158 Dr. Alarchant on the Oscillatory Discharge 



At time 0, i=0, and consequently 



* = ^£=tan(NKO) t 



at L 



R 



From therefore draw a short line OD, parallel to KN, 

 this will be the beginning of the current-curve since 



tan (I/OX) =tan (NKO) = j . 



It will be convenient to take a time-interval such that dt 

 is numerically some simple multiple o£ CR. Thus d£ = £CR. 

 The exact time-interval is unimportant, provided it be 

 sufficiently small. Since 



)idt = a.rea, of OLiXi = m l m 1 x OX 1 =m 1 m 1 x dt, 



\ idt dt , dt , . , ^idt 7 



RG ~ mimiX lC and RC bein S == ^ tpr=* XJW i ,w i- 



Plot downwards from X, therefore, a length /j X miwij to 

 N 19 and from Lj draw LjN/ parallel to OX to meet ON in X x . 



Then *W=i^ -*K«!)-i= ^— |g'-t. 



Producing I^N/ to meet KM in Mi and joining M^, 



tan N 1 M 1 N 1 / = * * = — therefore, from the equation above. 



JMjJNj at 



Draw from L, therefore a Hue LnL 2 parallel to MxNj, 



This represents the increase ot current during the second 



di 

 time-interval XiX 2 . To determine the value of -j at the 



time corresponding to L 2 draw the mean ordinate mgrns of 

 the trapezium LiXj L 2 X 2 ; the area = X 1 X 2 x m 2 m 2 , and is as 



before = \idt from X : to X 2 ; the value of ■L_ for this in- 

 J RC 



terval=&xm 2 m 2 as before. And plotting this length down- 

 wards from Nj to Xo the length XX 2 = the value of L- 



RC 

 from the beginning of the discharge. Continaing the con- 

 struction as before and joining M 2 X 2 the slope of this line 



represents the value of -j at the instant X 2 . 



Curves have been plotted for discharges through a circuit 

 having a capacity =1 m.f. and a self-induction = 1 henry 



