224 Flux of Light from the Electric Arc. 



crater separately, we obtain 



M.S.C.P.A.C. = iUc. (sin 2 « + sin 2 ft). 



wher 



_ b + x 



Hence 



M.S.C.P.CC. Ic.c. sin 2 a 



M.S.C.P.A.C. ~ l A .c. sin 2 « + sin 2 /3* 



Some experiments mentioned by Dr. Fleming and 

 Mr. Petavel give : — 



for a continuous-current arc 



l e = 910 when = 405°. 

 Hence Q 



and for a similar alternating-current arc 



I e — 300 when 6 = 60° for top crater, 



and 



I . = 370 when 6 = 63 for bottom crater. 



Hence as a mean 



Ia.c. = 707. 

 So we get 



1-99. 



Ic.c. 1400 



Ja.c. 707 

 Therefore 



M.S.C.P.c.c. = 1>99 sin 2 * 



M.S.C.P.A.C sin 2 a + sin 2 ft' 



Hence for equal efficiency we have 



1*99 sin 2 a = sin 2 a. + sin 2 ft. 



•99 sin 2 a. = sin 2 ft ; 



or putting in the values of the angles 



_-99(a + ^) 2 jb + x) 2 



(a + %y + m 25a 2 ~ {b + xf + '256 2 ' 



Now a = 10 mm. 



5 = 12 mm. 

 Hence 



•99(10 + ^) 2 = (12 + ^, 2 

 (10 + a?) 2 + 25 (12+ a?) 2 + 36' 



