32 ELECTROMOTIVE FORCE AND CURRENT. 



therefrom the angle of lag between the current and voltage 

 given by the alternator. 



(Note. In solving this problem remember that the value 

 of the voltage in the non-inductive resistance is the product 

 of current and resistance.) 



(2) An arc lamp is connected to a 100 volt alternating 

 supply in series with an inductive resistance. The arc is 

 found to take 30 volts and the inductive resistance 90 

 volts. Find by diagram the angle of lag of the circuit. 



(Note. The arc lamp behaves as a non-inductive 

 resistance.) 



Magnitude of Resultant Voltage. We are now in a position 

 to ascertain the relation between the resultant voltage in a 

 circuit and the current, resistance, and self-induction. 



Fig. 9- (page 24) shows the method of obtaining graphically 

 the magnitude and phase of the resultant voltage when the 

 component voltages are known. Since the two component 

 voltages spent in overcoming respectively the resistance^ and 

 inductance of the circuit are known to be always J period, or 

 90, apart in phase (one being in phase with the current and 

 the other J period in advance of it), the angle E r O E { will 

 always be a right angle. Consequently the length of the 

 diagonal representing the resul tant = *J sum of squares of 2 sides , 

 or 



E 1 = E* + E? 



or (resultant voltage) 2 = (resistance voltage) 2 + (inductance 



voltage)". 



We have seen that 



resistance voltage = C R. 



inductance voltage = lirnLC. (See page 16). 

 Hence if V = resultant voltage of the circuit 



F 2 = C 2 # 2 + (ZTrnLC 



or V = C V R 2 + (2 TT n Lf. 



This result is of the greatest importance. 



If the circuit has no self-induction, i.e., L = 0, 

 We have V = C *J R* = C R, as in the case of a direct- 

 current circuit. 



A further important result follows from Fig. 9. 



