Three-Wire Circuits

Almost all lighting is done with 115-volt or 120-volt lamps, whereas almost all motors of more than 1/4 hp are operated at 230 volts or more. It is sometimes desirable to distribute power for both lamps and motors over the same line. For this purpose, three-wire circuits are used, as shown in Fig. 3-5. The two 115-volt generators G₁ and G₂ are connected in series. This arrangement permits motors to be operated at 230 volts across the two outside wires while lamps may be used at 115 volts between either outside wire and the middle wire at the same time.

Furthermore, the power to the lamps is supplied with less power loss in the line wires than would be incurred with two-wire, 115-volt distribution and the use of the same amount of line copper. If, as in Fig. 3-5, the currents taken by the loads on the two parts of the line are unequal, the middle wire carries the difference between the currents in the outside wires. An attempt is always made to balance the loads on the two parts of the line as far as practicable.

Example 3-4. - A load of 20 amperes at 115 volts is to be supplied by a two-wire line each wire of which has a resistance of 0.5 ohm. Calculate the power loss in the line and the power in the load.

Solution. - By equation (2-2), the power loss in the line is

By equation (2-3), the power in the load is

If the load in the foregoing example were supplied by a three-wire line, it should be divided approximately equally between the two parts to avoid neutral-wire current as far as possible. The neutral wire should, in general, be of the same size as either outside wire in order that, if the load be removed from one side only, the middle wire will be as capable of carrying the remaining load as is the remaining outside wire because both then carry the same current.

Example 3-5. - If another similar wire be added to the line of Example 3-4, making a three-wire, 115-230-volt circuit, how much power could be delivered at the load point without having the loss in the line exceed 400 watts?

Solution. - For a given outside-wire current, maximum power can be delivered only to a load which is equally balanced between the two sides of the line. Then, no current will flow in the middle wire. With the same loss of 400 watts in two wires as in the line in Example 3-4, the outside-wire current must be 20 amp. The maximum power permissible in the load is

230 * 20 = 4600 watts

The preceding examples show that, by adding 50 per cent more copper, the power transmissible by a three-wire circuit with the same power loss as in a two-wire circuit is increased 100 per cent This fact demonstrates the economic advantage of three-wire distribution over two-wire distribution for providing 115-volt service as ordinarily used in home-lighting circuits. The greater the load unbalance, the less advantageous is the three-wire system. The load is carried with minimum line loss when the load is balanced between the two sides of the line and there is no neutral-wire current.