Nikola Tesla Books
HIGH FREQUENCY APPARATUS arithmetic. The subject of copper losses has been neglected solely because it introduces one more calculation that has no practical bearing upon the net results obtained. Let us assume, arbitrarily, that we wish a transformer of an efficiency approximating 93 per cent. Incidently, the transformer designed in this chapter is the one employed in the later chapter on experimental high frequency apparatus. We wish a secondary potential of 5,000 volts; a primary wound for 110-volt supply; and we wish to operate the instrument on a 60-cycle circuit. If we wish an output of ½ k.w. or 500 watts, and the efficiency is to be 93 per cent., it is obvious that we must have a greater input than 500 watts in order to compensate for the 7 per cent. loss. The input is calculated by dividing the output by the per cent. efficiency; thus: 500 .93 ==== 537.6344+ or we must have an input of 537.6344 watts in order to take out 500 watts. Core Volume.-The volume of the core receives our attention next. The initial step is to determine the watts loss in total and by subtracting 500 from 537.6344, or output from input, we find that the loss in the transformer is 37.6344 watts. This loss is made up of the I2R losses which are due to the heating effects in the copper of the windings, and the hysteresis and eddy current losses in the core; the latter are known as the core losses. The core losses constitute about 47 per cent. of the total of 37.6344 watts, or 17.688 watts. Approximately 20 per cent. of the total core loss will be the eddy current loss and the balance of 80 per cent. must, therefore, constitute the hysteresis loss. To determine the latter loss in our core, we take 80 per cent. of the core loss of 17.688 which gives us 14.1504 for the
