112 TREATISE ON ALTERNATING CURRENTS. 



Again, from equation (4) we get the resistances of the two circuits 

 to be 



75 

 ri = io^ 



= 075 ohm, nearly ..... (7) 



75 



= 100 2 

 = 0*0075 ohm, nearly ..... (8) 



Also taking p = 16 x 10" 7 , we have, by (1) 



P 



that is 



_ 0-0625 x 0-75 x 10 7 



T6~ 

 = 29297 centimetres, nearly ... (9) 



and 



_ (T)25 x 0-0075 X 10 7 



16 

 = 0-2930 centimetre, nearly . . . (10) 



We shall have to leave the determination of the number of 

 primary, NI, and secondary, N%, turns until the dimensions of the 

 iron core have been found. 



Summarizing the items already determined, we have 



i\ = 10 amperes 

 ? 2 = 100 amperes 



= amperes 

 = 0-0625 square centimetre 

 -2 = 0'625 square centimetre 



1 = 29297 centimetres 



2 = 2930 centimetres 



= 0*75 ohm 

 r 2 = 0-0075 ohm 



68. Iron Circuit. The dimensions of the iron circuit 

 are determined from the assumed iron losses. 



