ALTERNATING-CURRENT < IRCUITS 33 



and 



7 = * = E = E (13) 



VR* + XL* \ //,)* Z 



Z = \/R 2 + X L 2 is the impedance of the circuit and is ex- 

 <1 in ohms. It is ordinarily denoted by Z. Equation 

 corresponds to Ohm's law for the direct-current circuit. 

 The current in an alternating-current circuit is directly propor- 

 tional to the voltage across the circuit and inversely proportional 

 to the impedance of the circuit. That is, if the voltage in volts 

 he divided by the impedance in ohms, the value of the current in 

 umi>eres is obtained. 

 Also the voltage 



E = IZ. (14) 



An inspection of Fig. 32 shows that the angle 6 by which the 

 current lags the voltage may be determined as follows: 



IX L X L 2 7 ,/L 

 i&n 'TR --R -TT 



cos = IR = R = A> ( 16 ) 



" V(IR)* + (IX)* VR* + XL* / 

 Example. A circuit containing 0.1 henry inductance and 20 ohms resis- 



tance in x-ries is connected across 100-volt, 25-cycle mains. (/ What is 

 the impedance of the circuit? (h) What current flows? (c) What is the 

 je across the resistance? (d) What is the voltage across the induc- 

 '.' (e) Determine the angle by which the voltage leads the current. 

 A/ = 2r25 X 0.1 - 157 X 0.1 = 15.7 ohms 

 (a) Z = \/(20)' + (15.7) J = \/646 = 2:,. \ ohms. .1 



/ ' 7 ., I ' M 1 I 3.94 wnp. Ana. 



/ :,.. li; = 3.94 x 20 - 78.8 vo 



I X L = 3.94 X 15.7 - 61.8 volts. Ana. 



As x/(78.8) f 4- (61.8) - 100 volts. 



,, -' = 1- = 0.785. 

 /i 2() 



-r> 460, <?=-38.1. An*. 



17. Power. It h:is ;iln';idy ln-rn shnwn that a j>ure inductance 

 con-iinies no power, 'riien-fun-. tin- imluctan . lil con- 



snnio no [M.ucf. All the pmvrr expended in the circuit nn. 

 ; inted for in the re618t I'hat i> 



P = PI! 1 IE 

 1 1\ is nbvi)ii>!y e.iual t /' OOfl 1m. 32). 



