DESIGN OF TRANSFORMERS. 115 



If DC = 6#, and DA = 4x, then the winding space is (see 

 Fig. 39)- 



4-a 8 = 73 



whence a; = 4'27 centimetres. This w r ould give too small a cooling 

 surface, so we take x = 6 '5. 



Now, from equation (15) we get 



A = M 



= 281*4 square centimetres .... (16) 



therefore the length DE of the transformer is 



3 & centimetres 

 or 



60 centimetres 



.Again, the resistance of the secondary coil is 00075 ohm; 

 therefore the drop of volts due to secondary resistance from no- 

 load to full-load is 0*75 volt ; an extra turn on the secondary will 

 more than compensate for this. 



We have hitherto assumed that the primary applied P.D. and 

 the primary counter E.M.F. are the same, because it is the ratio of 

 the primary counter E.M.F. to the secondary induced E.M.F., that 

 is the same as the ratio of the turns. The drop of volts in the 

 primary coil itself is 



0-75 x 10 = 7-5 volts 



that is, 0'75 per cent., so that, neglecting magnetic leakage, the 

 secondary induced E.M.F. is given by 



4 4 () x 992-5 = 99-25 volts 



Taking into account the drop of volts in the secondary coil 

 itself, the secondary terminal P.D. is, neglecting leakage, 98'5 volts ; 

 and allowing 2 per cent, for magnetic leakage between the primary 

 and secondary coils, the secondary terminal P.D. becomes 96'5 

 volts. If we put 24 turns on the secondary, the terminal P.D. 

 becomes 103. 



69. The next step is to calculate the primary current on open 

 secondary circuit. This is 



