INDUCTANCE & CAPACITANCE CALCULATIONS

By
D.Rajesh Asst. Prof.
EEE Department
15 October 2020 D.Rajesh Asst.Prof. EEE Department 1

 Types of conductors, line parameters,
 Calculation of inductance and capacitance of
single and double circuit transmission lines,
Three phase lines with bundle conductors.
 Skin effect and
 proximity effect.

 Solid conductors
 Stranded conductors
 Hallow conductors
 Bundled conductors

 Solid conductors
1. Very costly & high tensile
strength
2. Difficult to string these
conductors between poles
3. For A.C. these exhibit more
skin effect
4. Less flexibility, so rarely used

 Stranded conductors
Three or more strands of conductor material is
twisted to form a single conductor.
They can carry high currents and are more flexible
than solid conductors.
Ex: ACSR,AAC,AAAC,ACAR

 Hallow conductors
Hollow conductors are used in the transmission
lines to:
(1) Improve heat dissipation
(2) Reduce corona loss
(3) Reduce skin effect
(4) Reduce the line inductance

 Bundled Conductors
 Bundled Conductors are used in transmission lines where the voltage exceeds
230 kV.
 At such high voltages, ordinary conductors will result in excessive corona
and noise which may affect communication lines.
 The increased corona will result in significant power loss. Bundle conductors
consist of three or four conductors for each phase.
 The conductors are separated from each other by means of spacers at regular
intervals. Thus, they do not touch each other.
 A bundled conductor reduces the reactance of the electric transmission line. It
also reduces voltage gradient, corona loss, radio interference, surge
impedance of the transmission lines. By making bundle conductor, the
geometric mean radius (GMR) of the conductor increased.

 (i) Resistance.
 (ii) Inductance.
 (iii) Capacitance.
A transmission line has resistance, inductance and
capacitance uniformly distributed along the whole
length of the line.

The resistance of transmission line conductors is the most important
cause of power loss in a transmission line. The resistance R of a line
conductor having resistivity ρ, length l and area of crossection a is given
by ;
R = ρ l/a
It is the opposition of line conductors to current flow. The resistance is
distributed uniformly along the whole length of the line as shown in
Fig. (i).
However, the performance of a transmission line can be analysed
conveniently if distributed resistance is considered as lumped as shown
Fig. (ii).

 When an alternating current flows through a conductor, a changing
flux is set up which links the conductor. Due to these flux linkages,
the conductor possesses inductance.
 Mathematically, inductance is defined as the flux linkages per
ampere i.e.,
Inductance, L =
ψ
I
henry
where ψ = flux linkages in weber-turns
I = current in amperes

 Any two conductors of an overhead transmission line are
separated by air which acts as an insulation, therefore,
capacitance exists between any two overhead line
conductors. The capacitance between the conductors is the
charge per unit potential difference i.e.,
 Capacitance, C =
𝑞
𝑣
farad
where q = charge on the line in coulomb
v = p.d. between the conductors in volts

When a conductor is carrying steady direct
current (d.c.), this current is uniformly
distributed over the whole X-section of the
conductor. However, an alternating current
flowing through the conductor does not
distribute uniformly, rather it has the
tendency to concentrate near the surface of the
conductor as shown in Fig. This is known as
skin effect.
The tendency of alternating current to
concentrate near the surface of a conductor is
known as skin effect.

 Due to skin effect, the effective area of cross-section of the conductor
through which current flows is reduced. Consequently, the
resistance of the conductor is slightly increased when carrying an
alternating current.
 The skin effect depends upon the following factors :
(i) Nature of material
(ii) Diameter of wire − increases with the diameter of wire.
(iii) Frequency − increases with the increase in frequency.
(iv) Shape of wire − less for stranded conductor than the solid
conductor.
 It may be noted that skin effect is negligible when the supply
frequency is low (< 50 Hz) and conductor diameter is small (< 1cm).

 The inductance of a circuit is defined as the flux
linkages per unit current.
1. Flux linkages due to a single current
carrying conductor
2. Flux linkages in parallel current
carrying conductors.

1. Flux linkages due to a single current carrying conductor
Consider a long straight cylindrical conductor of radirus r
metres and carrying a current I amperes (r.m.s.) as shown
in Fig.
This current will set up magnetic field. The magnetic lines
of force will exist inside the conductor aswell as outside the
conductor. Both these fluxes will contribute to the
inductance of the conductor.
(i) Flux linkages due to internal flux.
(ii) Flux linkages due to external flux.

(i) Flux linkages due to internal flux.
The magnetic field intensity at a point x metres from the centre is given by;
𝐻𝑥 =
𝐼𝑥
2π𝑥
[According to Ampere’s law, m.m.f. (ampere-turns) around any closed path equals
the current enclosed by the path. The current enclosed by the path is 𝐼𝑥 and m.m.f. =
𝐻𝑥 × 2π x.
∴ 𝐻𝑥 × 2π x = 𝐼𝑥.]
Assuming a uniform current density,
𝐼𝑥
π𝑥2
=
𝐼
π𝑟2
𝐼𝑥=
π𝑥2
π𝑟2 𝐼 =
𝑥2
𝑟2 𝐼
put 𝐼𝑥 in 𝐻𝑥
𝐻𝑥=
𝑥2
𝑟2 𝐼*
1
2π𝑥
=
𝑥
2π𝑟2 𝐼 𝐴𝑇/𝑚

 If μ (= 𝜇0𝜇𝑟 ) is the permeability of the conductor, then flux density at the considered point is
given by;
 𝐵𝑥 = 𝜇0𝜇𝑟𝐻𝑥 wb/𝑚2
= 𝜇0𝜇𝑟
𝑥
2π𝑟2 𝐼 =
𝜇0𝑥𝐼
2π𝑟2 𝑤𝑏/𝑚2 [∵ 𝜇𝑟 = 1 for non-magnetic material]
Now, flux dφ through a cylindrical shell of radial thickness dx and axial length 1 m is given by;
dφ= 𝐵𝑥 × 1 × dx=
𝜇0𝑥𝐼
2π𝑟2 dx 𝑤𝑒𝑏𝑒𝑟
This flux links with current 𝐼𝑥 (=
π𝑥2
π𝑟2 𝐼) only. Therefore, flux linkages per metre length of the
conductor is
𝑑ψ=
π𝑥2
π𝑟2 dφ
=
𝜇0𝑥3𝐼
2π𝑟4dx 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠
Total flux linkages from centre upto the conductor surface is
ψ𝑖𝑛𝑡 = න
0
𝑟
𝜇0𝑥3𝐼
2π𝑟4 dx
=
𝜇0𝐼
2π𝑟4
𝑥4
4
𝑟
0
=
𝜇0𝐼
2π𝑟4
𝑟4
4
=
𝜇0𝐼
8π
𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 per meter length
Internal inductance
is independent of
conductor radius.

 (ii) Flux linkages due to external flux.
Now let us calculate the flux linkages of the conductor due to external flux. The external flux
extends from the surface of the conductor to infinity.
The field intensity at a distance x metres (from centre) outside the conductor is given by ;
𝐻𝑥 =
𝐼
2π𝑥
AT / m
Flux density, 𝐵𝑥 = 𝜇0𝜇𝑟𝐻𝑥
=
𝜇0𝐼
2π𝑥
𝑤𝑏/𝑚2 [∵ 𝜇𝑟 = 1 for non-magnetic material]
Now, flux dφ through a cylindrical shell of thickness dx and axial length 1 metre is
dφ= 𝐵𝑥 × 1 × dx=
𝜇0𝐼
2π𝑥
𝑤𝑒𝑏𝑒𝑟
The flux dφ links all the current in the conductor once and only once.
∴ Flux linkages, dψ = dφ =
𝜇0𝐼
2π𝑥
dx 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠
Total flux linkages of the conductor from surface to infinity,
ψ𝑒𝑥𝑡 = න
𝑟
∞
𝜇0𝐼
2π𝑥
dx
∴ Overall flux linkages, ψ =ψ𝑖𝑛𝑡 + ψ𝑒𝑥𝑡 =
𝜇0𝐼
8π
+ ‫׬‬𝑟
∞ 𝜇0𝐼
2π𝑥
dx
=
𝜇0𝐼
2π
1
4
+ ‫׬‬
𝑟
∞ 1
𝑥
dx 𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 per meter length

2. Flux linkages in parallel current carrying conductors.
The conductors A,B, C etc. carrying currents 𝐼𝐴, 𝐼𝐵, 𝐼𝐶 etc. Let us consider the flux linkages
with one conductor, say conductor A.
There will be flux linkages with conductor A due to its own current and there will be flux
linkages with this conductor due to the mutual inductance effects of, 𝐼𝐵, 𝐼𝐶 , 𝐼𝐷, 𝐼𝐸 etc.
The total flux linkages with conductor A.
Flux linkages with conductor A due to its
own current
=
𝜇0𝐼𝐴
2π
1
4
+ ‫׬‬
𝑟
∞ 1
𝑥
dx …….(i)
Flux linkages with conductor A due to current 𝐼𝐵
=
𝜇0𝐼𝐵
2π ‫׬‬𝑑1
∞ 1
𝑥
dx …….(ii)
Flux linkages with conductor A due to current 𝐼𝐶
=
𝜇0𝐼𝐶
2π ‫׬‬
𝑑2
∞ 1
𝑥
dx …….. (iii)
∴ Total flux linkages with conductor A
= (i) + (ii) + (iii) + ......
=
𝜇0𝐼𝐴
2π
1
4
+ ‫׬‬
𝑟
∞ 1
𝑥
dx +
𝜇0𝐼𝐵
2π ‫׬‬
𝑑1
∞ 1
𝑥
dx +
𝜇0𝐼𝐶
2π ‫׬‬
𝑑2
∞ 1
𝑥
dx + ⋯

 Consider a single phase overhead line consisting of two parallel conductors A
and B spaced d metres apart.
 Conductors A and B carry the same amount of current (i.e. 𝐼𝐴 = 𝐼𝐵), but in the
opposite direction because one forms the return circuit of the other.
∴ 𝐼𝐴 + 𝐼𝐵 = 0
 In order to find the inductance of conductor A (or conductor B), we shall have
to consider the flux linkages with it.
 There will be flux linkages with conductor A due to its own current 𝐼𝐴 and
also due to the mutual inductance effect of current 𝐼𝐵 in the conductor B.
 Flux linkages with conductor A due to its own current
=
𝜇0𝐼𝐴
2π
1
4
+ ‫׬‬
𝑟
∞ 1
𝑥
dx …….(i)
=
𝜇0𝐼𝐵
2π ‫׬‬
𝑑
∞ 1
𝑥
dx …….(ii)

 Total flux linkages with conductor A is
 ψ𝐴 =
𝜇0𝐼𝐴
2π
1
4
+ ‫׬‬
𝑟
∞ 1
𝑥
dx +
𝜇0𝐼𝐵
2π ‫׬‬
𝑑
∞ 1
𝑥
dx
=
𝜇0
2π
1
4
+ ‫׬‬
𝑟
∞ 1
𝑥
dx 𝐼𝐴 + 𝐼𝐵 ‫׬‬
𝑑
∞ 1
𝑥
dx
=
𝜇0
2π
1
4
+ log𝑒 ∞ − log𝑒 𝑟 𝐼𝐴 + 𝐼𝐵 log𝑒 ∞ − log𝑒 𝑑
=
𝜇0
2π
𝐼𝐴
4
+ log𝑒 ∞ 𝐼𝐴 + 𝐼𝐵 − 𝐼𝐴 log𝑒 𝑟 − 𝐼𝐵 log𝑒 𝑑
=
𝜇0
2π
𝐼𝐴
4
− 𝐼𝐴 log𝑒 𝑟 − 𝐼𝐵 log𝑒 𝑑 ( ∴ 𝐼𝐴 + 𝐼𝐵 = 0)
=
𝜇0
2π
𝐼𝐴
4
− 𝐼𝐴 log𝑒 𝑟 + 𝐼𝐴 log𝑒 𝑑 (∴ 𝐼𝐴 = −𝐼𝐵)
=
𝜇0
2π
𝐼𝐴
4
+ 𝐼𝐴 log𝑒
𝑑
𝑟
=
𝜇0𝐼𝐴
2π
1
4
+ log𝑒
𝑑
𝑟
𝑤𝑒𝑏𝑒𝑟 − 𝑡𝑢𝑟𝑚𝑠 per meter length

 Inductance of conductor A,
𝐿𝐴 =
ψ𝐴
𝐼𝐴
=
𝜇0
2π
1
4
+ log𝑒
𝑑
𝑟
H/m
=
4Π∗10−7
2π
1
4
+ log𝑒
𝑑
𝑟
H/m
𝐿𝐴 =10−7 1
2
+ 2log𝑒
𝑑
𝑟
H/m ……..(i)
Loop inductance = 2 𝐿𝐴 H/m
=10−7
1 + 4log𝑒
𝑑
𝑟
H/m ………(ii)
Note that eq. (ii) is the inductance of the two-wire line and is
sometimes called loop inductance.
However, inductance given by eq. (i) is the inductance per conductor
and is equal to half the loop inductance.

 Expression in alternate form.
The expression for the inductance of a conductor can be put in a concise form.
𝐿𝐴 =
ψ𝐴
𝐼𝐴
=
𝜇0
2π
1
4
+ log𝑒
𝑑
𝑟
H/m =
4Π∗10−7
2π
1
4
+ log𝑒
𝑑
𝑟
H/m
=2* 10−7 1
4 + log𝑒
𝑑
𝑟
H/ = 2* 10−7
log𝑒 𝑒
1
4 + log𝑒
𝑑
𝑟
H/m
=2* 10−7
log𝑒
𝑑
𝑟𝑒
−
1
4
H/m ………(iii)
If we put 𝑟𝑒−
1
4=r′
The radius r′ is that of a fictitious conductor assumed to have no internal flux but with the same
inductance as the actual conductor of radius r. The quantity 𝑒−
1
4 = 0·7788 so that
r′ = r 𝑒−
1
4 = 0·7788 r
The term r′ (= r 𝑒−
1
4) is called geometric mean radius (GMR) of the wire. Note that eq. (iii) gives the same value
of inductance 𝐿𝐴 as eq. (i)
Loop inductance =2* 2* 10−7
log𝑒
𝑑
𝑟𝑒
−
1
4
H/m
Note that r′ = 0·7788 r is applicable to only solid round conductor.

Three conductors A, B and C of a 3-phase line carrying currents 𝐼𝐴, 𝐼𝐵, and 𝐼𝐶
respectively. Let 𝑑1, 𝑑2 and 𝑑3 be the spacings between the conductors as shown.
Let us further assume that the loads are balanced i.e. 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶 = 0.
Consider the flux linkages with conductor A. There will be flux linkages with
conductor A due to its own current and also due to the mutual inductance effects
of 𝐼𝐵 and 𝐼𝐶 .
Flux linkages with conductor A due to its own current
=
𝜇0𝐼𝐴
2π
1
4
+ ‫׬‬
𝑟
∞ 1
𝑥
dx …….(i)
=
𝜇0𝐼𝐵
2π ‫׬‬
𝑑3
∞ 1
𝑥
dx …….(ii)
Flux linkages with conductor A due to current 𝐼𝑐
=
𝜇0𝐼𝑐
2π ‫׬‬
𝑑2
∞ 1
𝑥
dx …….(ii)

 Total flux linkages with conductor A is
ψ𝐴 = (i) + (ii) + (iii)
=
𝜇0𝐼𝐴
2π
1
4
+ ‫׬‬
𝑟
∞ 1
𝑥
dx +
𝜇0𝐼𝐵
2π ‫׬‬
𝑑3
∞ 1
𝑥
dx+
𝜇0𝐼𝑐
2π ‫׬‬
𝑑2
∞ 1
𝑥
dx
= 𝜇0
2π
1
4
+ ‫׬‬
𝑟
∞ 1
𝑥
dx 𝐼𝐴 + 𝐼𝐵 ‫׬‬
𝑑3
∞ 1
𝑥
dx + 𝐼𝑐 ‫׬‬
𝑑2
∞ 1
𝑥
dx
=𝜇0
2π
1
4
+ log𝑒 ∞ − log𝑒 𝑟 𝐼𝐴 + 𝐼𝐵 log𝑒 ∞ − log𝑒 𝑑3 + 𝐼𝐵 log𝑒 ∞ − log𝑒 𝑑2
= 𝜇0
2π
𝐼𝐴
4
+ log𝑒 ∞ 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶 − 𝐼𝐴 log𝑒 𝑟 − 𝐼𝐵 log𝑒 𝑑3 − 𝐼𝐶 log𝑒 𝑑2
=
𝜇0
2π
𝐼𝐴
4
− 𝐼𝐴 log𝑒 𝑟 − 𝐼𝐵 log𝑒 𝑑3 − 𝐼𝐶 log𝑒 𝑑2 ( ∴ 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶= 0)
ψ𝐴=
𝜇0
2π
1
4
− log𝑒 𝑟 𝐼𝐴 − 𝐼𝐵 log𝑒 𝑑3 − 𝐼𝐶 log𝑒 𝑑2

 (i) Symmetrical spacing.
If the three conductors A, B and C are placed symmetrically at the corners of an equilateral
triangle of side d, then, 𝑑1=𝑑2=𝑑3 = 𝑑
ψ𝐴=
𝜇0
2π
1
4
− log𝑒 𝑟 𝐼𝐴 − 𝐼𝐵 log𝑒 𝑑 − 𝐼𝐶 log𝑒 𝑑
=
𝜇0
2π
1
4
− log𝑒 𝑟 𝐼𝐴 − (𝐼𝐵 + 𝐼𝐶) log𝑒 𝑑
=
𝜇0
2π
1
4
− log𝑒 𝑟 𝐼𝐴 − (−𝐼𝐴) log𝑒 𝑑 ( ∴ 𝐼𝐵 + 𝐼𝐶= −𝐼𝐴 )
=
𝜇0𝐼𝐴
2π
1
4
+ log𝑒
𝑑
𝑟
weber−turns/m
Inductance of conductor A,
𝐿𝐴 =
ψ𝐴
𝐼𝐴
H /m
=
𝜇0
2π
1
4
+ log𝑒
𝑑
𝑟
H /m
=
4Π∗10−7
2π
1
4
+ log𝑒
𝑑
𝑟
H/m
𝐿𝐴 =10−7 1
2
+ 2log𝑒
𝑑
𝑟
H/m
Derived in a similar way, the expressions for inductance are the same for conductors B and C.

 (ii) Unsymmetrical spacing.
When 3-phase line conductors are not equidistant from each other, the
conductor spacing is said to be unsymmetrical.
Under such conditions, the flux linkages and inductance of each phase are not
the same.
A different inductance in each phase results in unequal voltage drops in the
three phases even if the currents in the conductors are balanced.
Therefore, the voltage at the receiving end will not be the same for all phases. In
order that voltage drops are equal in all conductors, we generally interchange
the positions of the conductors at regular intervals along the line so that each
conductor occupies the original position of every other conductor over an equal
distance. Such an exchange of positions is known as transposition.

 A3-phase transposed line having unsymmetrical spacing. Let us assume that each of the three
sections is 1 m in length.
 Let us further assume balanced conditions i.e., 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶 = 0. . Let the line currents be :
 𝐼𝐴 =I∠00
= I (1+ j 0)
 𝐼𝐵 = I∠−1200
= I (− 0·5 − j 0·866)
 𝐼𝐶 = I∠−2400= I (− 0·5 + j 0·866)
 The total flux linkages per metre length of conductor A is
ψ𝐴=
𝜇0
2π
1
4
− log𝑒 𝑟 𝐼𝐴 − 𝐼𝐵 log𝑒 𝑑3 − 𝐼𝐶 log𝑒 𝑑2
Putting the values of 𝐼𝐴, 𝐼𝐵 and 𝐼𝐶 , we get,
ψ𝐴=
𝜇0
2π
1
4
− log𝑒 𝑟 𝐼 − I (− 0·5 − j 0·866) log𝑒 𝑑3 − I (− 0·5 + j 0·866) log𝑒 𝑑2
=
𝜇0
2π
𝐼
4
− 𝐼 log𝑒 𝑟 + 0·5 Ilog𝑒 𝑑3 + j 0·866 I log𝑒 𝑑3 + 0·5I log𝑒 𝑑2 −j 0·866I log𝑒 𝑑2
=
𝜇0
2π
𝐼
4
− 𝐼 log𝑒 𝑟 + 0·5 I(log𝑒 𝑑3 + log𝑒 𝑑2 )+ j 0·866 I( log𝑒 𝑑3 − log𝑒 𝑑2)
=
𝜇0
2π
𝐼
4
− 𝐼 log𝑒 𝑟 + 𝐼 log𝑒 𝑑2𝑑3 + j 0·866 I log𝑒
𝑑3
𝑑2
=
𝜇0
2π
𝐼
4
+ 𝐼 log𝑒
𝑑2𝑑3
𝑟
+ j 0·866 I log𝑒
𝑑3
𝑑2
=
𝜇0𝐼
2π
1
4
+ log𝑒
𝑑2𝑑3
𝑟
+ j 0·866 log𝑒
𝑑3
𝑑2

 ∴ Inductance of conductor A is
𝐿𝐴 =
ψ𝐴
𝐼𝐴
=
ψ𝐴
𝐼
H /m
=
𝜇0
2π
1
4
+ log𝑒
𝑑2𝑑3
𝑟
+ j 0·866 log𝑒
𝑑3
𝑑2
H /m
=
𝜇0
2π
1
4
+ log𝑒
𝑑2𝑑3
𝑟
+ j 0·866 log𝑒
𝑑3
𝑑2
H /m
=
4Π∗10−7
2π
1
4
+ log𝑒
𝑑2𝑑3
𝑟
+ j 0·866 log𝑒
𝑑3
𝑑2
H /m
= 10−7 1
2
+ 2 log𝑒
𝑑2𝑑3
𝑟
+ j 1.732 log𝑒
𝑑3
𝑑2
H /m
Similarly inductance of conductors B and C will be :
𝐿𝑩= 10−7 1
2
+ 2 log𝑒
𝑑3𝑑1
𝑟
+ j 1.732 log𝑒
𝑑1
𝑑3
H /m
𝐿𝑪=10−7 1
2
+ 2 log𝑒
𝑑1𝑑2
𝑟
+ j 1.732 log𝑒
𝑑2
𝑑1
H /m
Inductance of each line conductor
=
1
3
(𝐿𝐴+ 𝐿𝑩+ 𝐿𝑪)
= 10−7 1
2
+ 2 log𝑒
3
𝑑1𝑑2𝑑3
𝑟
H /m
If we compare the formula of
inductance of an unsymmetrically
spaced transposed line with that of
symmetrically spaced line, we find that
inductance of each line conductor in the
two cases will be equal if d = 3
𝑑1𝑑2𝑑3
The distance d is known as equivalent
equilateral spacing for unsymmetrically
transposed line.

The use of self geometrical mean distance
(abbreviated as self-GMD) and mutual
geometrical mean distance (mutual-GMD)
simplifies the inductance calculations, particularly
relating to multiconductor arrangements. The
symbols used for these are respectively 𝐷𝑠 and
𝐷𝑚.

 Inductance/conductor/m = 10−7 1
2
+ 2log𝑒
𝑑
𝑟
H/m
=
1
2
∗ 10−7 + 2 ∗ 10−7log𝑒
𝑑
𝑟
H/m
If we replace the original solid conductor by an equivalent hollow cylinder with extremely thin
walls, the current is confined to the conductor surface and internal conductor flux linkage would
be almost zero.
Consequently, inductance due to internal flux would be zero and the term
1
2
∗ 10−7
shall be
eliminated.
The radius of this equivalent hollow cylinder must be sufficiently smaller than the physical
radius of the conductor to allow room for enough additional flux to compensate for the absence
of internal flux linkage. It can be proved mathematically that for a solid round conductor of
radius r, the self-GMD or GMR = 0·7788 r.
Inductance/conductor/m
= 2 ∗ 10−7log𝑒
𝑑
𝐷𝑠
where 𝐷𝑠 = GMR or self-GMD = 0·7788 r
It may be noted that self-GMD of a conductor depends upon the size and shape of the conductor
and is independent of the spacing between the conductors.

 The mutual-GMD is the geometrical mean of the distances from one
conductor to the other and, therefore, must be between the largest
and smallest such distance.
 (a) The mutual-GMD between two conductors (assuming that
spacing between conductors is large compared to the diameter of
each conductor) is equal to the distance between their centres i.e.
𝐷𝑚 = spacing between conductors = d
 (b) For a single circuit 3-φ line, the mutual-GMD is equal to the
equivalent equilateral spacing i.e., 𝑑1𝑑2𝑑3
1
3
𝐷𝑚 = 𝑑1𝑑2𝑑3
1
3

 (c) The principle of geometrical mean distances can be most profitably employed to 3-φ
double circuit lines. Consider the conductor arrangement of the double circuit shown
in Fig. Suppose the radius of each conductor is r.
Self-GMD of conductor = 0·7788 r
Self-GMD of combination aa´ is
𝐷𝑠1 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư
𝑎 × 𝐷 ư
𝑎 ư
𝑎 × 𝐷 ư
𝑎𝑎 )
1
4
Self-GMD of combination bb´ is
𝐷𝑠2 = ( 𝐷𝑏𝑏 × 𝐷𝑏 ሗ
𝑏 × 𝐷 ሗ
𝑏 ሗ
𝑏 × 𝐷 ሗ
𝑏𝑏 )
1
4
Self-GMD of combination cc′ is
𝐷𝑠3 = ( 𝐷𝑐𝑐 × 𝐷𝑐 ư
𝑐 × 𝐷 ư
𝑐 ư
𝑐 × 𝐷 ư
𝑐𝑐 )
1
4
Equivalent self-GMD of one phase
𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3
1
3
The value of 𝐷𝑠 is the same for all the phases as each conductor has the same radius.

 Mutual-GMD between phases A and B is
𝐷𝐴𝐵 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ
𝑏 × 𝐷 ư
𝑎𝑏 × 𝐷 ư
𝑎 ሗ
𝑏 )
1
4
Mutual-GMD between phases B and C is
𝐷𝐵𝐶 = ( 𝐷𝑏𝑐 × 𝐷𝑏 ư
𝑐 × 𝐷 ሗ
𝑏𝑐 × 𝐷 ሗ
𝑏 ư
𝑐 )
1
4
Mutual-GMD between phases C and A is
𝐷𝐶𝐴 = ( 𝐷𝑐𝑎 × 𝐷𝑐 ư
𝑎 × 𝐷 ư
𝑐𝑎 × 𝐷 ư
𝑐 ư
𝑎 )
1
4
Equivalent mutual-GMD,
𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴
1
3
Mutual GMD depends only upon the spacing and is substantially independent
of the exact size, shape and orientation of the conductor.

(i) Single phase line
Inductance/conductor/m =2 ∗ 10−7log𝑒
𝐷𝑚
𝐷𝑠
𝐷𝑚 = Spacing between conductors = d
(ii) Single circuit 3-φ line
Inductance/phase/m = 2 ∗ 10−7log𝑒
𝐷𝑚
𝐷𝑠
𝐷𝑚 = 𝑑1𝑑2𝑑3
1
3
(iii) Double circuit 3-φ line
Inductance/phase/m = 2 ∗ 10−7
log𝑒
𝐷𝑚
𝐷𝑠
where 𝐷𝑠 = 𝐷𝑠1 ×𝐷𝑠2×𝐷𝑠3
1
3 𝐷𝑚 = 𝐷𝐴𝐵 ×𝐷𝐵𝐶×𝐷𝐶𝐴
1
3

 A single phase line has two parallel conductors 2 metres
apart. The diameter of each conductor is 1·2 cm.
Calculate the loop inductance per km of the line.
 Loop inductance per metre length of the line
=10−7
1 + 4log𝑒
𝑑
𝑟
H
= 10−7
1 + 4log𝑒
200
0.6
H
=24.23* 10−7
Loop inductance per km of the line
= 24.23* 10−7
*1000
=2.423mH

A single phase transmission line has two parallel conductors 3 m apart, the radius of each
conductor being 1 cm. Calculate the loop inductance per km length of the line if the
material of the conductor is (i) copper (ii) steel with relative permeability of 100.
 Loop inductance per metre length of the line
=10−7
𝜇𝑟 + 4log𝑒
𝑑
𝑟
H/m
(i) With copper conductors, 𝜇𝑟 = 1
∴ Loop inductance/m=10−7 1 + 4log𝑒
300
1
H/m
=23.8 * 10−7
H
= 23.8* 10−7*1000=2.38mH
(ii) With steel conductors, 𝜇𝑟 = 100
∴ Loop inductance/m=10−7
100 + 4log𝑒
300
1
H/m
=122.8 * 10−7H
= 122.8 * 10−7*1000=12.28mH

Find the inductance per km of a 3-phase transmission line using 1·24 cm
diameter conductors when these are placed at the corners of an
equilateral triangle of each side 2 m.
conductor spacing d = 2 m =200cm
conductor radius r = 1·24/2 = 0·62 cm.
Inductance/phase/m=10−7 1
2
+ 2log𝑒
𝑑
𝑟
H/m
=10−7 1
2
+ 2log𝑒
200
0.62
H/m
=12* 10−7 H
Inductance/phase/km = 12* 10−7
*1000H
=1.2mH

The three conductors of a 3-phase line are arranged at the corners of a triangle of
sides 2 m, 2·5 m and 4·5 m. Calculate the inductance per km of the line when the
conductors are regularly transposed. The diameter of each conductor is 1·24 cm.
Three conductors of a 3-phase line placed at the corners of a triangle of sides 𝐷12
= 2 m, 𝐷23 = 2·5 m and 𝐷31 = 4·5 m.
The conductor radius r = 1·24/2 = 0·62 cm.
Equivalent equilateral spacing,
𝐷𝑒𝑞=3
𝐷12 ∗ 𝐷23 ∗ 𝐷31
=
3
2 ∗ 2.5 ∗ 4.5=2.82m=282cm
Inductance/phase/m =10−7 1
2
+ 2log𝑒
𝐷𝑒𝑞
𝑟
H/m
=10−7 1
2
+ 2log𝑒
282
0.62
H/m
=12.74* 10−7
Inductance/phase/km = 12.74* 10−7 *1000H
=1.274mH

Calculate the inductance of each conductor in a 3-phase, 3-wire
system when the conductors are arranged in a horizontal plane
with spacing such that 𝐷31 = 4 m ; 𝐷12 = 𝐷23 = 2m. The
conductors are transposed and have a diameter of 2·5 cm.
The conductor radius r = 2·5/2 = 1·25 cm.
Equivalent equilateral spacing, 𝐷𝑒𝑞=3
𝐷12 ∗ 𝐷23 ∗ 𝐷31
𝐷𝑒𝑞=
3
2 ∗ 2 ∗ 4=2.52m=252cm
Inductance/phase/m =10−7 1
2
+ 2log𝑒
𝐷𝑒𝑞
𝑟
H/m
=10−7 1
2
+ 2log𝑒
252
1.25
H/m
=11.1* 10−7
Inductance/phase/km = 11.1* 10−7
*1000H
=1.11mH

Two conductors of a single phase line, each of 1 cm diameter, are arranged in a vertical plane with one conductor
mounted 1 m above the other. A second identical line is mounted at the same height as the first and spaced
horizontally 0·25 m apart from it. The two upper and the two lower conductors are connected in parallel. Determine
the inductance per km of the resulting double circuit line.
Conductors a, a′ form one connection and conductors b, b′ form the return connection. The conductor radius, r = 1/2
= 0·5 cm.
Inductance per conductor per metre= 2 ∗ 10−7
log𝑒
𝐷𝑚
𝐷𝑠
G.M.R. of conductor = 0·7788 r = 0·7788 × 0·5 = 0·389 cm
Self G.M.D. of aa′ combination is
𝐷𝑠 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư
𝑎 × 𝐷 ư
𝑎 ư
𝑎 × 𝐷 ư
𝑎𝑎 )
1
4
𝐷𝑠 = ( 0.389 × 100 ×0.389 × 100)
1
4 =6.23cm
Mutual G.M.D. between a and b is
𝐷𝑚 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ
𝑏 × 𝐷 ư
𝑎𝑏 × 𝐷 ư
𝑎 ሗ
𝑏 )
1
4
𝐷𝑎 ሗ
𝑏= 𝐷 ư
𝑎𝑏= 252 + 1002 = 103
𝐷𝑚 = ( 25 × 103 × 103 × 25 )
1
4 =50.47cm
Inductance per conductor per metre= 2 ∗ 10−7log𝑒
𝐷𝑚
𝐷𝑠
= = 2 ∗ 10−7
log𝑒
50.47
6.23
=0.42 × 10−6H
∴ Loop inductance per km of the line=2* 0.42 × 10−6
*1000=0.84mH

Find the inductance per phase per km of double circuit 3-phase line shown in Fig. The conductors are transposed
and are of radius 0·75 cm each. The phase sequence is ABC.
log𝑒
𝐷𝑚
𝐷𝑠
1
1
3
Self-GMD of conductor = 0·7788 r=0.7788*0.75=0.584cm
Self-GMD of combination aa´ is
𝑎 × 𝐷 ư
𝑎 ư
𝑎 × 𝐷 ư
𝑎𝑎 )
1
4
𝐷𝑎 ư
𝑎= 𝐷 ư
𝑎𝑎= 62 + 42=7.21m
𝐷𝑠1 = ( (0.584×10−2
)×7.21×(0.584×10−2
)×7.21)
1
4=0.205m
𝑏 × 𝐷 ሗ
𝑏 ሗ
𝑏 × 𝐷 ሗ
𝑏𝑏 )
1
4
= ((0.584×10−2
)× 5.5×(0.584×10−2
)×5.5)
1
4=0.18m
𝑐 × 𝐷 ư
𝑐 ư
𝑐 × 𝐷 ư
𝑐𝑐 )
1
4
= 𝐷𝑠1= 0.205m
1
3
𝐷𝑠 = 0.205×0.18×0.205
1
3
=0.195m

1
3
Mutual-GMD between phases A and B is
𝑏 × 𝐷 ư
𝑎𝑏 × 𝐷 ư
𝑎 ሗ
𝑏 )
1
4
𝐷𝑎 ሗ
𝑏= 𝐷 ư
𝑎𝑏= 32 + (4 + 0.75)2=5.62m
𝐷𝑎𝑏= 32 + (0.75)2=3.1m
𝐷𝐴𝐵 = ( 3.1 × 5.62 × 5.62× 3.1 )
1
4=4.17m
𝑐 × 𝐷 ሗ
𝑏 ư
𝑐 )
1
4
= 𝐷𝐴𝐵 =4.17m
𝑎 × 𝐷 ư
𝑐𝑎 × 𝐷 ư
𝑐 ư
𝑎 )
1
4
= ( 6 × 4 × 4 × 6 )
1
4=4.9m
1
3= 4.71×4.17×4.9
1
3=4.4m
log𝑒
𝐷𝑚
𝐷𝑠
= 2 ∗ 10−7
log𝑒
4.4
0.195
=0.623 ∗ 10−3mH
Inductance/phase/km= 0.623 ∗ 10−3
*1000=0.623mH

Calculate the inductance per phase per metre for a three-phase double-circuit line whose phase
conductors have a radius of 5·3 cm with the horizontal conductor arrangement as shown in Fig.
Inductance per conductor per metre= 2 ∗ 10−7log𝑒
𝐷𝑚
𝐷𝑠
1
1
3
Self-GMD of combination aa´ is0·7788 r = 0·7788 × 5·3 × 10−2
= 0·0413 m
𝑎 × 𝐷 ư
𝑎 ư
𝑎 × 𝐷 ư
𝑎𝑎 )
1
4
= (0·041× 24 ×0·041× 24 )
1
4=0.995m
𝑏 × 𝐷 ሗ
𝑏 ሗ
𝑏 × 𝐷 ሗ
𝑏𝑏 )
1
4
= (0·041× 24 ×0·041× 24 )
1
4=0.995m
𝑐 × 𝐷 ư
𝑐 ư
𝑐 × 𝐷 ư
𝑐𝑐 )
1
4
= (0·041× 24 ×0·041× 24 )
1
4=0.995m
Equivalent self-GMD of one phase
1
3=0.995m

1
3
Mutual-GMD between phases A and B is
𝑏 × 𝐷 ư
𝑎𝑏 × 𝐷 ư
𝑎 ሗ
𝑏 )
1
4
𝐷𝐴𝐵 = ( 8 × 32 × 16× 8 )
1
4=13.45m
𝑐 × 𝐷 ሗ
𝑏 ư
𝑐 )
1
4
= ( 8 × 32 × 16× 8 )
1
4=13.45m
𝑎 × 𝐷 ư
𝑐𝑎 × 𝐷 ư
𝑐 ư
𝑎 )
1
4
= ( 16 × 8 × 40 × 16 )
1
4=16.917m
1
3= 13.45×13.45×16.917
1
3=14.518m
log𝑒
𝐷𝑚
𝐷𝑠
= 2 ∗ 10−7log𝑒
14.518
0.995
=5.36 ∗ 10−7
H

In a single phase line (See. Fig.)conductors a and a′ in parallel form one circuit while
conductors b and b′ in parallel form the return path. Calculate the total inductance of the
line per km assuming that current is equally shared by the two parallel conductors.
Conductor diameter in 2·0 cm.
log𝑒
𝐷𝑚
𝐷𝑠
Mutual G.M.D., 𝐷𝑚 = ( 𝐷𝑎𝑏 × 𝐷𝑎 ሗ
𝑏 × 𝐷 ư
𝑎𝑏 × 𝐷 ư
𝑎 ሗ
𝑏 )
1
4
= ( 120 × 140 × 100× 120 )
1
4=119cm
Self G.M.D., 𝐷𝑠 = ( 𝐷𝑎𝑎 × 𝐷𝑎 ư
𝑎 × 𝐷 ư
𝑎 ư
𝑎 × 𝐷 ư
𝑎𝑎 )
1
4
𝐷𝑎𝑎= 𝐷 ư
𝑎 ư
𝑎=0.7788*r=0.7788*
2
2
=0.7788cm
𝐷𝑠 = (0.7788 × 20 × 0.7788× 20 )
1
4=3.94cm
log𝑒
𝐷𝑚
𝐷𝑠
= 2 ∗ 10−7
log𝑒
119
3.94
=6.81* 10−7
H
Loop Inductance/phase/km=2*6.81* 10−7
*1000H
=1.363mH

Any two conductors of an overhead transmission line are
separated by air which acts as an insulation, therefore,
capacitance exists between any two overhead line
conductors. The capacitance between the conductors is the
charge per unit potential difference i.e.,
Capacitance, C =
𝑞
𝑣
farad
where q = charge on the line in coulomb
v = p.d. between the conductors in volts

 Electric Potential
The electric potential at a point due to a charge is the
work done in bringing a unit positive charge from
infinity to that point.
(i) Potential at a charged single conductor
(ii) Potential at a conductor in a group of charged
conductors.

(i) Potential at a charged single conductor.
Consider a long straight cylindrical conductor A of radius r metres. Let the conductor operate at
such a potential (𝑉𝐴) that charge 𝑄𝐴 coulombs per metre exists on the conductor.
The electric intensity E at a distance x from the centre of the conductor in air is given by:
𝐸 =
𝑄𝐴
2𝜋𝑥𝜀0
volts/m
where 𝑄𝐴 = charge per metre length
𝜀0 = permittivity of free space
As x approaches infinity, the value of E approaches zero. Therefore, the potential difference
between conductor A and infinity distant neutral plane is given by :
Work is done in bringing a unit positive charge against E from infinity to conductor surface.
𝑉𝐴 = න
𝑟
∞
𝐸 𝑑𝑥
𝑉𝐴 = න
𝑟
∞
𝑄𝐴
2𝜋𝑥𝜀0
𝑑𝑥 =
𝑄𝐴
2𝜋𝜀0
න
𝑟
∞
1
𝑥
𝑑𝑥

(ii)Potential at a conductor in a group of charged conductors.
Consider a group of long straight conductors A, B, C etc. operating at potentials
such that charges 𝑄𝐴, 𝑄𝐵, 𝑄𝐶 etc. coulomb per metre length exist on the
respective conductors
The potential at A
Potential at A due to its own charge (i.e. 𝑄𝐴)
= න
𝑟
∞
𝑄𝐴
2𝜋𝑥𝜀0
𝑑𝑥
Potential at conductor A due to charge 𝑄𝐵
= න
𝑑1
∞
𝑄𝐵
2𝜋𝑥𝜀0
𝑑𝑥
Potential at conductor A due to charge 𝑄𝐶
= න
𝑑2
∞
𝑄𝐶
2𝜋𝑥𝜀0
𝑑𝑥

 Overall potential difference between conductor A and
infinite neutral plane is
𝑉𝐴 = (i) + (ii) + (iii) + .......
= ‫׬‬
𝑟
∞ 𝑄𝐴
2𝜋𝑥𝜀0
𝑑𝑥+ ‫׬‬
𝑑1
∞ 𝑄𝐵
2𝜋𝑥𝜀0
𝑑𝑥 +‫׬‬
𝑑2
∞ 𝑄𝐶
2𝜋𝑥𝜀0
𝑑𝑥+………
=
1
2𝜋𝜀0
‫׬‬
𝑟
∞ 𝑄𝐴
𝑥
dx + ‫׬‬
𝑑1
∞ 𝑄𝐵
𝑥
𝑑𝑥 + ‫׬‬
𝑑2
∞ 𝑄𝐶
𝑥
𝑑𝑥 + ⋯ … . .
=
1
2𝜋𝜀0
𝑄𝐴 log𝑒 ∞ − log𝑒 𝑟 + 𝑄𝐵 log𝑒 ∞ − log𝑒 𝑑1 + 𝑄𝐶 log𝑒 ∞ − log𝑒 𝑑2 + ⋯ .
= 1
2𝜋𝜀0
log𝑒 ∞ 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶 + 𝑄𝐴 log𝑒
1
𝑟
+ 𝑄𝐵 log𝑒
1
𝑑1
+ 𝑄𝐶 log𝑒
1
𝑑2
+ ⋯ .
Assuming balanced conditions i.e., 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶=0
𝑉𝐴=
1
2𝜋𝜀0
𝑄𝐴 log𝑒
1
𝑟
+ 𝑄𝐵 log𝑒
1
𝑑1
+ 𝑄𝐶 log𝑒
1
𝑑2
+ ⋯ .

Consider a single phase overhead transmission line consisting of two parallel
conductors A and B spaced d metres apart in air. Suppose that radius of each
conductor is r metres. Let their respective charge be + Q and − Q coulombs per
metre length.
The total p.d. between conductor A and neutral “infinite” plane is
𝑉𝐴 = ‫׬‬
𝑟
∞ Q
2𝜋𝑥𝜀0
𝑑𝑥+ ‫׬‬
𝑑
∞ −𝑄
2𝜋𝑥𝜀0
𝑑𝑥
=
1
2𝜋𝜀0
Q log𝑒 ∞ − log𝑒 𝑟 − Q log𝑒 ∞ − log𝑒 𝑑
=
Q
2𝜋𝜀0
log𝑒 𝑑 − log𝑒 𝑟 =
Q
2𝜋𝜀0
log𝑒
𝑑
𝑟
volts

The total p.d. between conductor B and neutral “infinite” plane is
𝑉𝐵 = ‫׬‬
𝑟
∞ −Q
2𝜋𝑥𝜀0
𝑑𝑥+ ‫׬‬
𝑑
∞ 𝑄
2𝜋𝑥𝜀0
𝑑𝑥
=
1
2𝜋𝜀0
−Q log𝑒 ∞ − log𝑒 𝑟 +Q log𝑒 ∞ − log𝑒 𝑑
=
−Q
2𝜋𝜀0
log𝑒 𝑑 − log𝑒 𝑟 =
−Q
2𝜋𝜀0
log𝑒
𝑑
𝑟
volts
Both these potentials are w.r.t. the same neutral plane. Since the unlike charges
attract each other, the potential difference between the conductors is
𝑉𝐴𝐵 = 2𝑉𝐴=
2Q
2𝜋𝜀0
log𝑒
𝑑
𝑟
∴ Capacitance, 𝐶𝐴𝐵 =
Q
𝑉𝐴𝐵
=
Q
2Q
2𝜋𝜀0
log𝑒
𝑑
𝑟
𝐶𝐴𝐵 =
𝜋𝜀0
log𝑒
𝑑
𝑟
F/m

 Capacitance to neutral.
Since potential of the mid-point between the conductors is zero,
the potential difference between each conductor and the ground
or neutral is half the potential difference between the conductors.
Thus the capacitance to ground or capacitance to neutral for the
twowire line is twice the line-to-line capacitance
∴ Capacitance to neutral
𝐶𝑁 = 𝐶𝐴𝑁 = 𝐶𝐵𝑁 = 2𝐶𝐴𝐵
𝐶𝑁=
2𝜋𝜀0
log𝑒
𝑑
𝑟
F/m

 In a 3-phase transmission line, the capacitance of each conductor is
considered instead of capacitance from conductor to conductor.
 Here, again two cases arise viz., symmetrical spacing and
unsymmetrical spacing.
 (i) Symmetrical Spacing.
The three conductors A, B and C of the 3-phase overhead transmission
line having charges 𝑄𝐴, 𝑄𝐵, and 𝑄𝐶 per metre length respectively. Let the
conductors be equidistant (d metres) from each other.

 overall potential difference between conductor A and infinite neutral plane is
 𝑉𝐴 = ‫׬‬𝑟
∞ 𝑄𝐴
2𝜋𝑥𝜀0
𝑑𝑥+ ‫׬‬𝑑
∞ 𝑄𝐵
2𝜋𝑥𝜀0
𝑑𝑥 +‫׬‬𝑑
∞ 𝑄𝐶
2𝜋𝑥𝜀0
𝑑𝑥
= 1
2𝜋𝜀0
‫׬‬
𝑟
∞ 𝑄𝐴
𝑥
dx + ‫׬‬
𝑑
∞ 𝑄𝐵
𝑥
𝑑𝑥 + ‫׬‬
𝑑
∞ 𝑄𝐶
𝑥
𝑑𝑥
=
1
2𝜋𝜀0
𝑄𝐴 log𝑒 ∞ − log𝑒 𝑟 + 𝑄𝐵 log𝑒 ∞ − log𝑒 𝑑 + 𝑄𝐶 log𝑒 ∞ − log𝑒 𝑑
= 1
2𝜋𝜀0
log𝑒 ∞ 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶 + 𝑄𝐴 log𝑒
1
𝑟
+ (𝑄𝐵 + 𝑄𝐶 ) log𝑒
1
𝑑
Assuming balanced conditions i.e., 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶=0
𝑄𝐵 + 𝑄𝐶= −𝑄𝐴
∴ 𝑉𝐴=
1
2𝜋𝜀0
𝑄𝐴 log𝑒
1
𝑟
+ (−𝑄𝐴 ) log𝑒
1
𝑑
𝑉𝐴 =
𝑄𝐴
2𝜋𝜀0
log𝑒
𝑑
𝑟
F/m
∴ Capacitance of conductor A w.r.t neutral,
𝐶𝐴 =
𝑄𝐴
𝑉𝐴
=
𝑄𝐴
𝑄𝐴
2𝜋𝜀0
log𝑒
𝑑
𝑟
𝐶𝐴 =
2𝜋𝜀0
log𝑒
𝑑
𝑟
F/m
Note that this equation is identical to capacitance to neutral for two-wire line. Derived in a
similar manner, the expressions for capacitance are the same for conductors B and C.

 (ii) Unsymmetrical spacing.
 Assuming balanced conditions i.e., 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶=0
 Potential of 1st position,
𝑉1=
1
2𝜋𝜀0
𝑄𝐴 log𝑒
1
𝑟
+ 𝑄𝐵 log𝑒
1
𝑑3
+ 𝑄𝐶 log𝑒
1
𝑑2
 Potential of 2nd position,
𝑉2=
1
2𝜋𝜀0
𝑄𝐴 log𝑒
1
𝑟
+ 𝑄𝐵 log𝑒
1
𝑑1
+ 𝑄𝐶 log𝑒
1
𝑑3
 Potential of 3rd position,
𝑉3=
1
2𝜋𝜀0
𝑄𝐴 log𝑒
1
𝑟
+ 𝑄𝐵 log𝑒
1
𝑑2
+ 𝑄𝐶 log𝑒
1
𝑑1
 Average voltage on condutor A is
𝑉𝐴 =
1
3
(𝑉1+ 𝑉2+ 𝑉3)
=
1
3∗2𝜋𝜀0
𝑄𝐴 log𝑒
1
𝑟3 + (𝑄𝐵+ 𝑄𝐶) log𝑒
1
𝑑1𝑑2𝑑3

 As 𝑄𝐴 + 𝑄𝐵 + 𝑄𝐶=0, therefore 𝑄𝐵 + 𝑄𝐶= −𝑄𝐴
=
1
2∗3𝜋𝜀0
𝑄𝐴 log𝑒
1
𝑟3 + (−𝑄𝐴) log𝑒
1
𝑑1𝑑2𝑑3
=
1
3
∗
𝑄𝐴
2𝜋𝜀0
log𝑒
𝑑1𝑑2𝑑3
𝑟3
=
𝑄𝐴
2𝜋𝜀0
log𝑒
𝑑1𝑑2𝑑3
𝑟3
1
3
=
𝑄𝐴
2𝜋𝜀0
log𝑒
𝑑1𝑑2𝑑3
1
3
𝑟
∴ Capacitance from conductor to neutral is
𝐶𝐴 =
𝑄𝐴
𝑉𝐴
=
𝑄𝐴
𝑄𝐴
2𝜋𝜀0
log𝑒
3 𝑑1𝑑2𝑑3
𝑟
=
2𝜋𝜀0
log𝑒
3 𝑑1𝑑2𝑑3
𝑟

A single-phase transmission line has two parallel conductors 3
metres apart, radius of each conductor being 1 cm. Calculate the
capacitance of the line per km. Given that 𝜀0 =8·854 × 10−12
F/m.
Conductor radius, r = 1 cm
Spacing of conductors, d = 3 m = 300 cm
capacitance of the line =
𝜋𝜀0
log𝑒
𝑑
𝑟
F/m
=
𝜋∗8·854 × 10−12
log𝑒
300
1
=0.485* 10−11
F/m
capacitance of the line per km= 0.485* 10−11
*1000
= 0.485* 10−8
F/km
= 0.485* 10−2
μF/km

A 3-phase overhead transmission line has its conductors arranged at the
corners of an equilateral triangle of 2 m side. Calculate the capacitance
of each line conductor per km Given that diameter of each conductor is
1·25 cm. 𝜀0 =8·854 × 10−12
F/m.
Conductor radius, r = 1·25/2 = 0·625 cm
Spacing of conductors, d = 2 m = 200 cm
Capacitance of each line conductor=
2𝜋𝜀0
log𝑒
𝑑
𝑟
F/m
==
2𝜋8·854 × 10−12
log𝑒
200
0.625
F/m=0.0096* 10−9 F/m
capacitance of the line per km 0.0096* 10−9 *1000
0.0096* 10−6
F/km
= 0.0096 μF/km

A 3-phase, 50 Hz, 66 kV overhead line conductors are placed in a horizontal plane as
shown in Fig. The conductor diameter is 1·25 cm. If the line length is 100 km, calculate (i)
capacitance per phase, (ii) charging current per phase, assuming complete transposition of
the line. 𝜀0 =8·854 × 10−12 F/m.
Conductor radius, r = 1·25/2 = 0·625 cm
The equivalent equilateral spacing is
𝑑 = 3
𝑑1𝑑2𝑑3
=
3
2 × 2 ⋅ 5 × 4 ⋅5 = = 2·82 m
Conductor spacing , d = 2·82 m = 282 cm
(i) Line to neutral capacitance==
2𝜋𝜀0
log𝑒
𝑑
𝑟
F/m
==
2𝜋8·854 × 10−12
log𝑒
282
0.625
F/m=0.0091* 10−9 F/m
capacitance of the line per km 0.0091* 10−9 *1000
0.0091* 10−6
F/km
= 0.0091 μF/km
∴ Line to neutral capacitance for 100 km line is
C = 0·0091 × 100 = 0·91 μF

 (ii) Charging current per phase is
𝐼𝑐 =
𝑉𝑝ℎ
𝑋𝑐
=
𝑉𝑝ℎ
1
2𝜋𝑓𝑐
=
66000
3
∗ 2𝜋*50*0·91 μ
=10.9A

A 3-phase, 50 Hz, 132 kV overhead line has conductors placed in a horizontal plane 4 m
apart. Conductor diameter is 2 cm. If the line length is 100 km, calculate the charging
current per phase assuming complete transposition. 𝜀0 =8·854 × 10−12
F/m.
The diameter of each conductor is 2 cm
so that conductor radius r =2/2 = 1 cm = 1× 10−2 m.
Now 𝑑1 = AB = 4m; 𝑑2 = BC = 4 m; 𝑑3 = AC = 8 m
The equivalent equilateral spacing is
𝑑 = 3
𝑑1𝑑2𝑑3
=
3
4×4×8 = 5.04 m
(i) Line to neutral capacitance==
2𝜋𝜀0
log𝑒
𝑑
𝑟
F/m
==
2𝜋8·854 × 10−12
log𝑒
5.04
1× 10−2
F/m=0.0085* 10−9
F/m
capacitance of the line per km= 0.0085* 10−9
*1000
= 0.0085* 10−6 F/km
= 0.0085 μF/km
∴ Line to neutral capacitance for 100 km line is
C = 0.0085 × 100 = 0·885 μF

 Charging current per phase is
𝐼𝑐 =
𝑉𝑝ℎ
𝑋𝑐
=
𝑉𝑝ℎ
1
2𝜋𝑓𝑐
=
132000
3
∗ 2𝜋*50*0·885 μ
=21.18A

 Definition:
Proximity Effect is the phenomena of non-uniform current
distribution on the surface of adjacent current carrying
conductor due to the effect of another current carrying
conductor in its proximity.
Since in cables, the conductors are very near to each other,
this effect is dominant. Whereas in overhead lines as lines
are far apart when compared with cable, this effect can be
neglected.

 Cause of Proximity Effect:
 The main reason for proximity effect is production of
magnetic field in the surrounding of a current carrying
conductor.
 We know that any current carrying conductor produces
magnetic field in its surrounding as per Biot Savart Law.
 When this magnetic field links with the adjacent conductor, it
gives rise to a circulating or eddy current in it. Because of this
eddy current, the current distribution on the nearby becomes
non-uniform.
 This leads to additional resistance to the flow of current in the
nearby conductor. Hence due to proximity effect, the net ac
resistance of conductor increases.

 Two conductors carrying current in the same direction. The magnetic field
created by conductor A will link with conductor B. Similarly, the magnetic
field created by conductor B will link with conductor A.
 The direction of magnetic field at any point P inside the conductor B will in
going inside the plane of device.
 Since current is flowing in upward direction, this means that free electrons
are moving in downward direction. The direction of force on the free
electrons due to the magnetic field of conductor A will be given by the cross
vector product of velocity vector of electron and magnetic field vector at
point P.

 From the cross vector product, the free electrons of conductor B will be
drifted toward the inner surface i.e. toward conductor A and hence the
current will flow mostly in the farthest surface of conductor B.
 The same reasoning when applied to conductor A, it can be seen that current
in the conductor A will mostly flow in the farthest surface. This effect of re-
distribution of current on the conductor surface is called the proximity effect.
This is so called proximity effect as this is due to the effect of proximity
(nearby) conductor.

 when two current conductors carry current in the same
direction, the current gets distributed to the farthest half of
both the conductor.

 Similarly when two conductors carry current in opposite
direction, the current in conductors gets concentrated toward
the inner facing surface of both the conductors

 Factors affecting Proximity Effect:
 The proximity effect mainly depends on the factors like conductor material,
conductor diameter, frequency and conductor structure. The factors are
explained below in details
 Frequency – The proximity increases with the increases in the frequency.
This effect is not observed in case of DC. In case of DC, as the current is
uniformly distributed over the entire cross-section of conductor, the magnetic
force on the free electrons inside conductor is less prominent. Hence this effect
is not observed.
 Diameter – The proximity effect increases with the increase in the
conductor.
 Structure – This effect is more on the solid conductor as compared to the
stranded conductor (i.e., ASCR) because the surface area of the stranded
conductor is smaller than the solid conductor.
 Material – If the material is made up of high ferromagnetic material then the
proximity effect is more on their surface.

 How to reduce Proximity Effect?
 The proximity effect can be reduced by using the ACSR
(Aluminum Core Steel Reinforced) conductor. In ACSR
conductor the steel is placed at the centre of the
conductor and the aluminium conductor is positioned
around steel wire.
 The steel increased the strength of the conductor but
reduced the surface area of the conductor. Thus, the
current flow mostly in the outer layer of the conductor
and no current is carried in the centre of the conductor.
Thus, reduced the proximity effect on the conductor.

INDUCTANCE & CAPACITANCE CALCULATIONS

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