244 ALTERNATING CURRENTS 



During the first stage (Fig. 150) the total instantaneous current 

 at P is given by 



i = #! - ^1 J = l m (cOS - fc) 



while during the second stage (Fig. 151) it is given by 



i' = 4*i + Fm = 4I(cos + 4) 



The squares of the currents during the two stages are given by 

 i 2 = Ilm 2 (cos 2 + I - cos 0) = i-I m 2 (f + i cos 20 - cos 0) * 



and 



* = J 4 Im 2 (I + cos 20 + cos 0) 

 respectively. 



In order to find the mean value of the square of the current 

 which determines the heating effect at P we may suppose the square 

 of the total current at P plotted as a function of 9. By determining 

 the area of the curve, and dividing this by the base (TT), we obtain 

 the mean value of the ordinate, i.e. the mean square of the current 

 at P. The determination of the area is most easily accomplished 

 by a simple integration. Taking first the portion of the curve corre- 

 sponding to the first stage, during which > 6 > 0, we have, fidO 

 representing a narrow strip of the area 



<t> ^ 



area corresponding to first stage = / i*dO = ll m z I (- + cos 29 cos 6)dB 



o o 



= Fm 2 [f + i sin 20 - sin 0f 



o 



= 4l 2 (!0 + i sin 20 - sin $) 

 Similarly, we find 



r* 

 area corresponding to second stage = I i'*dO = |I m 2 [|0+isin 20 + sin 0] 



* * 



= ilm 2 (|ir - $<jt - \ sin 20 - sin 0) 

 The total area is thus 



ilm\f TT - 2 sin 0) 

 and the mean ordinate, i.e. the mean square of the current at P, is 



1T 2 /3 2 . \ 

 * m \4 ~ T Sm V 



* Since cos 2 = J(l + cos 26) 



