FALL OF POTENTIAL ALONG REACTIVE CIUCUITS. 71 



Then 



si = 2ir X 100 X 5 = 3141-6 



2 



" 2?r X 100 X 10~ 6 "" 

 whence 



+ si 2 = 3142 nearly 



y/rf + s 2 2 = 3183 nearly 

 and 



\(i+rtf + (si + s 2 = 44 nearly 

 Therefore 



vi : v 2 : F = 3142 : 3183 : 44 approximately 

 That is, if 



V = 100 volts 

 then 



vi = 7141 volts approximately 

 and 



#2 = 7234 volts approximately 



This shows that, even if low voltage mains are used, dangerous 

 potential differences between portions of reactive circuits may exist. 



PROBLEMS ON CHAPTER VIII. 



1. What is the phase difference between the current in, and potential dif- 

 ference between the terminals of, a coil whose resistance is 10 ohms, and self- 

 induction 0-075 henry, when the frequency is 50 periods per second ? 



Answer. 67 nearly. 



2. What frequency would make the phase difference 45 with the coil in 

 Question 1 ? 



Answer. 21 '22 periods per second. 



3. What self-induction must a coil whose resistance is 7'5 ohms have in 

 order that, when placed in a circuit of frequency 100, the phase difference 

 between the P.D. between its terminals and the current flowing in it may be 60 ? 



Ansiver. 0'02074 henry. 



4. What resistance must be placed in series with a self-induction of 0'025 

 henry, in a circuit of frequency 75 periods per second, in order that the 

 phase difference between the current and P.D. between the terminals of the 

 arrangement may be 30 ? 



Answer. 20'4 ohms. 



5. Calculate the wattless and energy currents in Questions 1, 3, 4, in terms of 

 the total current i. 



