60 Dr. McLachlan on Effective Inductance, Effective 



Assuming for simplicity thai the voltage follows a sine 

 law from zero to its first maximum, we have 



?9 = fi)Uv2 cos cot and Jrvs = ^- r . 



v 2 



The loss in joules is W = IitMs 2 Re2 x £, where f = time for the 



voltage to attain its maximum value 



The energy stored in the condenser when the voltage is a 



maximum = ^02 V 2 2 , and this is dissipated chiefly in the spark. 



Hence total energy in secondary is TTh- -IC2V2 2 . Thus 

 in V 2 



effloie »°y =w+ii.w 



The maximum or peak voltage which would have occurred 

 if there had been no loss, is obtained from the relationship 



y, -=[ 2 5t v ']* 



As an example take the following data : — 



Y 2 = 10 4 volts. 



2 = 55 picofarads, including leads. 

 R e2 = 2 x 10 5 ohms. 



y = 6 x 10 3 ~ (mean value for the two oscillations 

 after break)*. 



Then 



W = Ikms IW 



= &) 2 C 2 2 V 2 2 B e2 */2 



= 1-8 xlO" 3 joule. 

 Actual energy obtained 1 _ in v 2 



from magneto ln > „ . 



electrostatic form. * = 2 ' 8 x 10 J oule - 



Treating the primary circuit in a similar manner and 

 assuming as a first approximation that the ratio of the 

 peak voltages is equal to the ratio of the turns, we find 

 the loss is 1*3 x 10~ 3 joule. Hence, neglecting the loss 

 in the primary condenser, the total loss prior to the passage 

 of the spark is 3*1 x 10~ 3 joule. 



If this energy had been transferred to the secondary, the 



* This frequency is assumed for the period prior to sparking. Its 

 magnitude is suggested by the curve of fig. 2 in Dr. Campbell's paper 

 (loc. cit.). 



