POWER SUPPLY SYSTEMS& DISTRIBUTION SYSTEMS

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

 Unit-i Electric Power Generation& Economic
Considerations
 Unit-ii Power Supply Systems& Distribution
Systems
 Unit-iii Inductance & Capacitance Calculations
 Unit-iv Performance Of Transmission Lines
 Unit-v Overhead Line Insulators

POWER SUPPLY SYSTEMS& DISTRIBUTION
SYSTEMS:
 Transmission and distribution systems- D.C 2-wire
and 3- wire systems, A.C single phase , three phase
and 4-wire systems, comparison of copper efficiency.
 Primary and secondary distribution systems,
concentrated & uniformly distributed loads on
distributors fed at both ends, ring distributor,
voltage drop and power loss calculation,
 Kelvin’slaw.

Power
Transmission
D.C
D.C. two-wire
D.C. two-wire
with mid-point
earthed
D.C. three-wire
A.C
Single-phase A.C.
system
Single-phase two-
wire
Single-phase two-
wire with mid-
point earthed
Single-phase
three-wire
Two-phase A.C.
system
Two-phase four-
wire
Two-phase three
wire.
Three-phase A.C.
system
Three-phase
three-wire
Three-phase four-
wire

Overhead
Under ground

(i) Same power (P watts) transmitted by each system.
(ii) The distance (l metres) over which power is
transmitted remains the same.
(iii) The line losses (W watts) are the same in each
case.
(iv) The maximum voltage between any conductor
and earth (𝑉
𝑚) is the same in each case.

Max. voltage between conductors = 𝑉
𝑚 ,
Power to be transmitted = P= 𝑉
𝑚𝐼1 1
∴ Load current, 𝐼1 = P/ 𝑉
𝑚
If 𝑅1 is the resistance of each line conductor, then,
𝑅1 = ρ l/ 𝑎1
where 𝑎1 is the area of X-section of the conductor.
copper losses in line W = 𝐼1
2
𝑅1 + 𝐼1
2
𝑅1
=2𝐼1
2
𝑅1
1
1
2

The maximum voltage between any conductor and earth is 𝑉
𝑚 so that maximum
voltage between conductors is 2 𝑉
𝑚. 𝑅2 is the resistance of each line conductor
Power to be transmitted
P= 𝑉
𝑚𝐼2+ 𝑉
𝑚𝐼2=2 𝑉
𝑚𝐼2
Load current, 𝐼2 = P/2 𝑉
𝑚
Since from assumption 1 the power transmitted is same
for all systems so from eq 1& 3
2𝑉
𝑚𝐼2= 𝑉
𝑚𝐼1
𝐼2 =
𝐼1
2
2
𝑅2+ 𝐼2
2
𝑅2 = 2 𝐼2
2
𝑅2
From eq 4
W=2(
𝐼1
2
)2
𝑅2 =
𝐼1
2
2
𝑅2
3
4
5

 Compare eq 2 & 5
2𝐼1
2
𝑅1 = (
𝐼1
2
2
) 𝑅2
𝑅1
𝑅2
=
1
4
𝑎1
𝑎2
=4 (where R = ρ l/ 𝑎)
𝑎1=X-area of conductor in 2 wire dc system
𝑎2= X-area of conductor in 2 wire dc system with mid-point earthed
Then
volume of copper wire Two−wire D.C system with mid−point earthed
volume of copper wire Two−wire D.C system
=
𝑎2∗𝑙2
𝑎1∗𝑙1
=
1
4
𝑎1∗2𝑙
𝑎1∗2𝑙
=
1
4
where 𝑙1 = 𝑙2=l+l
Hence, the volume of conductor material required in Two−wire D.C system with
mid−point earthed is one-fourth of that required in a two-wire D.C system with
one conductor earthed.

If the load is balanced, the current in the neutral wire is zero. Assuming balanced loads,
P= 𝑉
𝑚𝐼3+ 𝑉
𝑚𝐼3=2 𝑉
𝑚𝐼3 6
2𝑉
𝑚𝐼3= 𝑉
𝑚𝐼1
𝐼3 =
𝐼1
2
7
Let 𝑅3 is the resistance of each line conductor,
the area of X-section of neutral wire to be half that of the outer wire,
2
𝑅3+ 𝐼3
2
𝑅3 = 2𝐼3
2
𝑅3 8
From eq 7
W=2(
𝐼1
2
)2
𝑅3 =
𝐼1
2
2
𝑅3 9

2𝐼1
2
𝑅1 =
𝐼1
2
2
𝑅3
𝑅1
𝑅3
=
1
4
𝑅3 = ρ l/ 𝑎3 where 𝑎3 is the area of X-section of the conductor.
𝑎1
𝑎3
=4
Then
volume of copper wire Three−wire D.C
=
𝑎3∗𝑙3
𝑎1∗𝑙1
=
1
4
𝑎1∗2.5𝑙
𝑎1∗2𝑙
=
2.5
8
where 𝑙3=l+l+0.5l
=0.3125
Hence, the volume of conductor material required in Three−wire D.C system is 0.3125
times of that required in a two-wire D.C system with one conductor earthed.

The maximum voltage between conductors is 𝑉
𝑚, so that r.m.s. value of voltage
between them is
𝑉𝑚
2
. Assuming the load power factor to be cos φ, load current 𝐼4
Power supplied P=
𝑉𝑚
2
𝐼4 cos φ 10
𝑉𝑚
2
𝐼4 cos φ= 𝑉
𝑚𝐼1
𝐼4=
2𝐼1
cos φ
11
Let 𝑅4 is the resistance of each line conductor
2
𝑅4+ 𝐼4
2
𝑅4 = 2𝐼4
2
𝑅4
=2(
2𝐼1
cos φ
)2 𝑅4 =4
𝐼1
2
𝑐𝑜𝑠2φ
𝑅4 12

 Comparing copper losses with 2 wire d.c system
4
𝐼1
2
𝑐𝑜𝑠2φ
𝑅4= 2𝐼1
2
𝑅1
𝑅4
𝑅1
=
𝑐𝑜𝑠2φ
2
𝑎4
𝑎1
=
2
𝑐𝑜𝑠2φ
13
volume of copper wire 1−phase 2− wire
=
𝑎4𝑙4
𝑎1𝑙1
=
2
𝑐𝑜𝑠2φ𝑎12𝑙
𝑎12𝑙
=
2
𝑐𝑜𝑠2φ
Hence, the volume of conductor material required in this system is
2
𝑐𝑜𝑠2φ
times
that of 2-wire d.c. system with the one conductor earthed.

 The two wires possess equal and opposite voltages to earth (i.e., 𝑉
𝑚).
 Therefore, the maximum voltage between the two wires is
𝑉𝑚
2
. The
r.m.s. value of voltage between conductors is =2
𝑉𝑚
2
= 2 𝑉
𝑚.
Assuming the power factor of the load to be cos φ, load current 𝐼5
 Power supplied P= 2 𝑉
𝑚𝐼5 cos φ 14
2 𝑉
𝑚𝐼5 cos φ= 𝑉
𝑚𝐼1
𝐼5=
𝐼1
2 cos φ
15

2
𝑅5+ 𝐼5
2
𝑅5 = 2𝐼5
2
𝑅5
=2(
𝐼1
2 cos φ
)2 𝑅5 =
𝐼1
2
𝑐𝑜𝑠2φ
𝑅5
Comparing copper losses with 2 wire d.c system
𝐼1
2
𝑐𝑜𝑠2φ
𝑅5 =2𝐼1
2
𝑅1
𝑅5
𝑅1
= 2𝑐𝑜𝑠2φ
𝑎5
𝑎1
=
1
2𝑐𝑜𝑠2φ
volume of copper wire 1−phase 2− wire 𝑤𝑖𝑡ℎ 𝑚𝑖𝑑 − 𝑝𝑜𝑖𝑛𝑡 𝑒𝑎𝑟𝑡ℎ𝑒𝑑
𝑎5𝑙5
𝑎1𝑙1
=
1
2𝑐𝑜𝑠2φ 𝑎12𝑙
𝑎12𝑙
=
1
2𝑐𝑜𝑠2φ
Hence the volume of conductor material required in this system is
1
2𝑐𝑜𝑠2φ
times that of 2-wire d.c. system with one conductor earthed.

 If the load is balanced, the current through the neutral wire is zero.
Assuming balanced load,
 The two wires possess equal and opposite voltages to earth (i.e., 𝑉
𝑚).
 Therefore, the maximum voltage between the two wires is
𝑉𝑚
2
. The r.m.s.
value of voltage between conductors is 2
𝑉𝑚
2
= 2 𝑉
𝑚. Assuming the
power factor of the load to be cos φ, load current 𝐼6
 Power supplied P= 2 𝑉
𝑚𝐼6 cos φ 16
2 𝑉
𝑚𝐼6 cos φ= 𝑉
𝑚𝐼1
𝐼6=
𝐼1
2 cos φ
17

2
𝑅6+ 𝐼6
2
𝑅6 = 2𝐼6
2
𝑅6
=2(
𝐼1
2 cos φ
)2 𝑅6 =
𝐼1
2
𝑐𝑜𝑠2φ
𝑅6
𝐼1
2
𝑐𝑜𝑠2φ
𝑅6 =2𝐼1
2
𝑅1
𝑅6
𝑅1
= 2𝑐𝑜𝑠2φ
𝑎6
𝑎1
=
1
2𝑐𝑜𝑠2φ
volume of copper wire 1−phase3−wire system
𝑎6𝑙6
𝑎1𝑙1
=
1
2𝑐𝑜𝑠2φ 𝑎12.5𝑙
𝑎12𝑙
=
0.625
𝑐𝑜𝑠2φ
Hence the volume of conductr material required in this system is
0.625
𝑐𝑜𝑠2φ
times that of 2-wire d.c. system with one conductor earthed.

 This system can be considered as two independent single-
phase systems, each transmitting one half of the total power.
 As the voltage is same (𝑉
𝑚) as in a Single phase 2-wire system
with mid-point earthed, so the current will be half of the
Single phase system
 i.e. 𝐼7=
𝐼5
2
=
𝐼1
2 cos φ
2
=
𝐼1
2 2 cos φ
 If the current density remains constant the
X-area will be half of the single-phase system
 i.e. 𝑎7=
𝑎5
2
=
1
2𝑐𝑜𝑠2φ 𝑎1
2
=
𝑎1
4𝑐𝑜𝑠2φ

 Since wires are four
 Then

volume of copper wireTwo phase, 4−wire a.c. system
𝑎7𝑙7
𝑎1𝑙1
=
1
4𝑐𝑜𝑠2φ 𝑎14𝑙
𝑎12𝑙
=
1
2𝑐𝑜𝑠2φ
Hence the volume of conductor material required in this system
is
1
2𝑐𝑜𝑠2φ
times that of 2-wire d.c. system with one conductor
earthed.

The third or neutral wire is taken from the junction of two-phase windings whose
voltages are in quadrature with each other. Obviously, each phase transmits one
half of the total power. The R.M.S. voltage between outgoing conductor and
neutral =
𝑉𝑚
2
, Let 𝑅8 is the resistance of each line conductor
Total power transmitted=2
𝑉𝑚
2
𝐼8 cos φ
Comparing the power with 2-wire D.C
2
𝑉𝑚
2
𝐼8 cos φ= 𝑉
𝑚𝐼1
𝐼8 =
𝐼1
2cos φ
The middle wire will act as a common return and the current through it will be
𝐼8
2
+ 𝐼8
2
= 2 𝐼8

 Assuming the current density to be constant, the area of X-section of the neutral wire will be
2 times that of either of the outers.
 ∴ Resistance of neutral wire=
𝑅8
2
 Copper losses in this system
W= 𝐼8
2
𝑅8+ 𝐼8
2
𝑅8 + ( 2 𝐼8)2𝑅8
2
= 2𝐼8
2
𝑅8+ 2𝐼8
2
𝑅8=(2+ 2) 𝐼8
2
𝑅8
Where 𝐼8 =
𝐼1
2cos φ
W=(2+ 2) (
𝐼1
2cos φ)2 𝑅8=(2+ 2)
𝐼1
2𝑅8
2𝑐𝑜𝑠2φ
(2+ 2)
𝐼1
2𝑅8
2𝑐𝑜𝑠2φ
= 2𝐼1
2
𝑅1
𝑅8
𝑅1
=
4𝑐𝑜𝑠2φ
(2+ 2)
𝑎8
𝑎1
=
(2+ 2)
4𝑐𝑜𝑠2φ
𝑎8𝑙8
𝑎1𝑙1
=
(2+ 2)
4𝑐𝑜𝑠2φ 𝑎1(2+ 2)𝑙
𝑎12𝑙
=
1.457
𝑐𝑜𝑠2φ
Hence, the volume of conductor material required for this system is
1.457
𝑐𝑜𝑠2φ times that of 2-wire d.c.
system with one conductor earthed.

This system is almost universally adopted for transmission of electric power.
The 3-phase, 3- wire system may be star connected or delta connected.
Maximum voltage between each conductor and neutral =𝑉
𝑚
R.M.S. voltage per phase =
𝑉𝑚
2
Power supplied=3(
𝑉𝑚
2
I9cos φ )
 Comparing the power with 2-wire D.C
3
𝑉𝑚
2
𝐼9 cos φ= 𝑉
𝑚𝐼1
𝐼9 =
2𝐼1
3cos φ

 Total copper losses= 3𝐼9
2
𝑅9 =3 (
2𝐼1
3cos φ
)2 𝑅9=
2𝐼1
2
𝑅9
3𝑐𝑜𝑠2φ
2𝐼1
2
𝑅9
3𝑐𝑜𝑠2φ
= 2𝐼1
2
𝑅1
𝑅9
𝑅1
= 3𝑐𝑜𝑠2φ
𝑎9
𝑎1
=
1
3𝑐𝑜𝑠2φ
volume of copper wire3−Phase, 3−wire system
=
𝑎9𝑙9
𝑎1𝑙1
=
1
3𝑐𝑜𝑠2φ𝑎13𝑙
𝑎12𝑙
=
0.5
𝑐𝑜𝑠2φ
0.5
𝑐𝑜𝑠2φ times that required for 2-wire d.c. system with one conductor earthed.

In this case, 4th or neutral wire is taken from the neutral point. The area of X-
section of the neutral wire is generally one-half that of the line conductor. If the
loads are balanced, then current through the neutral wire is zero. Assuming
balanced loads and p.f. of the load as cos φ,
𝑉𝑚
2
Power supplied=3(
𝑉𝑚
2
I10cos φ )
3
𝑉𝑚
2
𝐼10 cos φ= 𝑉
𝑚𝐼1
𝐼10 =
2𝐼1
3cos φ

2
𝑅10 =3 (
2𝐼1
3cos φ
)2 𝑅10=
2𝐼1
2
𝑅10
3𝑐𝑜𝑠2φ
2𝐼1
2
𝑅10
3𝑐𝑜𝑠2φ
= 2𝐼1
2
𝑅1
𝑅10
𝑅1
= 3𝑐𝑜𝑠2φ
𝑎10
𝑎1
=
1
3𝑐𝑜𝑠2φ
=
𝑎10𝑙10
𝑎1𝑙1
=
1
3𝑐𝑜𝑠2φ𝑎13.5𝑙
𝑎12𝑙
=
0.583
𝑐𝑜𝑠2φ
0.583
𝑐𝑜𝑠2φ times that required for 2-wire d.c. system with one conductor earthed.

In the under-ground system the maximum stress exists
on the insulation between the conductors.
(i) Same power (P watts) transmitted by each system.
(ii) The distance (l meters) over which power is
transmitted remains the same.
(iii) The line losses (W watts) are the same in each case.
(iv) The maximum voltage between conductors (𝑉
𝑚) is
the same in each case.

Max. voltage between conductors = 𝑉
𝑚 ,
Power to be transmitted = P= 𝑉
𝑚𝐼1 1
∴ Load current, 𝐼1 = P/ 𝑉
𝑚
If 𝑅1 is the resistance of each line conductor, then,
𝑅1 = ρ l/ 𝑎1
2
𝑅1 + 𝐼1
2
𝑅1
=2𝐼1
2
𝑅1
1
1
2

Maximum voltage between conductors is 𝑉
𝑚. 𝑅2 is the resistance of each line
conductor
P= 𝑉
𝑚𝐼2
Load current, 𝐼2 = P/ 𝑉
𝑚
𝑉
𝑚𝐼2= 𝑉
𝑚𝐼1
𝐼2 = 𝐼1
2
𝑅2+ 𝐼2
2
𝑅2 = 2 𝐼2
2
𝑅2
From eq 4
W=2( 𝐼1)2 𝑅2 =2𝐼1
2
𝑅2
3
4
5

2𝐼1
2
𝑅1 = (𝐼1
2
) 𝑅2
𝑅1
𝑅2
=1
𝑎1
𝑎2
=1 (where R = ρ l/ 𝑎)
𝑎1=X-area of conductor in 2 wire dc system
𝑎2= X-area of conductor in 2 wire dc system with mid-point earthed
Then
volume of copper wire Two−wire D.C system with mid−point earthed
=
𝑎2∗𝑙2
𝑎1∗𝑙1
=
𝑎1∗2𝑙
𝑎1∗2𝑙
=1 where 𝑙1 = 𝑙2=l+l
Hence, the volume of conductor material required in this system is the same as
that for 2-wire d.c. system.

If the load is balanced, the current in the neutral wire is zero. Assuming balanced loads,
P= 𝑉
𝑚𝐼3 6
𝑉
𝑚𝐼3= 𝑉
𝑚𝐼1
𝐼3 = 𝐼1 7
Let 𝑅3 is the resistance of each line conductor,
the area of X-section of neutral wire to be half that of the outer wire,
2
𝑅3+ 𝐼3
2
𝑅3 = 2𝐼3
2
𝑅3 8
From eq 7
W=2( 𝐼1)2
𝑅3 =2𝐼1
2
𝑅3 9

2𝐼1
2
𝑅1 = 2(𝐼1
2
) 𝑅3
𝑅1
𝑅3
=1
𝑎1
𝑎3
=1 𝑅3 = ρ l/ 𝑎3
Then
volume of copper wire Three−wire D.C
=
𝑎3∗𝑙3
𝑎1∗𝑙1
=
𝑎1∗2.5𝑙
𝑎1∗2𝑙
=
2.5
2
where 𝑙3=l+l+0.5l
=1.25
Hence, the volume of conductor material required in the system is 1·25 times
that required for 2-wire d.c. system.

The maximum voltage between conductors is 𝑉
𝑚, so that r.m.s. value of voltage
between them is
𝑉𝑚
2
. Assuming the load power factor to be cos φ, load current 𝐼4
Power supplied P=
𝑉𝑚
2
𝐼4 cos φ 10
𝑉𝑚
2
𝐼4 cos φ= 𝑉
𝑚𝐼1
𝐼4=
2𝐼1
cos φ
11
2
𝑅4+ 𝐼4
2
𝑅4 = 2𝐼4
2
𝑅4
=2(
2𝐼1
cos φ
)2 𝑅4 =4
𝐼1
2
𝑐𝑜𝑠2φ
𝑅4 12

4
𝐼1
2
𝑐𝑜𝑠2φ
𝑅4= 2𝐼1
2
𝑅1
𝑅4
𝑅1
=
𝑐𝑜𝑠2φ
2
𝑎4
𝑎1
=
2
𝑐𝑜𝑠2φ
13
volume of copper wire 1−phase 2− wire
=
𝑎4𝑙4
𝑎1𝑙1
=
2
𝑎12𝑙
=
2
𝑐𝑜𝑠2φ
2
𝑐𝑜𝑠2φ
times
that of 2-wire d.c. system.

The maximum value of voltage between the outers is 𝑉
𝑚.
Therefore, the r.m.s. value of voltage is
𝑉𝑚
2
Assuming the power factor of the load to be cos φ, load current 𝐼5
Power supplied P=
𝑉𝑚
2
𝐼5 cos φ 14
𝑉𝑚
2
𝐼5 cos φ= 𝑉
𝑚𝐼1
𝐼5=
2𝐼1
cos φ
15

2
𝑅5+ 𝐼5
2
𝑅5 = 2𝐼5
2
𝑅5
=2(
2𝐼1
cos φ
)2 𝑅5 =
4𝐼1
2
𝑐𝑜𝑠2φ
𝑅5
4𝐼1
2
𝑐𝑜𝑠2φ
𝑅5 =2𝐼1
2
𝑅1
𝑅5
𝑅1
=
𝑐𝑜𝑠2φ
2
𝑎5
𝑎1
=
2
𝑐𝑜𝑠2φ
𝑎5𝑙5
𝑎1𝑙1
=
2
𝑐𝑜𝑠2φ 𝑎12𝑙
𝑎12𝑙
=
2
𝑐𝑜𝑠2φ
2
𝑐𝑜𝑠2φ
times that required in 2-wire d.c. system.

 If the load is balanced, the current through the neutral wire is zero.
Assuming balanced load,
 The maximum value of voltage between the outers is 𝑉
𝑚. Therefore, the
r.m.s. value of voltage is
𝑉𝑚
2
.
 Assuming the power factor of the load to be cos φ, load current 𝐼6
 Power supplied P=
𝑉𝑚
2
𝐼6 cos φ 16
𝑉𝑚
2
𝐼6 cos φ= 𝑉
𝑚𝐼1
𝐼6=
2𝐼1
cos φ
17

2
𝑅6+ 𝐼6
2
𝑅6 = 2𝐼6
2
𝑅6
=2(
2𝐼1
cos φ
)2 𝑅6 =
4𝐼1
2
𝑐𝑜𝑠2φ
𝑅6
4𝐼1
2
𝑐𝑜𝑠2φ
𝑅6 =2𝐼1
2
𝑅1
𝑅6
𝑅1
=
𝑐𝑜𝑠2φ
2
𝑎6
𝑎1
=
2
𝑐𝑜𝑠2φ
𝑎6𝑙6
𝑎1𝑙1
= =
2
𝑐𝑜𝑠2φ 𝑎12.5𝑙
𝑎12𝑙
=
2.5
𝑐𝑜𝑠2φ
 Hence, the volume of conductor material required in this system is
2.5
𝑐𝑜𝑠2φ
times that required in 2-wire d.c. system.

 This system can be considered as two independent single-phase
systems, each transmitting one half of the total power.
 Current (𝐼7) in each conductor will be half that in single-phase 2-
wire system. Consequently, area of X-section of each conductor is
also half but as there are four wires, so volume of conductor material
used is the same as in a single phase, 2-wire system
 i.e. 𝐼7=
𝐼5
2
=
2𝐼1
cos φ
2
=
𝐼1
2 cos φ
 If the current density remains constant the
X-area will be half of the single-phase system
 i.e. 𝑎7=
𝑎5
2
=
2
𝑐𝑜𝑠2φ
2
=
1
𝑐𝑜𝑠2φ

 Since wires are four
 Then

𝑎7𝑙7
𝑎1𝑙1
=
1
𝑐𝑜𝑠2φ 𝑎14𝑙
𝑎12𝑙
=
2
𝑐𝑜𝑠2φ
Hence, volume of conductor material required in this
system is
2
𝑐𝑜𝑠2φ
times that required in 2- wire d.c. system.

 Let us suppose that maximum voltage between the outers is 𝑉
𝑚. Then
maximum voltage between either outer and neutral wire is
𝑉𝑚
2
 R.M.S. voltage between outer and neutral wire=
𝑉𝑚
2
2
=
𝑉𝑚
2
 Total power transmitted=2
𝑉𝑚
2
𝐼8 cos φ = 𝑉
𝑚 𝐼8 cos φ
𝑉
𝑚 𝐼8 cos φ= 𝑉
𝑚𝐼1
𝐼8 =
𝐼1
cos φ
The middle wire will act as a common return and the current through it will be
𝐼8
2
+ 𝐼8
2
= 2 𝐼8

 Assuming the current density to be constant, the area of X-section of the neutral wire will be
2 times that of either of the outers.
 ∴ Resistance of neutral wire=
𝑅8
2
 Copper losses in this system
W= 𝐼8
2
𝑅8+ 𝐼8
2
𝑅8 + ( 2 𝐼8)2𝑅8
2
= 2𝐼8
2
𝑅8+ 2𝐼8
2
𝑅8=(2+ 2) 𝐼8
2
𝑅8
Where 𝐼8 =
𝐼1
cos φ
W=(2+ 2) (
𝐼1
cos φ)2
𝑅8=(2+ 2)
𝐼1
2𝑅8
𝑐𝑜𝑠2φ
(2+ 2)
𝐼1
2𝑅8
𝑐𝑜𝑠2φ
= 2𝐼1
2
𝑅1
𝑅8
𝑅1
=
2𝑐𝑜𝑠2φ
(2+ 2)
𝑎8
𝑎1
=
(2+ 2)
2𝑐𝑜𝑠2φ
𝑎8𝑙8
𝑎1𝑙1
=
(2+ 2)
2𝑐𝑜𝑠2φ 𝑎1(2+ 2)𝑙
𝑎12𝑙
=
2.914
𝑐𝑜𝑠2φ
2.914
𝑐𝑜𝑠2φ times that of 2-wire d.c.
system.

This system is almost universally adopted for transmission of electric power. The 3-phase,
3- wire system may be star connected or delta connected.
Maximum voltage between conductors =𝑉
𝑚, Let 𝑅9 is the resistance of each line conductor
Then maximum voltage between each phase and neutral is
𝑉𝑚
3
𝑉𝑚
3
2
=
𝑉𝑚
6
Power supplied=3(
𝑉𝑚
6
I9cos φ )
3
𝑉𝑚
6
𝐼9 cos φ= 𝑉
𝑚𝐼1
𝐼9 =
6𝐼1
3cos φ

2
𝑅9 =3 (
6𝐼1
3cos φ
)2 𝑅9=
2𝐼1
2
𝑅9
𝑐𝑜𝑠2φ
2𝐼1
2
𝑅9
𝑐𝑜𝑠2φ
= 2𝐼1
2
𝑅1
𝑅9
𝑅1
= 𝑐𝑜𝑠2φ
𝑎9
𝑎1
=
1
𝑐𝑜𝑠2φ
=
𝑎9𝑙9
𝑎1𝑙1
=
1
𝑎12𝑙
=
1.5
𝑐𝑜𝑠2φ
1.5
𝑐𝑜𝑠2φ times that required for 2-wire d.c. system.

In this case, 4th or neutral wire is taken from the neutral point. The area of X-
section of the neutral wire is generally one-half that of the line conductor. If the
loads are balanced, then current through the neutral wire is zero. Assuming
balanced loads and p.f. of the load as cos φ,
𝑉𝑚
6
Power supplied=3(
𝑉𝑚
6
I10cos φ )
3
𝑉𝑚
6
𝐼10 cos φ= 𝑉
𝑚𝐼1
𝐼10 =
6𝐼1
3cos φ

2
𝑅10 =3 (
6𝐼1
3cos φ
)2 𝑅10=
2𝐼1
2
𝑅10
𝑐𝑜𝑠2φ
2𝐼1
2
𝑅10
𝑐𝑜𝑠2φ
= 2𝐼1
2
𝑅1
𝑅10
𝑅1
= 𝑐𝑜𝑠2φ
𝑎10
𝑎1
=
1
𝑐𝑜𝑠2φ
=
𝑎10𝑙10
𝑎1𝑙1
=
1
𝑐𝑜𝑠2φ𝑎13.5𝑙
𝑎12𝑙
=
1.75
𝑐𝑜𝑠2φ
Hence, the volume of conductor material required for this system
is
1.75
𝑐𝑜𝑠2φtimes that required for 2-wire d.c. system.

System Same maximum
voltage to earth
Same maximum
voltage
between conductors
1. D.C. system
(i) Two-wire
(ii) Two-wire mid-point earthed
(iii) 3-wire
1
0.25
0.3125
1
1
1.25
2. Single phase system
(i) Two-wire
(ii) Two-wire mid-point earthed
(iii) 3-wire
2
𝑐𝑜𝑠2φ
0.5
𝑐𝑜𝑠2φ
1.625
𝑐𝑜𝑠2φ
2
𝑐𝑜𝑠2φ
0.5
𝑐𝑜𝑠2φ
2.5
𝑐𝑜𝑠2φ

System Same maximum voltage to
earth
Same maximum voltage
between conductors
3. Two-phase system
(i) 2–phase 4-wire
(ii)2-phase 3-wire
0.5
𝑐𝑜𝑠2φ
1.457
𝑐𝑜𝑠2φ
2
𝑐𝑜𝑠2φ
2.914
𝑐𝑜𝑠2φ
4. Three-phase system
(i) 3-phase 3-wire
(ii)3-phase 4-wire
,
0.5
𝑐𝑜𝑠2φ
0.583
𝑐𝑜𝑠2φ
1.5
𝑐𝑜𝑠2φ
1.75
𝑐𝑜𝑠2φ

What is the percentage saving in feeder copper if
the line voltage in a 2-wire d.c. system is raised
from 200 volts to 400 volts for the same power
transmitted over the same distance and having the
same power loss ?

 (i) Conductors
 (ii) Step-up and step-down transformers
 (iii) Line insulators
 (iv) Support
 (v) Protective devices
 (vi) Voltage regulating devices

The following two fundamental economic
principles which closely influence the electrical
design of a transmission line
(i) Economic choice of transmission voltage
(ii) Economic choice of conductor size

 The effect of increased voltage level of
transmission on
1. Volume of conductor material
used for transmission
2. Line Drop

Let a 3-phase A.C. system is used for the transmission, The resistance per conductor is given by
R= ρ l/A
Power transmitted P= 3𝑉𝐼cos φ
I=
𝑃
3𝑉cos φ
Total copper losses 𝑃𝑐𝑢= 3𝐼2R =3(
𝑃
3𝑉cos φ
)2ρ l
𝐴
A=
𝑃2ρ l
𝑃𝑐𝑢𝑉2𝑐𝑜𝑠2φ
The volume of copper used is V=3(A*l)
Volume of copper=
3𝑃2ρ 𝑙2
𝑃𝑐𝑢𝑉2𝑐𝑜𝑠2φ
Volume of copper 𝛼
1
𝑉2𝑐𝑜𝑠2φ
Thus greater is the
transmission voltage level,
lesser is the volume of the
copper required.

Line Drop=IR=I ρ l/A
Current density(J)=
𝐼
𝐴
Then Line Drop=
𝐼ρ l
𝐼
𝐽
= Jρ l
%Line Drop=
Jρ l
𝑉
*100
Higher the transmission voltage level, then lesser the
%Line Drop

 The most economical area of conductor is that
for which the total annual cost of transmission
line is minimum . This is known as Kelvin’s
Law
 The total annual cost of transmission line can
be divided broadly into two parts viz.,
 Annual charge on capital outlay
 Annual cost of energy wasted in the conductor.

 This is on account of interest and depreciation on the capital cost of
complete installation of transmission line. In case of overhead system,
it will be the annual interest and depreciation on the capital cost of
conductors, supports and insulators and the cost of their erection.
 Now, for an overhead line, insulator cost is constant, the conductor
cost is proportional to the area of X-section and the cost of supports
and their erection is partly constant annual charge.
 Transmission line can be expressed as :
𝐴𝑛𝑛𝑢𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒 = 𝑃1 + 𝑃2𝑎
𝑃1 constant depending on the copper conductor
𝑃2 constant depending on the X-area of the conductor
a is the area of X-section of the conductor

 This is on account of energy lost mainly in the conductor due to 𝐼2R losses.
 Assuming a constant current in the conductor throughout the year,
 The energy lost in the conductor is proportional to resistance.
where R = ρ l/ 𝑎
 The energy lost in the conductor is inversely proportional to area of X-section.
energy lost α
1
𝑎
 Thus, the annual cost of energy wasted in an overhead transmission line can
be expressed as :
Annual cost of energy wasted =
𝑃3
𝑎
where P3 is a constant.

 Total annual cost, C=𝑃1 + 𝑃2𝑎+
𝑃3
𝑎
only area of X-section a is variable. Therefore, the total annual cost of
transmission line will be minimum if differentiation of C w.r.t. a is zero i.e.
ⅆ𝐶
ⅆ𝑎
=0
ⅆ
ⅆ𝑎
(𝑃1 + 𝑃2𝑎+
𝑃3
𝑎
)=0
𝑃2𝑎 −
𝑃3
𝑎2=0
𝑃2𝑎 =
𝑃3
𝑎2
i.e. Variable part of annual charge = Annual cost of energy wasted
Therefore Kelvin’s Law can also be stated
The most economical area of conductor is that for which the variable part of
annual charge is equal to the cost of energy losses per year.

 The straight line shows the relation between the
annual charge (i.e., 𝑃1 + 𝑃2𝑎) and the area of X-
section a of the conductor
 The rectangular hyperbola
gives the relation between
annual cost of energy wasted (
𝑃3
𝑎
)
and X-sectional area a.
 Curve 3 shows the relation between total annual
cost of transmission line and area of X-section a.

 (i) It is not easy to estimate the energy loss in the line without actual load
curves, which are not available at the time of estimation.
 (ii) The assumption that annual cost on account of interest and
depreciation on the capital outlay is in the form 𝑃1 + 𝑃2𝑎 is strictly
speaking not true. For instance, in cables neither the cost of cable dielectric
and sheath nor the cost of laying vary in this manner.
 (iii) This law does not take into account several physical factors like safe
current density, mechanical strength, corona loss etc.
 (iv) The conductor size determined by this law may not always be
practicable one because it may be too small for the safe carrying of
necessary current.
 (v) Interest and depreciation on the capital outlay cannot be determined
accurately.

A 2-conductor cable 1 km long is required to supply a constant current
of 200 A throughout the year. The cost of cable including installation is
Rs. (20 a + 20) per meter where ‘a’ is the area of X-section of the
conductor in 𝑐𝑚2
. The cost of energy is 5P per kWh and interest and
depreciation charges amount to 10%. Calculate the most economical
conductor size. Assume resistivity of conductor material to be 1·73 μ Ω
cm.
economical conductor size a =?
Energy lost per annum=2𝐼2
𝑅𝑡
Resistance of one conductor=ρ l/ 𝑎
=
1.73∗10−6∗105
𝑎
=
0.173
𝑎
Ω
Energy lost per annum= 2∗ 2002
∗
0.173
𝑎
∗ 8760
=
121238.4
𝑎
kWh

 Annual cost of energy lost = Cost per kWh × Annual energy loss
=Rs 0.05*
121238.4
𝑎
kWh=Rs
6062
𝑎
The capital cost (variable) of the cable is given to be Rs 20 a per metre.
Therefore, for 1 km length of the cable, the capital cost (variable) is Rs. 20 a ×
1000 = Rs. 20,000 a.
Variable annual charge = Annual interest and depreciation on
capital cost (variable) of cable
= Rs 0·1 × 20,000 a
= Rs 2000 a
According to Kelvin’s law, for most economical X-section of the conductor,
Variable annual charge = Annual cost of energy lost
2000 a =
6062
𝑎
a=
6062
2000
=1.74 𝑐𝑚2

A 2-wire feeder carries a constant current of 250 A throughout the year.
The portion of capital cost which is proportional to area of X-section is
Rs 5 per kg of copper conductor. The interest and depreciation total
10% per annum and the cost of energy is 5P per kWh. Find the most
economical area of X-section of the conductor. Given that the density of
copper is 8·93gm/ 𝑐𝑚3
and its specific resistance is 1·73 × 10−8
Ω m.
Consider 1 metre length of the feeder. Let a be the most economical
area of X-section of each conductor in 𝑚2
Energy lost per annum=2𝐼2
𝑅𝑡
Resistance of one conductor=ρ l/ 𝑎
=
1.73∗10−8∗1
𝑎
=
1.73∗10−8
𝑎
Ω
Energy lost per annum= 2∗ 2502
∗
1.73∗10−8
𝑎
∗ 8760
=
18,94,350
𝑎
∗ 10−8
kWh

 Annual cost of energy lost = Cost per kWh × Annual energy loss
=Rs 0.05*
18,94,350
𝑎
∗ 10−8
kWh=Rs
94,717.5
𝑎
∗ 10−8
Mass of 1 metre feeder = 2 (Volume × density)
= 2 × a × 1 × 8·93 × 103kg
= 17·86 × 103 a kg
Capital cost (variable) = Rs 5 × 17·86 × 103
a
= Rs 89·3 × 103
a
Variable annual charge = 10% of capital cost (variable)
= 0·1 × 89·3 × 103 a = Rs 8930 a
Variable annual charge = Annual cost of energy lost
8930 a =
94,717.5
𝑎
∗ 10−8
a=
94,717.5
8930
∗ 10−8
=3.25∗ 10−4
𝑚2
= 3.25 𝑐𝑚2

Determine the most economical cross-section for a 3-phase
transmission line, 1 km long to supply at a constant voltage of
110 kV for the following daily load cycle :
6 hours 20 MW at p.f. 0·8 lagging
The line is used for 365 days yearly. The cost per km of line
including erection is Rs (9000 +6000 a) where ‘a’ is the area of X-
section of conductor in cm2. The annual rate of interest and
depreciation is 10% and the energy costs 6P per kWh. The
resistance per km of each conductor is0·176/a.
Power transmitted P= 3𝑉𝐼cos φ
Energy lost per annum=3R(𝐼2
𝑡)

 The load currents at various loads are :
 At 20 MW, 𝐼1=
𝑃
3𝑉cos φ
=
20∗106
3∗110∗103∗0.8
= 131.2𝐴
 At 5 MW, 𝐼1=
𝑃
3𝑉cos φ
=
5∗106
3∗110∗103∗0.8
= 32.8𝐴
 At 6 MW, 𝐼1=
𝑃
3𝑉cos φ
=
6∗106
3∗110∗103∗0.8
= 39.36𝐴
 Energy loss per day in 3-phase line
=3*
0.176
𝑎
(131.22
∗ 6+ 32.82
∗ 12+ 36.362
∗ 6)
=
66.26
𝑎
kWh
Energy lost per annum =
66.26
𝑎
*365= =
24,184.9
𝑎
kWh
Annual cost of energy=Rs
6
100
∗
24,184.9
𝑎
= =
1451.09
𝑎
Variable annual charge = 10% of capital cost (variable) of line
= Rs 0·1 × 6000 a = Rs 600 a
Variable annual charge = Annual cost of energy
600 a=
1451.09
𝑎
a=
1451.09
600
=1.56 𝑐𝑚2

The conveyance of electric power from a power station to
consumers’ premises is known as electric supply system.
 Electric supply system consist of three principle components:
1) Power station
2) Transmission lines
3) Distribution system
 The electric supply system can be broadly classified into:
1) DC or AC system
2)Overhead or underground system

Different blocks of typical acpowersupply scheme are asper
following::
1)Generating Station:
Ingenerating station power is producedby three phase alternators
operating inparallel.
The usual generation voltage is 11kV.
Foreconomy in the transmission of electric power, the generation
voltage is stepped upto 132 kVor more at generating station with the
help of three phase transformer.

2)Primarytransmission:
Theelectric power at 132 kVis transmitted by 3-phase,3 wire overhead system to the out
3)Secondary transmission:
Atthe receiving station the voltage is reduced to33kV by step down transformer.
4)Primary distribution:
Atthe substation voltage is reduced 33kV to 11kV.The11kV line run along important
roadsides to city. This forms the primary distribution.

5)Secondarydistribution:
Theelectric power form primary distribution line is delivered to distribution substation. The
substation is located nearthe consumers location and step down the voltage to 400V, 3-
phase ,4-wire for secondary distribution.
The voltage between any two phases is 400V and between any phase and neutral is 230V.

Basis for Comparison AC Transmission Line DC Transmission Line
Definition The ac transmission line transmit the
alternating current.
The dc transmission line is used for
transmitting the direct current.
Number of Conductors Three Two
Inductance & surges Have Don’t Have
Voltage drop High Low
Skin Effect Occurs Absent
Need of Insulation More Less
Interference Have Don’t Have
Corona Loss Occur Don’t occur
Dielectric Loss Have Don’t have
Synchronizing and Stability Problem No difficulties Difficulties
Cost Expensive Cheap
Length of conductors Small More
Repairing and Maintenance Easy and Inexpensive Difficult and Expensive
Transformer Requires Not Requires

Which distributes
electric power for
local use is known as
distribution system.

 (i) Feeders. A feeder is a
conductor which connects the
sub-station to the area where
power is to be distributed.
 (ii) Distributor. A distributor is
a conductor from which
tappings are taken for supply
to the consumers.
 (iii) Service mains. A service
mains is generally a small cable
which connects the distributor
to the consumers’ terminals.

Primary
distribution system
Secondary
distribution system

2-wire d.c. system
3 -wire d.c. system

 (i) Radial System.
In this system, separate feeders radiate from a single substation and feed
the distributors at one end only.

 Limitations
 (a) The end of the distributor nearest to the feeding point will be
heavily loaded.
 (b) The consumers are dependent on a single feeder and single
distributor. Therefore, any fault on the feeder or distributor cuts off
supply to the consumers who are on the side of the fault away from
the substation.
 (c) The consumers at the distant end of the distributor would be
subjected to serious voltage fluctuations when the load on the
distributor changes.
Due to these limitations, this system is used for short
distances only.

 (ii) Ring main system.
In this system, the primaries of distribution transformers form a loop. The
loop circuit starts from the substation bus-bars, makes a loop through the area to be served,
and returns to the substation.
✓ substation supplies to the
closed feeder LMNOPQRS.
✓ The distributors are tapped
from different points M, O and Q
of the feeder through distribution
transformers.
➢ There are less voltage fluctuations
at consumer’s terminals.
➢ The system is very reliable as each
distributor is fed via two feeders.
➢ In the event of fault
on any section of the feeder,
the continuity of supply is maintained.

 (iii) Interconnected system. When the feeder ring is energized by two or
more than two generating stations or substations, it is called inter-connected
system.
(a) It increases the service reliability.
(b) Any area fed from one generating station during peak load hours can be fed
from the other generating station. This reduces reserve power capacity and
increase efficiency of the system.

 (i) Proper voltage.
 One important requirement of a distribution system is that voltage
variations at consumer’s terminals should be as low as possible.
 The changes in voltage are generally caused due to the variation of
load on the system.
 Low voltage causes loss of revenue, inefficient lighting and possible
burning out of motors.
 High voltage causes lamps to burn out permanently and may cause
failure of other appliances.
 Thus, if the declared voltage is 230 V, then the highest voltage of the
consumer should not exceed 244 V while the lowest voltage of the
consumer should not be less than 216 V.

 (ii) Availability of power on demand. Power must be
available to the consumers in any amount that they
may require from time to time.
 (iii) Reliability. Modern industry is almost dependent
on electric power for its operation.
 The reliability can be improved to a considerable extent
by
(a) interconnected system
(b) reliable automatic control system
(c) providing additional reserve facilities.

(i) Distributor fed at one end
(ii) Distributor fed at both ends
(iii) Distributor fed at the centre
(iv) Ring distributor.

 (i) Distributor fed
at one end
 (ii) Distributor
fed at both ends

 (iii) Distributor
fed at the centre
 (iv) Ring
distributor

(i)
Concentrated
loading
(ii) Uniform
loading
(iii) Both
concentrated
and uniform
loading.

 2-wire d.c. distributor AB fed at one end A and having concentrated loads 𝐼1,
𝐼2, 𝐼3 and 𝐼4 tapped off at points C, D, E and F respectively.
 Let 𝑟1 , 𝑟2, 𝑟3 𝑎𝑛𝑑 𝑟4 be the resistances of both wires (go and return) of the sections AC,
CD, DE and EF of the distributor respectively.
 Current fed from point A = 𝐼1 + 𝐼2 + 𝐼3 + 𝐼4 Current in section AC = 𝐼1 + 𝐼2 + 𝐼3 + 𝐼4
 Current in section CD = 𝐼2 + 𝐼3 + 𝐼4 Current in section DE = 𝐼3 + 𝐼4
 Current in section EF = 𝐼4
 Voltage drop in section AC = 𝑟1 (𝐼1 + 𝐼2 + 𝐼3 + 𝐼4)
 Voltage drop in section CD = 𝑟2 (𝐼2 + 𝐼3 + 𝐼4) Voltage drop in section DE = 𝑟3 (𝐼3 + 𝐼4)
 Voltage drop in section EF = 𝑟4 𝐼4
 ∴ Total voltage drop in the distributor
=𝑟1 (𝐼1 + 𝐼2 + 𝐼3 + 𝐼4)+𝑟2 (𝐼2 + 𝐼3 + 𝐼4)+𝑟3 (𝐼3 + 𝐼4)+ 𝑟4 𝐼4

 2-wire d.c. distributor AB fed at one end A and
loaded uniformly with i amperes per metre length.
 The load tapped is i amperes. Let 𝑙 metres be the
length of the distributor and r ohm be the
resistance per metre run.

 Consider a point C on the distributor at a distance x metres from the feeding point A
 Then current at point C is
= i 𝑙− i x amperes = i (𝑙− x) amperes
Now, consider a small length dx near point C. Its resistance is r dx and the voltage drop over
length dx is
dv = i (𝑙 − x) r dx = i r (𝑙 − x) dx
Total voltage drop in the distributor upto point C is
v =‫׬‬0
𝑥
ir (l − x)dx =ir 𝑙𝑥 −
𝑥2
2
The voltage drop upto point B (i.e. over the whole distributor) can be obtained by putting x = 𝑙 in
the above expression.
∴ Voltage drop over the distributor AB
= ir 𝑙 ∗ 𝑙 −
𝑙2
2
=
1
2
𝑖𝑟𝑙2
=
1
2
𝑖𝑙 𝑟𝑙 =
1
2
𝐼𝑅
Thus, in a uniformly loaded distributor fed at one end, the total voltage drop is equal to that
produced by the whole of the load assumed to be concentrated at the middle point.

 The two ends of the distributor may be supplied
with (i) equal voltages (ii) unequal voltages.

 Consider a distributor AB fed at both ends with equal voltages V volts and having
concentrated loads 𝐼1, 𝐼2, 𝐼3, 𝐼4 and 𝐼5 at points C, D, E, F and G respectively
 As we move away from one of the feeding points, say A, p.d. goes on decreasing till it
reaches the minimum value at some load point, say E, and then again starts rising and
becomes V volts as we reach the other feeding point B.
 All the currents tapped off between points A and E (minimum p.d. point) will be
supplied from the feeding point A while those tapped off between B and E will be
supplied from the feeding point B.
 The current tapped off at point E itself will be partly supplied from A and partly from
B. If these currents are x and y respectively, then,
x+y= 𝐼3
 Therefore, we arrive at a very important conclusion that at the point of minimum
potential, current comes from both ends of the distributor.
 Voltage drop from feeding point A to the point of minimum potential difference is
equal to the voltage drop from feeding point B to the point of minimum potential
difference

 Distributor AB fed with unequal voltages ; end A being fed at 𝑉1volts
and end B at 𝑉2 volts.
 In order to determine the point of minimum potential, assume of 𝐼𝐴
amperes is supplied from feeding point A
 Let r be the resistance of each conductor per unit length(l=1) for both go
& return
 Resistance of portions AC,CD, DE,EF,FB are 𝑟1 , 𝑟2, 𝑟3 , 𝑟4, 𝑎𝑛𝑑 𝑟5
respectively
 Voltage drop between A and B = Voltage drop over AB
𝑉1 − 𝑉2 = 𝐼𝐴𝑟1+(𝐼𝐴- 𝐼1) 𝑟2 + (𝐼𝐴- 𝐼1- 𝐼2) 𝑟3 + (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3) 𝑟4+ (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3 − 𝐼4) 𝑟5
The load point at which currents are coming from both sides of the
distributor will be the point of minimum potential

 A 2-wire d.c. street mains AB, 600 m long is fed from both ends at 220 V. Loads of 20
A, 40 A, 50 A and 30 A are tapped at distances of 100m, 250m, 400m and 500 m from
the end A respectively. If the area of X-section of distributor conductor is 1𝑐𝑚2
, find
the minimum consumer voltage. Take ρ = 1·7 × 10−6 Ω cm.
R= ρ l/A
 Resistance of section AC of distributor=2×
1·7 × 10−6∗100∗102
1
= 3·4 × 10−2 Ω
 Resistance of section CD of distributor =0.051 Ω
 Resistance of section DE of distributor =0.051 Ω
 Resistance of section EF of distributor =0.034 Ω
 Resistance of section FB of distributor =0.034 Ω
 𝑉𝐴 − 𝑉𝐵 = 𝐼𝐴𝑟1+(𝐼𝐴- 𝐼1) 𝑟2 + (𝐼𝐴- 𝐼1- 𝐼2) 𝑟3 + (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3) 𝑟4+ (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3 − 𝐼4) 𝑟5
 220-200=0.034 𝐼𝐴+(𝐼𝐴-20)0.051+ (𝐼𝐴-60)0.051+(𝐼𝐴-110)0.034+(𝐼𝐴-140)0.034
𝐼𝐴=61.7A

 Actual distribution of currents in the various sections of the
distributor is
 Current in section EF, 𝐼𝐸𝐹 = 𝐼𝐴−110 =61·7−110 =−48·3A from E to F
= 48·3 A from F to E
 Current in section DE, 𝐼𝐷𝐸 = 𝐼𝐴 −60=61.7-60=1.7A from D to E
It is clear that currents are coming to load point E from both sides i.e.
from point D and point F. Hence, E is the point of minimum potential.
∴ Minimum consumer voltage,
𝑉𝐸 = 𝑉𝐴-(𝐼𝐴𝐶 𝑟1+ 𝐼𝐶𝐷 𝑟2+ 𝐼𝐷𝐸 𝑟3)
=220-(61·7 × 0·034 + 41·7 × 0·051 + 1·7 × 0·051]
=215.69V

A two-wire d.c. distributor AB, 600 metres long is loaded as under :
Distance from A (metres) : 150 300 350 450
Loads in Amperes : 100 200 250 300
The feeding point A is maintained at 440 V and that of B at 430 V. If
each conductor has a resistance of 0·01 Ω per 100 metres, calculate :
(i) the currents supplied from A to B,
(ii) the power dissipated in the distributor.
Let 𝐼𝐴amperes be the current supplied from the feeding point A. Then currents in the
various sections of the distributor are
Resistance of 100 m length of distributor (both wires)
= 2 × 0·01 = 0·02 Ω
Resistance of section AC, 𝑟1 = 0·02 × 150/100 = 0·03 Ω
Resistance of section CD, 𝑟2 = 0·02 × 150/100 = 0·03 Ω
Resistance of section DE, 𝑟3= 0·02 × 50/100 = 0·01 Ω
Resistance of section EF, 𝑟4= 0·02 × 100/100 = 0·02 Ω
Resistance of section FB, 𝑟5= 0·02 × 150/100 = 0·03 Ω

 𝑉𝐴 − 𝑉𝐵 = 𝐼𝐴𝑟1+(𝐼𝐴- 𝐼1) 𝑟2 + (𝐼𝐴- 𝐼1- 𝐼2) 𝑟3 + (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3) 𝑟4+ (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3 − 𝐼4) 𝑟5
 440-430=0.03 𝐼𝐴+(𝐼𝐴-100)0.03+ (𝐼𝐴-300)0.01+(𝐼𝐴-550)0.02+(𝐼𝐴-850)0.03
𝐼𝐴=437.5A
 The actual distribution of currents in the various sections of the distributor are
 Current supplied from end A, 𝐼𝐴 = 𝐼𝐴𝐶 = 437·5 A
 Current supplied from end B, 𝐼𝐵 = 𝐼𝐹𝐵 = 𝐼𝐴 -850= - 412·5 A
 𝐼𝐶𝐷 = 337.5𝐴 𝐼𝐷𝐸 = 137.5𝐴 𝐼𝐸𝐹 = −112.5𝐴
 Therefore E is the point of minimum potential.
 Power loss in the distributor
=𝐼𝐴𝐶
2
𝑟1+ 𝐼𝐶𝐷
2
𝑟2 + 𝐼𝐷𝐸
2
𝑟3+ 𝐼𝐸𝐹
2
𝑟4+ 𝐼𝐹𝐵
2
𝑟5
=(437.5)2
∗ 0.03 + 337.5 2
∗ 0.03 + 137.5 2
∗ 0.01 + 112.5 2
∗ 0.02 + (412.5)2
∗ 0.03
=14.705kW

 (i) Distributor fed at both ends with equal voltages.
 Consider a distributor AB of length 𝑙 metres, having resistance r
ohms per metre run and with uniform loading of 𝑖 amperes per
metre run.
 Let the distributor be fed at the feeding points A and B at equal
voltages, say V volts.
 The total current supplied to the distributor is 𝑖𝑙.
 As the two end voltages are equal, therefore, current supplied from
each feeding point is
𝑖𝑙
2
i.e.

 Consider a point C at a distance x metres from the feeding point A. Then
current at point C is
=
𝑖𝑙
2
- 𝑖𝑥= 𝑖(
𝑙
2
-𝑥)
Now, consider a small length dx near point C. Its resistance is r dx and the
voltage drop over length dx is
dv = i (
𝑙
2
− x) r dx = i r (
𝑙
2
− x) dx
Total voltage drop in the distributor upto point C is
v =‫׬‬
0
𝑥
ir (
𝑙
2
− x)dx =ir
𝑙𝑥
2
−
𝑥2
2
=
𝑖𝑟
2
(𝑙𝑥- 𝑥2)
Obviously, the point of minimum potential will be the mid-point. Therefore,
maximum voltage drop will occur at mid-point i.e. where x =
𝑙
2
Max. voltage drop=
𝑖𝑟
2
(𝑙𝑥- 𝑥2)=
𝑖𝑟
2
(𝑙
𝑙
2
- (
𝑙
2
)2)=
1
8
𝑖𝑟𝑙2=
1
8
𝑖𝑙 𝑟𝑙 =
1
8
IR
Minimum voltage=V-
1
8
IR volts

 (ii) Distributor fed at both ends with unequal voltages.
 Consider a distributor AB of length l metres having resistance r
ohms per metre run and with a uniform loading of i amperes
permetre run. Let the distributor be fed from feeding points A and B
at voltages 𝑉𝐴 and 𝑉𝐵 respectively.
 Suppose that the point of minimum potential C is situated at a
distance x metres from the feeding point A. Then current supplied
by the feeding point A will be i x.
 ∴ Voltage drop in section AC=
1
2
𝑖𝑟𝑥2 volts
 As the distance of C from feeding point B is (l − x), therefore, current
fed from B is i (l − x).
 ∴ Voltage drop in section BC =
1
2
𝑖𝑟(𝑙 − 𝑥)2
volts

 Voltage at point C, 𝑉𝐶 = 𝑉𝐴 − Drop over AC
= 𝑉𝐴 -
1
2
𝑖𝑟𝑥2
Voltage at point C, 𝑉𝐶 = 𝑉𝐵 − Drop over BC
= 𝑉𝐵 -
1
2
From above equations we get,
𝑉𝐴 -
1
2
𝑖𝑟𝑥2 = 𝑉𝐵 -
1
2
𝑉𝐴- 𝑉𝐵 =
1
2
𝑖𝑟𝑥2
-
1
2
=
1
2
𝑖𝑟𝑥2
-
1
2
𝑖𝑟𝑙2
-
1
2
𝑖𝑟𝑥2
+
1
2
𝑖𝑟2𝑙𝑥
= 𝑖𝑟𝑙 {𝑥-
1
2
𝑙 }
𝑥=
𝑉𝐴− 𝑉𝐵
𝑖𝑟𝑙
+
𝑙
2
The point on the distributor where minimum potential occurs can be calculated

 Expression for the power loss in a uniformly loaded distributor fed at both ends
 current supplied from each feeding point is
𝑖𝑙
2
 Consider a small length dx of the distributor at point P which is at a distance x from
the feeding end A.
 Resistance of length dx = r dx
 Current in length dx =
𝑖𝑙
2
- 𝑖𝑥= 𝑖(
𝑙
2
-𝑥)
 Power loss in length dx
= (current in dx)2 × Resistance of dx
= 𝑖(
𝑙
2
−𝑥)
2
×rdx
 Total power loss in the distributor is=‫׬‬
0
𝑙
𝑖(
𝑙
2
−𝑥)
2
×rdx
=𝑖2
𝑟 ‫׬‬
0
𝑙 𝑙2
4
− 𝑙𝑥 + 𝑥2
𝑑𝑥
= 𝑖2𝑟
𝑙2𝑥
4
−
𝑙𝑥2
2
+
𝑥3
3
𝑙
0
P=
𝑖2𝑟𝑙3
12

A 2-wire d.c. distributor AB 500 metres long is fed from both ends and is loaded uniformly
at the rate of 1·0 A/metre. At feeding point A, the voltage is maintained at 255 V and at B
at 250 V. If the resistance of each conductor is 0·1 Ω per kilometre, determine :
(i) the minimum voltage and the point where it occurs
(ii) the currents supplied from feeding points A and B
Let the minimum potential occur at a point C distant x metres from the feeding point A.
𝑥=
𝑉𝐴− 𝑉𝐵
𝑖𝑟𝑙
+
𝑙
2
Resistance of distributor/m, r = 2 × 0·1/1000 = 0·0002 Ω
𝑥=
255− 250
1∗0.0002∗500
+
500
2
=300m
i.e. minimum potential occurs at 300 m from point A.
Minimum voltage, 𝑉𝐶 = 𝑉𝐴 -
1
2
𝑖𝑟𝑥2 = 𝑉𝐵 -
1
2
255-
1∗0.0002∗ (300)2
2
=246V
Current supplied from A = i x = 1 × 300 = 300 A
Current supplied from B = i (𝑙 − x) = 1 (500 − 300)
= 200 A

A 800 metres 2-wire d.c. distributor AB fed from both ends is uniformly
loaded at the rate of 1·25 A/metre run. Calculate the voltage at the
feeding points A and B if the minimum potential of 220 V occurs at
point C at a distance of 450 metres from the end A. Resistance of each
conductor is 0·05 Ω/km.
 Minimum voltage, 𝑉𝐶 = 𝑉𝐴 -
1
2
𝑖𝑟𝑥2
= 𝑉𝐵 -
1
2
 For 𝑉𝐴
𝑉𝐶 = 𝑉𝐴 -
1
2
𝑖𝑟𝑥2
𝑉𝐴=220+
1.25∗0.0001∗ (450)2
2
=232.65V
 For 𝑉𝐵
𝑉𝐶 = 𝑉𝐵 -
1
2
𝑉𝐵 =220+
1.25∗0.0001∗ (800−450)2
2
=227.65V

 (i) A uniformly loaded distributor is fed at the centre. Show that maximum
voltage drop = I R/8 where I is the total current fed to the distributor and R
is the total resistance of the distributor.
 (ii) A 2-wire d.c. distributor 1000 metres long is fed at the centre and is
loaded uniformly at the rate of 1·25 A/metre. If the resistance of each
conductor is 0·05 Ω/km, find the maximum voltage drop in the distributor.
 ∴ Max. voltage drop = Voltage drop in half distributor
=
1
2
(
𝑖𝑙
2
)(
𝑟𝑙
2
) =
1
8
𝑖𝑙 𝑟𝑙 =
1
8
IR
where i l = I, the total current fed to the distributor
r l = R, the total resistance of the distributor
 (ii) Total current fed to the distributor is
 I = i l = 1·25 × 1000 = 1250 A
 Total resistance of the distributor is
 R = r l = 2 × 0·05 × 1 = 0·1 Ω
 Max. voltage drop =
1
8
IR
=
1
8
*1250*0.1
=15.62V

 A two-wire d.c. distributor cable 1000 metres long is loaded with 0·5
A/metre. Resistance of each conductor is 0·05 Ω/km. Calculate the
maximum voltage drop if the distributor is fed from both ends with
equal voltages of 220 V. What is the minimum voltage and where it
occurs ?
 Current loading, i = 0·5 A/m
 Resistance of distributor/m, r = 2 × 0·05/1000 = 0·1 × 10−3
Ω
 Length of distributor, l = 1000 m
 Total current supplied by distributor, I = i l = 0·5 × 1000 = 500 A
 Total resistance of the distributor, R = r l = 0·1 × 10−3 × 1000 = 0·1 Ω
 ∴ Max. voltage drop =
1
8
IR
= 6·25 V
 Minimum voltage will occur at the mid-point of the distributor and
its value is
 = 220 − 6·25 = 213·75 V

The total drop over any section of the distributor
is equal to the sum of drops due to concentrated
and uniform loading in that section.

Two conductors of a d.c. distributor cable AB 1000 m long have a total
resistance of 0·1 Ω. The ends A and B are fed at 240 V. The cable is
uniformly loaded at 0·5 A per metre length and has concentrated loads
of 120 A, 60 A, 100 A and 40 A at points distant 200 m, 400 m, 700 m and
900 m respectively from the end A. Calculate (i) the point of minimum
potential (ii) currents supplied from ends A and B (iii) the value of
minimum potential.
Distributor resistance per metre length, r = 0·1/1000 = 10−4
Ω
(i) Point of minimum potential. The point of minimum potential is not
affected by the uniform loading of the distributor. Therefore, let us
consider the concentrated loads.
Suppose the current supplied by end A is 𝐼𝐴. Then currents in the
various sections will be

 𝑉𝐴 − 𝑉𝐵 = 𝐼𝐴𝑟1+(𝐼𝐴- 𝐼1) 𝑟2 + (𝐼𝐴- 𝐼1- 𝐼2) 𝑟3 + (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3)
𝑟4+ (𝐼𝐴- 𝐼1- 𝐼2- 𝐼3 − 𝐼4) 𝑟5
 0= 10−4(
)
200 𝐼𝐴+(𝐼𝐴−120)200+
(𝐼𝐴−180)300+(𝐼𝐴−280)200+(𝐼𝐴−320)100
𝐼𝐴=166A
The actual distribution of currents in the various sections of the
distributor due to concentrated loading.
It is clear from this figure that D is the point of minimum
potential.

(ii) The feeding point A will supply 166 A due to concentrated loading
plus 0·5 × 400 = 200 A due to uniform loading.
∴ Current supplied byA, 𝐼𝐴 = 166 + 200 = 366 A
The feeding point B will supply a current of 154 A due to concentrated
loading plus 0·5 × 600 = 300 A due to uniform loading.
∴ Current supplied by B, 𝐼𝐵 = 154 + 300 = 454 A
(iii) ∴ Minimum potential, 𝑉𝐷 = 𝑉𝐴 − Drop in AD due to conc. loading –
Drop in AD due to uniform loading 𝑉𝐷
Now, Drop in AD due to conc. loading = 𝐼𝐴𝐶 𝑅𝐴𝐶 + 𝐼𝐶𝐷 𝑅𝐶𝐷
= 166 × 10−4
× 200 + 46 × 10−4
× 200
= 3·32 + 0·92 = 4·24 V
Drop in AD due to uniform loading =
1
2
𝑖𝑟𝑙2
=
1
2
0.5 ∗ 10−4
∗ (400)2
=4V
∴ 𝑉𝐷 = 240 − 4·24 − 4 = 231·76 V

A d.c. 2-wire distributor AB is 500m long and is fed at
both ends at 240 V. The distributor is loaded as shown
in Fig. The resistance of the distributor (go and return)
is 0·001Ω per metre. Calculate (i) the point of minimum
voltage and (ii) the value of this voltage.
Let D be the point of minimum potential and let x be the
current flowing in section CD. Then current supplied by
end B will be (60 − x).

 (i) If r is the resistance of the distributor (go and return) per metre length,
then,
Voltage drop in length AD = 𝐼𝐴𝐶 𝑅𝐴𝐶 + 𝐼𝐶𝐷 𝑅𝐶𝐷
=(100+x)*100r+x*150r
Voltage drop in length BD=
1
2
𝑖𝑟𝑙2
+(60-x)*250r
=
1
2
1 ∗ 𝑟 ∗ (200)2 +(60-x)*250r
Since the feeding points A and B are at the same potential
(100+x)*100r+x*150r=
1
2
1 ∗ 𝑟 ∗ (200)2 +(60-x)*250r
X=50A
The actual directions of currents in the various sections of the distributor are
currents supplied by A and B meet at D. Hence point D is the point of minimum

 (ii) Total current = 160 + 1 × 200 = 360 A
 Current supplied by A, 𝐼𝐴 = 100 + x = 100 + 50 = 150 A
 Current supplied by B, 𝐼𝐵 = 360 − 150 = 210 A
 Minimum potential, 𝑉𝐷 = 𝑉𝐴 − (𝐼𝐴𝐶 𝑅𝐴𝐶 + 𝐼𝐶𝐷 𝑅𝐶𝐷 )
= 240 − 150 × (100 × 0·001) − 50 × (150 × 0·001)
= 240 − 15 − 7·5 = 217·5 V

 A distributor arranged to form a closed loop and
fed at one or more points is called a ring
distributor.
 For the purpose of calculating voltage distribution,
the distributor can be considered as consisting of a
series of open distributors fed at both ends.
 The principal advantage of ring distributor is that
by proper choice in the number of feeding points,
great economy in copper can be affected.

 Sometimes a ring distributor has to
serve a large area. In such a case,
voltage drops in the various
sections of the distributor may
become excessive.
 In order to reduce voltage drops in
various sections, distant points of
the distributor are joined through a
conductor called interconnector.

 The ring distributor ABCDEA. The points B and D of the ring distributor are joined
through an interconnector BD.
 Network can be obtained by applying Thevenin’s theorem.
 (i) Consider the interconnector BD to be disconnected and find the potential difference
between B and D. This gives Thevenin’s equivalent circuit voltage 𝐸0.
 (ii) Next, calculate the resistance viewed from points B and D of the network composed
of distribution lines only. This gives Thevenin’s equivalent circuit series resistance 𝑅0.
 (iii) If 𝑅𝐵𝐷 is the resistance of the interconnector BD, then Thevenin’s equivalent circuit
will be
∴ Current in interconnector BD =
𝐸0
𝑅0+𝑅𝐵𝐷
 Therefore, current distribution in each section and the voltage of load points can be
calculated.

A 2-wire d.c. distributor ABCDEA in the form of a ring main is fed at
point A at 220 V and is loaded as under :
10A at B ; 20A at C ; 30A at D and 10 A at E.
The resistances of various sections (go and return) are : AB = 0·1 Ω ; BC
= 0·05 Ω ; CD = 0·01 Ω ; DE = 0·025 Ω and EA = 0·075 Ω. Determine :
 (i) the point of minimum potential
 (ii) current in each section of distributor
Let us suppose that current I flows in section AB of the distributor. Then
currents in the various sections of the distributor are
According to Kirchhoff’s voltage law, the voltage drop in the closed
loop ABCDEA is zero i.e.
𝐼𝐴𝐵𝑅𝐴𝐵+𝐼𝐵𝐶𝑅𝐵𝐶+ 𝐼𝐶𝐷𝑅𝐶𝐷+ 𝐼𝐷𝐸𝑅𝐷𝐸 + 𝐼𝐸𝐴𝑅𝐸𝐴 =0
0.1𝐼+(𝐼-10)0.05+ (𝐼-30)0.01+(𝐼-60)0.025+(𝐼-70)0.075=0
0.26 𝐼 = 7.55
𝐼 = 29.04A

 (ii) Current in section AB = I = 29·04 A from A to B
 Current in section BC = I − 10 = 29·04 − 10 = 19·04 A from B to C
 Current in section CD = I − 30 = 29·04 − 30 = − 0·96 A = 0·96 A from D to C
 Current in section DE = I − 60 = 29·04 − 60 = − 30·96 A = 30·96 A from E to D
 Current in section EA = I − 70 = 29·04 − 70 = − 40·96 A = 40·96 A from A to E
∴ C is the point of minimum potential.

 A d.c. ring main ABCDA is fed from point A from a 250 V supply and the resistances
(including both lead and return) of various sections are as follows : AB = 0·02 Ω ; BC =
0·018 Ω ; CD = 0·025 Ω and DA = 0·02 Ω. The main supplies loads of 150 A at B ; 300 A
at C and 250 A at D. Determine the voltage at each load point. If the points A and C are
linked through an interconnector of resistance 0·02 Ω, determine the new voltage at
each load point.
 Without Interconnector.
 Let us suppose that a current I flows in section AB of the distributor. Then currents in
various sections of the distributor will be
 According to Kirchhoff’s voltage law, the voltage drop in the closed loop ABCDA is
zero i.e.
 𝐼𝐴𝐵𝑅𝐴𝐵+𝐼𝐵𝐶𝑅𝐵𝐶+ 𝐼𝐶𝐷𝑅𝐶𝐷+ 𝐼𝐷𝐴𝑅𝐷𝐴 =0
 0.02𝐼+(𝐼-150)0.018+ (𝐼-450)0.025+(𝐼-700)0.02=0
0·083 I = 27·95
∴ I = 27·95/0·083 = 336·75 A

The actual distribution of currents are ..
 Current in section AB = I = 336.75 A from A to B
 Current in section BC = I − 150 = 336.75 − 150 = 186.75 A from B to C
 Current in section CD = I − 450 = 336.75 − 450 = − 113.25 A = 113.25 A from D to C
 Current in section DA = I − 700 = 336.75 − 700 = − 363.25A = 363.25 A from A to D
Voltage drop in AB = 336·75 × 0·02 = 6·735 V
Voltage drop in BC = 186·75 × 0·018 = 3·361 V
Voltage drop in CD = 113·25 × 0·025 = 2·831 V
Voltage drop in DA = 363·25 × 0·02 = 7·265 V
∴ Voltage at point B = 250 − 6·735 = 243·265 V
Voltage at point C = 243·265 − 3·361 = 239·904 V
Voltage at point D = 239·904 + 2·831 = 242·735 V

 With Interconnector.
 The ring distributor with interconnector AC. The current in the interconnector can be
found by applying Thevenin’s theorem.
𝐸0 = Voltage between points A and C
= 250 − 239·904 = 10·096 V
𝑅0 = Resistance viewed from points A and C
=
0.02+0.018 0.02+0.025
0.02+0.018 + 0.02+0.025
= 0.02Ω
𝑅𝐴𝐶 = Resistance of interconnector = 0·02 Ω
 From Thevenin’s equivalent circuit Current in interconnector AC is
𝐸0
𝑅0 + 𝑅𝐴𝐶
=
10.096
0.02 + 0.02
= 252.4𝐴 𝑓𝑟𝑜𝑚 𝐴 𝑡𝑜 𝐶
Let us suppose that current in section AB is 𝐼1 . Then current in section BC will be 𝐼1 − 150.
As the voltage drop round the closed mesh ABCA is zero,
0.02𝐼1+(𝐼1-150)0.018-0.02*252.4=0
0.038 𝐼1=7.748
𝐼1=203.15A

The actual distribution of currents are ..
 Current in section AB = 𝐼1 = 203.15A from A to B
 Current in section BC = 𝐼1 − 150 = 203.15 − 150 = 53.15 A from B to C
 Current in section CD = 𝐼1 − 450+252.4 = 203.15− 450 +252.4= 5.55A C to D
 Current in section DA = 𝐼1 − 700 = 203.15+252.5 − 700 = − 244.45A = 244.45A
from A to D
 Drop in AB = 203·15 × 0·02 = 4·063 V
 Drop in BC = 53·15 × 0·018= 0·960 V
 Drop in AD = 244·45 × 0·02 = 4·9 V
 ∴ Potential of B = 250 − 4·063= 245·93 V
 Potential of C = 245·93 − 0·96 = 244·97 V
 Potential of D = 250 − 4·9 = 245·1 V
It may be seen that with the use of interconnector, the voltage drops in the various
sections of the distributor are reduced.

 A.C. distribution calculations differ from those of d.c. distribution in the
following respects :
 (i) In case of d.c. system, the voltage drop is due to resistance alone.
However, in a.c. system, the voltage drops are due to the combined effects of
resistance, inductance and capacitance.
 (ii) In a d.c. system, additions and subtractions of currents or voltages are
done arithmetically but in case of a.c. system, these operations are done
vectorially.
 (iii) In an a.c. system, power factor (p.f.) has to be taken into account. Loads
tapped off form the distributor are generally at different power factors. There
are two ways of referring power factor viz
(a) It may be referred to supply or receiving end voltage which is
regarded as the reference vector.
(b) It may be referred to the voltage at the load point itself.

 The power factors of load currents may be
given
(i) w.r.t. receiving or sending end voltage
or
(ii) w.r.t. to load voltage itself.

 (i) Power factors referred to receiving end voltage.
Consider an a.c. distributor AB with concentrated loads of 𝐼1and 𝐼2 tapped off at
points C and B. Taking the receiving end voltage 𝑉𝐵 as the reference vector, let
lagging power factors at C and B be cos φ1 and cos φ2 w.r.t. 𝑉𝐵 . Let 𝑅1, 𝑋1 and
𝑅2, 𝑋2 be the resistance and reactance of sections AC and CB of the distributor.
Impedance of section AC, 𝑍𝐴𝐶= 𝑅1 + j 𝑋1
Impedance of section CB,𝑍𝐶𝐵 = 𝑅2 + j 𝑋2
Load current at point C, 𝐼1 = 𝐼1 (cos φ1 − j sin φ1)
Load current at point B, 𝐼2 = 𝐼2 (cos φ2 − j sin φ2)
Current in section CB, 𝐼𝐶𝐵= 𝐼2 = 𝐼2 (cos φ2 − j sin φ2)
Current in section AC, 𝐼𝐴𝐶 = 𝐼1+ 𝐼2
=𝐼1 (cos φ1 − j sin φ1)+𝐼2 (cos φ2 − j sin φ2)

φ1
 Voltage drop in section CB,
 𝑉𝐶𝐵= 𝐼𝐶𝐵 𝑍𝐶𝐵= 𝐼2 (cos φ2 − j sin φ2) (𝑅2 + j 𝑋2 )
 Voltage drop in section AC,
 𝑉𝐴𝐶= 𝐼𝐴𝐶 𝑍𝐴𝐶=(𝐼1+ 𝐼2) 𝑍𝐴𝐶
 =𝐼1 (cos φ1 − j sin φ1)+𝐼2 (cos φ2 − j sin φ2)(𝑅1 + j 𝑋1 )
 Sending end voltage, 𝑉𝐴= 𝑉𝐵 + 𝑉𝐶𝐵 + 𝑉𝐴𝐶
 Sending end current, 𝐼𝐴= 𝐼1+ 𝐼2
𝑉𝑩
𝐼𝟐
𝐼𝟏
𝐼𝑨𝑪
φ2
θ

φ1
 (ii) Power factors referred to respective load voltages.
 Suppose the power factors of loads are referred to their respective load voltages. Then φ1 is the phase angle
between V𝐶 and I1 and φ2 is the phase angle between V𝐵 and I2. The vector diagram under these conditions is
 Voltage drop in section CB,
 𝑉𝐶𝐵= 𝐼2 𝑍𝐶𝐵= 𝐼2 (cos φ2 − j sin φ2) (𝑅2 + j 𝑋2 )
 Voltage at point C = 𝑉𝐵 + Drop in section CB = V𝐶 ∠ α (say)
 Now 𝐼1 = 𝐼1 ∠ − φ1 w.r.t. voltage V𝐶
 ∴ . 𝐼1 = 𝐼1 ∠ − (φ1 − α) w.r.t. voltage V𝐵
 i.e. 𝐼1 = 𝐼1 [cos (φ1 − α) − j sin (φ1 − α)]
 Now 𝐼𝐴𝐶 = 𝐼1+ 𝐼2
=𝐼1 [cos (φ1 − α) − j sin (φ1 − α)]+ 𝐼2 (cos φ2 − j sin φ2)
 Voltage drop in section AC,
𝑉𝐴𝐶= 𝐼𝐴𝐶 𝑍𝐴𝐶
 ∴ Voltage at point A = V𝐵 + Drop in CB + Drop in AC
𝑉𝑩
𝐼𝟐
𝐼𝟏
𝐼𝑨𝑪
φ2
α

A single phase a.c. distributor AB 300 metres long is fed from end A and is loaded as under :
(i) 100 A at 0·707 p.f. lagging 200 m from point A (ii) 200 A at 0·8 p.f. lagging 300 m from point A
The load resistance and reactance of the distributor is 0·2 Ω and 0·1 Ω per kilometre. Calculate the
total voltage drop in the distributor. The load power factors refer to the voltage at the far end.
Impedance of distributor/km = (0·2 + j 0·1) Ω
Impedance of section AC, 𝑍𝐴𝐶 = (0·2 + j 0·1) × 200/1000 = (0·04 + j 0·02) Ω
Impedance of section CB, 𝑍𝐶𝐵 = (0·2 + j 0·1) × 100/1000 = (0·02 + j 0·01) Ω
Taking voltage at the far end B as the reference vector, we have, Load current at point B,
𝐼2 = 𝐼2 (cos φ2 − j sin φ2) = 200 (0·8 − j 0·6) = (160 − j 120) A
Load current at point C, 𝐼1 = 𝐼1 (cos φ1 − j sin φ1) = 100 (0·707 − j 0·707) = (70·7− j 70·7) A
Current in section CB,𝐼𝐶𝐵= 𝐼2 = (160 − j 120) A
Current in section AC,𝐼𝐴𝐶= 𝐼1+ 𝐼2
= (70·7 − j 70·7) + (160 − j 120)= (230·7 − j 190·7) A
Voltage drop in section CB, 𝑉𝐶𝐵= 𝐼𝐶𝐵 𝑍𝐶𝐵 = (160 − j 120) (0·02 + j 0·01) = (4·4 − j 0·8) volts
Voltage drop in section AC, 𝑉𝐴𝐶= 𝐼𝐴𝐶 𝑍𝐴𝐶 = (230·7 − j 190·7) (0·04 + j 0·02) = (13·04 − j 3·01) volts
Voltage drop in the distributor = 𝑉𝐴𝐶 + 𝑉𝐶𝐵 = (13·04 − j 3·01) + (4·4 − j 0·8) = (17·44 − j 3·81) volts
Magnitude of drop = 17.44 2 + 3.81 2 = 17·85 V

 A single phase ring distributor ABC is fed at A. The loads at B
and C are 20 A at 0.8 p.f. lagging and 15 A at 0.6 p.f. lagging
respectively ; both expressed with reference to the voltage at
A. The total impedance of the three sections AB, BC and CA
are (1 + j 1), (1+ j2) and (1 + j3) ohms respectively. Find the
total current fed at A and the current in each section. Use
Thevenin’s theorem to obtain the results.

POWER SUPPLY SYSTEMS& DISTRIBUTION SYSTEMS

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