It is more economical to transmit 10,000 volts a long distance than 1,000 volts, because the lower the pressure, or the voltage, the larger must be the conductor to avoid loss. It is for this reason that 500 volts, or more, are used on electric railways. For electric light purposes, where the current goes into dwellings, even this is too high, so a transformer is used to take a high-voltage current from the main line and transform it into a low voltage. This is done by means of two distinct coils of wire, wound upon an iron core.

Fig. 115. A Transformer

In Fig. 115 the core is O-shaped, so that a primary winding (A), from the electrical source, can be wound upon one limb, and the secondary winding (B) wound around the other limb. The wires, to supply the lamps, run from the secondary coil. There is no electrical connection between the two coils, but the action from the primary to the secondary coil is solely by induction. When a current passes through the primary coil, the surging movement, heretofore explained, is transmitted to the iron core, and the iron core, in turn, transmits this electrical energy to the secondary coil.

## How The Voltage Is Determined

The voltage produced by the secondary coil will depend upon several things, namely, the strength of the magnetism transmitted to it; the rapidity, or periodicity of the current, and the number of turns of wire around the coil. The voltage is dependent upon the length of the winding. But the voltage may also be increased, as well as decreased. If the primary has, we will say, 100 turns of wire, and has 200 volts, and the secondary has 50 turns of wire, the secondary will give forth only one-half as much as the primary, or 100 volts.

If, on the other hand, 400 volts would be required, the secondary should have 200 turns in the winding.

## Voltage And Amperage In Transformers

It must not be understood that, by increasing the voltage in this way, we are getting that much more electricity. If the primary coil, with 100 turns, produces a current of 200 volts and 50 amperes, which would be 200 × 50 = 10,000 watts, and the secondary coil has 50 turns, we shall have 100 volts and 100 amperes: 100 (V.) × 100 (A.) = 10,000 watts. Or, if, on the other hand, our secondary winding is composed of 200 turns, we shall have 400 volts and 25 amperes, 400 (volts) × 25 (amperes) also gives 10,000 watts.

Necessarily, there will be some loss, but the foregoing is offered as the theoretical basis of calculation.