1. The power (P) transmitted by all the systems is same.
2. The distance ( l ) over which the power is transmitted is same.
3. The power loss (W) in all the systems are same.
4. The maximum voltage (Vm) exists between any conductor and the earth, in all the systems.
Based on these assumptions, let us compare the various types of systems for the volume of copper required.
1.1 Two Wire D.C. System With One Line Earthed
The system is represented in the Fig. 1.
 |
| Fig. 1 |
The maximum voltage between the conductors is Vm, as one material is earthed.
Where P = Power transmitted
Let R = resistance of each conductor
Total copper losses in both the lines are,
Volume of copper required is
The volume of copper required for other systems is compared by
taking volume of copper required for this system as base. Let it be
constant K and volume of copper required for other systems can be
expressed interms of K.
1.2 Two Wire D.C. System With Midpoint Earthed
The system is represented in the Fig. 2.
 |
| Fig. 2 |
As power transmitted is same as P, the current in each conductor is,
The total copper loss in both the lines is,
where A = area of cross-section of each line.
The total volume of copper required is,
Thus the volume of copper required in this system is one fourth
the volume of copper required for two wire d.c. system with one line
earthed.
1.3 Three Wire D.C. System
The system is represented as shown in the Fig. 3.
 |
| Fig. 3 |
When the load is balanced, current through the third neutral wire is zero.
Let A = Cross-section of outer conductors
The total copper loss is,
Let area of cross-section of the middle neutral wire is half of the area of cross-section of the outer conductor.
Hence the total volume of copper
= Volume of copper for outer wires + volume of copper for neutral wire
Thus the volume of copper required in this system is 0.3125 times
the volume of copper required for two wire d.c. system with one line
earthed.
1.4 Three Phase Three Wire A.C. System
This is most commonly used system for the transmission. The three
phase three wire star connected system with neutral earthed is shown in
the Fig. 4.
 |
| Fig. 4 |
The maximum voltage between each line conductor and the neutral is Vm as shown in the Fig. 4.
The R.M.S. value of the voltage per phase given by,
Then total power transmitted is P watts hence per phase power transmission is,
Let cosΦ be the load power factor
Hence the total copper loss is,
The volume of the copper required is
Thus the volume of copper required depends on power factor of the load and it is 0.5/cos2Φ
times the volume of copper required by twp wire d.c. system with one
line earthed. This system may be delta connected but irrespective of the
method of connection star or delta, the result derived remains same.
1.5 Three Phase Four Wire A.C. System
This system is popularly used for secondary distribution. The
neutral is also made available for the connection of the load. The
system is shown in the Fig. 4.
 |
| Fig. 5 |
Assuming the load balanced, there is no current flowing through the neutral.
The cross-section area of neutral is half the cross-section of
each conductor i.e. 0.5 A where A is cross-section of each conductor.
The maximum voltage between any conductor and the neutral is Vm hence r.m.s. voltage per phase is,
The power transmitted per phase is,
Hence all the calculations upto the copper losses and expression
of A remain same as derived for three phase three wire system.
The total volume of copper requires is,
= Volume of copper for 3 lines + copper required for neutral
= Volume of copper for 3 lines + copper required for neutral
Thus the volume of copper required is 0.583/cos2Φ times the volume of copper required by two wire d.c. system with one line earthed.
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