34 ALTERNATING CURRENTS 



Therefore, the power 



p = I(IR) = IE cos 6 = El cos 



As has already been shown, cos 6 is the power-factor of the 

 circuit and is equal to the true power divided by the volt-amperes 

 or apparent power. 



PF = 

 EI 



Obviously the power-factor can never exceed 1.0. It is usually 

 less than 1.0. 



Example. How much power is consumed in the foregoing circuit and 

 what is the power-factor? 



p = PR = (3.94) 2 X 20 = 310 watts. Ans. 

 P *^10 



Also cos = P. F. = = jL = 0.787. Ans. 



i ZO.* 



18. Circuit Containing Resistance and Capacitance in Series. 

 Figure 33 shows a circuit containing a resistance R and a con- 



i 



FIG. 33. Circuit containing FIG. 34. Vector diagram for circuit 



resistance and capacitance in containing resistance and capacitance in 

 series. series. 



densive reactance X c in series. An alternating voltage E, of 

 frequency/ cycles per second, is impressed across this circuit and 

 a current / flows. Let it be required to determine the relation 

 existing among E, /, R and X c . 



The current / is the same in both R and X c and is laid off 

 horizontal in the vector diagram, Fig. 34. The voltage E R 

 across the resistance is in phase with the current. The voltage 

 EC across the condensive reactance lags the current / by 90 

 (see Fig. 29), page 30. The line voltage E is obviously the vector 

 sum of IR and IX c and is therefore the hypotenuse of the right 

 triangle having these two voltages as sides. Obviously 



E =V(/#) 2 + (ix j* = i VR* + x c 2 = iz 



where Z is the impedance of the circuit. 



