 log G J mH/km

H r¢ K



\



la = 0.4605 Ia log



Therefore,



La =



la

Ia



...(2.58)



...(2.59)



Because of symmetry, la = lb = lc and hence three inductances are identical, i.e.,

Lb = Lc = La.



2.10.1



Inductance of Three Phase Transmission Lines with

Asymmetrical Spacing



In actual practice, the conductors of a three phase

transmission line are not at the corners of an equilateral

triangle because of construction considerations. Therefore

with asymmetrical spacing, even with balanced currents,

the flux linkages and inductance of each phase are not the

same. A different inductance in each phase, resulting in

unbalanced receiving-end voltages even when sendingend voltages and line currents are balanced. Figure 2.8

Fig. 2.8: Three phase line with

shows the conductors of a three phase transmission line

asymmetrical spacing.

with asymmetrical spacing.

Using eqn. (2.46) will result in the following flux linkages.



R 1 + I log 1 + I log 1 U

S r¢

V

D

D W

T

R 1 + I log 1 + I log 1 U

l = 0.4605 SI log

V

r¢

D W

T D

R 1 + I log 1 + I log 1 U

l = 0.4605 SI log

V

D

r¢ W

T D

LM log 1 log 1 log 1 OP LI O

LMl OP

MM r¢ D D PP MM PP

MMl PP = 0.4605 Mlog 1 log 1 log 1 P MI P

MM PP

MM D r¢ D PP MM PP

MNl PQ

MMlog D1 log D1 log r1¢ PP MN I PQ

Q

N

la = 0.4605 I a log



b



a



c



a



ab



ca



a



or in matrix form



b



b



c



ab



b



b



ab



ab



ca



c



bc



ca



bc



ca



bc



Therefore symmetrical inductance matrix L is given by



...(2.61)



bc



c



...(2.62)



a



b



c



c



...(2.60)



...(2.63)



Resistance and Inductance of Transmission Lines 29



LM log 1

MM r¢

L = 0.4605 Mlog 1

MM D

MMlog D1

N



log



1

1

log

Dab

Dca



log



ab



log



ca



1

r¢



1

D bc



log



1

D bc



log



1

r¢



OP

PP

PP

PP

PQ



mH/km



For balanced three-phase currents with Ia as reference, we have

Ib = a2 Ia

Ic = a Ia

where the complex operator a = 1 120° and a2 = 1 240° . Note that a3 = 1.

Using eqns. (2.63) and (2.65), we get



FG

IJ

H

K

F 1 + log 1 + a log 1 IJ

= 0.4605 G a log

r¢

D K

H D

F

1

1

1I

= 0.4605 G a log

+ a log

+ log J

D

r¢ K

H D



...(2.64)



...(2.65)



La =



la

1

1

1

= 0.4605 log + a 2 log

+ a log

Ia

r¢

Dab

D ca



...(2.66)



Lb =



lb

Ib



...(2.67)



Lc =



lc

Ic



2



ab



bc



2



ca



bc



...(2.68)



Equations (2.66), (2.67) and (2.68) show that the phase inductances are not equal and due

to mutual inductance they contain imaginary terms.



2.11 TRANSPOSE TRANSMISSION LINE

As mentioned in the previous section, asymmetrical spacing gives complex values of phase

inductances, which makes the study of power system difficult. However, one way to regain

symmetry in good measure and obtain a per phase model by exchanging the positions of the

conductors at regular intervals along the line such that each conductor occupies the original

position of every other conductor. Such an exchange of conductor positions is called transposition.

The transposition is usually carried out at switching stations. A complete transposition cycle is

shown in Fig. 2.9. This arrangement causes each conductor to have the same average inductance



Fig. 2.9: Transposition cycle of three-phase line.