**We will now compare the 2-wire and 3-wire systems from the point of view of economy of conductor material.**

For this comparison, it will be assumed that—

- the amount of power transmitted is the same in both cases;
- the distance of transmission is the same;
- the efficiency of transmission (and hence losses) is the same;
- voltage at consumer’s terminals is the same;
- the 3-wire system is balanced; and
- in the 3-wire system, the mid-wire is of half the cross-section of each outer.

Let W be the transmitted power in watts and V the voltage at the consumer's terminals.

Also, let

R2 = resistance in ohms of each wire of 2-wire system.

R3 = resistance in ohms of each outer in 3-wire system.

The current in 2-wire system is W/V and the losses are 2 *(W/V)sqr*R2.

In the case of 3-wire system, voltage between outers is 2 V, so that current through outers is

(W/2V), because there is no current in the neutral according to our assumption (5) above. Total losses

in the two outers are 2*(W/2V)sqr*R3.

Since efficiencies are the same, it means the losses are also the same.

∴ 2(W/V)sqr*R2 = 2(W/2V)sqr*R3 or

R3/R2=4/1

Since the cross-section and hence the volume of a conductor of given length, is inversely propor-

tional to its resistance,

∴ (volume of each 3-wire conductor)/(volume of each 2-wire conductor) = 1/4

Let us represent the volume of copper in the 2-wire system by 100 so that volume of each

conductor is 50.

Then, volume of each outer in 3-wire system = 50/4 = 12.5

volume of neutral wire ,, ,, = 12.5/2 = 6.25

∴ total volume of copper in 3-wire system = 12.5 + 6.25 + 12.5 = 31.25

∴ (total coper vol. in 3-wire feeder)/(total copper vol. in 2-wire feeder) = 31.25/100 6=5/6

Hence, a 3-wire system requires only 5/16th (or 31.25%) as much copper as a 2-wire system.

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