70 POWER AND POWER-FACTOR. 



The energy voltage represented by this lost power = 



- = -=-=- = 16 '8. The watts taken by the arc, 

 current 10 



which acts like a non-inductive resistance, = (current X 

 voltage) = 10 x 40 = 400 watts, hence, total power 

 given to circuit = 400 + 168 = 568 watts. 



The total energy voltage = 16 -8 + 40 = 5 6 '8. 



The total voltage applied to the circuit = 100, hence, 

 the idle voltage, or voltage spent in overcoming the self- 

 induction of the coil, will be the third side of a right- 

 angled triangle, having its hypotenuse 100 and one side 

 56'8. This is shown in Fig. 31. 



The remaining voltage may be got by measurement 

 or calculation. Its value = E o =V 100* - (56'8) 3 = 82-2 

 volts. The total voltage across the coil is the resultant 

 of this voltage and the energy voltage of the coil 16 '8. 



Total volts across coil = E l = V(82'2) 2 + (16'8) 2 = 

 84 volts. 



This could also have been obtained by direct 

 measurement from the figure as shown. The angle of 

 lag for the whole circuit is the angle between the energy 

 voltage A D and the total voltage A B. Referring to the 



t n . o 



figure, cos <f> = -rrr = '568 which is the power factor of 



the circuit. The angle of lag for the coil alone is the 



1 fi'X 

 angle DCS. For the coil, cos 9 ,=-=^- = '2. 



OT* 



The coil has consequently to be designed so that the 

 back electromotive force produced in it by self-induction 

 is 82-2 volts. From the formula given previously, 



V = 4e-4,4,ZNnlQ- 8 . 



Where V = back voltage due to self-induction. 

 Z = number of magnetic lines in core. 

 jV = number of windings. 

 n = frequency of current. 



Hence the product Z N can be calculated. Usually the 

 value of Z will be determined by the size of core it is 

 desired to use, and the magnetic density found desirable 

 to avoid excessive iron losses. 



Suppose in the present instance periodicity = 50. 



09.9 v i o 8 

 Then ZN = 4.44 x 50 = 37,000,000. 



