g_B] CIRCLE DIAGRAM. 283 



scaled value of amperes, or can determine a separate scale for 

 watts. In the present example, every ampere (vertical) repre- 

 sents 321 watts. To get horse-power, we have I horse- 

 power = 746 watts. 



Certain quantities, as power factor, efficiency and slip de- 

 pend upon ratios and are, accordingly, independent of scale. 



12. Input. For any load, the line MP is the power com- 

 ponent of current and is, therefore, proportional to the total 

 input in watts. The maximum* input is seen to be at P 7 where 

 a tangent to the circle is parallel to the diameter AP" f . 



13. Output. The useful output is proportional to KP, 

 which is equal to the input MP less the total losses, MK, dis- 

 cussed in the next paragraph. The maximum output is seen to 

 be at P 5 , where a tangent to the circle is parallel to AP". 



14. Losses. The total losses MK, are equal to the no-load 

 losses MN, increased by the " added " losses NK, which are cop- 

 per losses due to the added current 7 (2 ) and vary as 7 ( 2 2) . In the 

 construction of Fig. I, K is located on a straight line connecting 

 A and P". The point K moves toward P" as the load increases, 

 and, from the construction, f NK oc 7 ( 2 2) . 



15. The added losses NK consist of the secondary^ copper 

 loss SK, and the primary copper loss NS, which is practically || 



*(i2a). Neglecting the no-load losses, the maximum input is the 

 radius of the circle O'Pi, which may be taken as E -?- 2.X; see 34. The 

 input, and hence the output, of a motor on a given voltage is accordingly 

 limited solely by the leakage reactance, which in design is made as small 

 as possible. 



f (i4a). In Fig. I, the triangle ANP is similar to the triangle (not 

 shown) APP'". Accordingly, AN :AP : :AP \AP'" and AN oc (AP) 2 , since 

 AP'" is constant. Knowing that AP = I<. 2 ) and that NK is proportional to 

 A N, we have NK <x / 2 {2) . Q. E. D. 



$ It will be remembered that R 2 and 7< 2 > are the values of secondary 

 resistance and current, respectively, in terms of the primary. 



II ( J 5a). The primary copper loss at no load is RJ 2 . For a primary 

 current A, at any load, the primary copper loss is RJi 2 = RJ<? -}- RJ w 2 

 2RJJ 2 cosPAO. Strictly speaking, therefore, the last two terms are the 

 "added" primary copper loss due to 7< 2 >. We may consider, either (i) 



