High Frequency Apparatus Page 29

THE HIGH POTENTIAL TRANSFORMER 29 larger current value in the primary. Taking the apparent amperes as 4.88, we may multiply by 1.15 (assuming a power factor of 85 per cent.) and we get 5.61, or 5.61 actual amperes in the primary winding. Allowing here, also, 600 circular mils, we find that the primary conductor must have an area of 600 times 5.61, or 3366 circular mils. The wire table tells us that No. 15 has an area of 3,257 while No. 14 has an area of 4,107 circular mils. Following the rule of plenty, we may adopt the latter as the correct conductor to use for the primary. We now come to the point that has puzzled more amateur experimenters than almost any other, i. e., the calculation of primary turns. Of course, once this number is known, the determination of the secondary turns is a simple matter. The formula for the primary is not complex, and its working requires only the application of ordinary arithmetic. The maximum flux is equal to the density multiplied by the area of the core in square inches. The e.m.f. generated in the primary winding is: Ер == 4.44 N Tp n 108 N-maximum flux. Tp-primary turns. n-frequency. where Ep-impressed primary voltage, therefore Тр == Primary voltage x 108 4.44 x N x n Working this formula, we first determine the maximum flux. As the section of the core is 2 inches, we square this to get the area, or 4 inches. Multiplying the area by the density per square inch, we find that 4 times 30,000 will