INDUCTIVE CIRCUITS. 57 



. We shall frequently make use of the vector equations (1) and 

 (3), so they should be thoroughly understood ; in fact, to proceed 

 further without first mastering them would be useless. 



43. To find the Equivalent Resistance, Self- 

 induction, and Impedance of n Inductive 

 Circuits connected in Series. 



Let TI, r%, r 3 , , . . r n be the ohmic resistances of the several 

 circuits. 



Let LI, LZ, Z 3 , . . . L n be their inductances. 



01, e & e s> - e n be the vectors representing the E.M.S. 

 potential differences between their respective terminals. 



Let i be the vector representing the E.M.S. current flowing 

 through the series circuit. 



Let p = 27rn, where n is the frequency of the current. 



Then, the vector e representing the potential difference 

 between the extreme terminals of the series circuit is given by 

 the vector equation 



e = ei -h e 2 + e B + . . . -f e n 



But by equation (1) 



e l = TI% 4- 

 e% = rgi 4- 

 3 = r& 4- JcpL 



Now, all the ri's are in the same direction, viz. along the 

 current vector, as also are all the JcpLi's, viz. leading a right angle 

 in front of the current vector ; they can, therefore, be respectively 

 added together numerically, and we get 



e = GI 4 e% 4 3 + . . . + e n 

 = O'i + r 2 + r 8 + . . . + r n )i 



4- Z 2 + Z 3 + . . . + L u )i . . (4) 



But if R and L are the equivalent resistances and self-induction 

 of the complete circuit, we have by equation (1) 



e = Ei 4- kpLi ....... (5) 



Comparing equations (4) and (5), we see that 



E = n 4- r 2 4- rz 4 . . . + >' 



