Nikola Tesla Books
28 HIGH FREQUENCY APPARATUS section is 2 times 2 or 4 square inches. The length, 25, multiplied by the section, 4, gives us 100 cubic inches for the volume of this core. As it is always well to err on the right side, this core may be taken as quite satisfactory. The computations for the windings which follow will show that its proportions are just right. To determine the weight of the core we multiply its volume, 100 cu. in. by .25, as each cubic inch of laminated silicon steel weighs approximately 4 lb. This gives us 25 pounds as the weight of the iron in the core. We may next figure the current in the secondary winding at full load. As the potential is to be 5,000 volts, and the output 500 watts, we may divide secondary watts by secondary volts to find secondary amperes, which in this case will, of course, be .1 ampere. In power transformer work it is customary to allow at least 1000 circular mils of area in the conductor for each ampere of current to be carried. For our purposes, however, the transformer is to be used but a short time when it is permitted to cool and in practice a density of 600 circular mils per ampere has been found quite satisfactory and safe. As the secondary current is .1 ampere, we find that our secondary conductor must have an area of 600 times .1 or 60 circular mils in order that it may safely carry the current. In the back of the book, we find tables giving the area of copper wires in circular mils. No. 32 is found to have an area of 63.21 and this wire is accordingly quite suitable. The primary current is next in order. At a unity power factor, the primary watts divided by primary volts gives primary amperes. This we find to be 537.63110, or 4.88 amperes. As the power factor of this type of transformer may be assumed to be in the vicinity of 85 per cent., we shall have to compensate for this by using a somewhat
