444 Induction of Currents in Linear Circuits [CH. xiv 



value, as given by Ohm's Law, the number of tubes through circuit 2 will no 

 longer vary with the time, so that there will be no electromotive force in 

 circuit 2, and the galvanometer will shew no current. If we break the 

 circuit 1, there is again a change in the number of tubes of induction passing 

 through the second circuit, so that the galvanometer will again shew a 

 momentary current. 



GENERAL EQUATIONS OF INDUCTION IN LINEAR CIRCUITS. 



512. Let us suppose that we have any number of circuits 1, 2, 



Let their resistances be R 1} R 2 ,..., let them contain batteries of electro- 

 motive forces E lt E 2) ..., and let the currents flowing in them at any instant 

 be i 1} i,, .... 



The numbers of tubes of induction N ly A 7 2 , ... which cross these circuits 

 are given by (cf. equations (429)), 



N! = L n ii + L lz i a + L 13 i 3 + . . . , etc. 



In circuit 1 there is an electromotive force E l due to the batteries, and an 



dN 



electromotive force r- 1 due to induction. Thus the total electromotive 



dt 



force at any instant is E 1 -~ , and this, by Ohm's Law, must be equal to 



(tt 



R^. Thus we have the equation 



Similarly for the second circuit, 



d 

 E*--i (Lvi l + L. a i9 + L & i 9 +...) = R 3 is (432), 



and so on for the other circuits. 



Equations (431), (432), ... may be regarded as differential equations 

 from which we can derive the currents i\, i z , ... in terms of the time and the 

 initial conditions. We shall consider various special cases of this problem. 



INDUCTION IN A SINGLE CIRCUIT. 



513. If there is only a single circuit, of resistance R and self-induction L, 

 equation (431) becomes 



E-(Li 1 ) = Ri 1 (433). 



