Measurements oj Inductance. 523 



z are all zero, provided P : Q = R : S, so that L=KPS, 



Ci + c 2 + <?3 = c x a + c 2 ft + c 3 y = c x a 2 + c 2 ft 2 + c 3 7 2 = 0, 



or ci=c 2 =e 3 =0. In this measurement, then, if a balance 

 is attained at all, it is necessarily continuous. It will save 

 time in dealing with other methods if we notice that this is 

 because z is initially zero; that z and z are so is of the 

 essence of the mode of measurement, of course. The reason 

 why we get a cubic in D is that the network comprises, in 

 different arms, three repositories of energy (static or kinetic) ; 

 the capacity and inductance of the connecting wires being- 

 ignored. When the conditions PS = QR and L = KPS are 

 fulfilled the cubic breaks up into 



(KQSD + Q + S){^D + G + R)(KPSD + P + R) + PS-R 2 }=0. 



The roots «, ft, 7, with signs reversed, are of course the 



reciprocals of three time-constants, of which one is *X , 



the time-constant of K discharging itself through Q and S in 

 parallel, or of L discharging itself through R and the short- 



circuit. The other two lie between ~ r and 7- — ~> with 



(JT + Jtt S>+ b 



which, however, they coincide if Q = R = ^/PS. 



At first sight it seems odd that G and g should play any 

 part in determining the rates of decay in the other arms, 

 although no current flows through G_, but this is explained 

 later. We may now write 



x = a^ + a 2 e& + a z e^ + =- 1= ; 



Jr + xx 



in y and u we must replace a 1? a 2 , a B by 61, b 2 , b 3 or d lf d 2 , d B 

 respectively, and the constant term by 



R Lf Lf 



Q'P + R ° r Q' 



To evaluate the constants we have the initial values of at, ?/, u 

 and their two derivatives, viz. 



0'f- E /- P % P + R R P + R (P + R)» 



The proportionality shows that - 



a L : b } : d 1 = a 2 : 6 2 : d 2 — a ?> : h 3 : d 3 =Q : — R : —(P + R). 



