ALTERNATING-CURRENT CIRCUITS 35 



Solving the above for the current /, 



- (18) 



The power taken by the circuit is obviously 

 p = PR = I(IR) 



as the net power taken by the condenser is zero. 

 IR = E cos 6 



Therefore P = El cos 6, which is the same expression for 

 power as with inductance and resistance in circuit. 



The angle 9 may be determined as follows: 



cos0 = - R JL = **=P.F. 



C must be expressed in farads. 



-A capacitance of -jo microfarads and a resistance of 100 ohms 

 rinected in scrips across TJO-volt, (iO-eycle mains. Dctennir 

 The impedance of the circuit. (/) The current flowing in the circuit, (c) 

 The voltage across the resistance, (d) The voltage across the capacitance. 

 (e) The angle between the voltage and the current. (/) The power, (g) 

 The : tor of the circuit. 



20 mf. = 0.000020 farads. 



0.000020. 



(a) Z = v/dOOi'- 1 ! , . \ J, mo lf,c, ohms. Ana. 



I \^. 0.728 amp. Ans. 



//,' 0.723 X 100 = 7-J.n volt- i 

 /A- 0.728 X 133 = 96.2 volt 



ti.2) - 120 volts ( 



!,;;; 



e - 63.1. Ana. 

 I /.' - (0.723) 1 X 100 - is. Ans. 



(g) co80 " 0.602. 



/ 



ck). 



