CURRENT VALUE. 93 



voltage is 50. The maximum current, taken at any time by 



50 

 the wire, will be KKK = -166 amps., since 50 volts is the 



maximum value of the voltage. The maximum rate at 

 which heat is given to the wire, is, therefore, ( 166) 2 x 300 = 

 8-33 joules per second. 



The average rate at which heat is communicated to the 

 wire may be calculated roughly from the curve in the following 

 manner. Dividing the complete cycle into six equal parts 

 and estimating the average value of the electromotive force 

 during each of these intervals, we may tabulate our results 

 as follows : 



Interval. Average E.M.F. C C 2 R 



0-30 ... 12-9 ... -043 ... -555 



30- 60 ... 35-4 .. -118 ... 4-18 



60 -90 ... 48-3 ... -161 ... 7-78 



90- 120 ... 48-3 ... -161 ... 7-78 



120- 150 ... 35-4 ... -118 ... 4-18 



150 -180 . 12-9 -043 "555 



Mean value of C 2 R for - cycle = 4-172 joules per sec. 

 4 



Since the curve is symmetrical arid repeats itself, the 

 average rate at which heat is given to the voltmeter wire 

 when connected to the supply will be 4-172 joules per second 

 and this will determine the deflection. 



Comparing this value with the maximum rate of 8-33 

 joules per second, it is seen that the average rate of generation 

 of heat in the wire is almost exactly one-half the maximum rate. 

 (If the average value had been determined more accurately 

 it would have been found to be exactly one-half.) Thus 

 the deflection of an alternating-current voltmeter or ammeter 

 is exactly one-half the deflection which would be observed 

 if the value of the voltage or current were maintained constant 

 at its maximum value. 



In order to ascertain the direct-current voltage which 

 would produce the same reading we must write 



VI R - J-(F AMAX .) 9 JB 



F * T7 



r o -' ^2 A MAX 



where V D = the value of the equivalent direct voltage. 

 F A MAX. = maximum value of alternating voltage. 



