(575 ) 
2,153 _ 0,1906 
f= Dt oe ee ee 
0,102“ (0,102)? Bl 
y =81,508 — 1,462 A — 18,320 A?, 
or 
putting — A for kw (A being negative). 
The maximum, which is to be found in the Pangan branch, may 
b 
be calculated from the first of these equations (En = =) 
6 
En = ond x 0,102 = 5,65 x 0,102 Volts, 
1. 6. 
En = 5,65 X 500 = 2820, 
Further we find for 
b? 
Ym == 4 + Ac 
(2,153)2 
n= 20400 oe = 20,456 —— 6,080 = 37 54: 
ph ‚to T 73<0,1906 2) + ’ 31,54 
In the descending branch no maximum is to be found, because a 
maximum requires there a negative value of A, and A is here positive. 
We see, that the maximum (Z = 2820) does not coincide with the 
point, where w=0 (Z= 3020, as we found above). The difference 
is not great, but still (6,04—5,65) X 0,102 Volts = 40 millivolts. And 
in other instances it may be greater of course. It depends wholly 
on the value of A. 
The figure represents the accurate course of the two parts of parabolae. 
The dotted curves indicate, how the course should have been, if the 
branches had been continued on the other side of A= 0. 
The abscissae are the electromotive forces EZ of the inserted cell, 
and increase with 500 = 0,102 Volts; so they are respectively 1 < 500, 
2 500, 3x 500 etc. The ascending branch is AP, and would 
have been continued in P4’, if the coefficients remained the same 
after A = 0. The maximum is to be found at M, somewhat to the 
left of P, the point which separates the two different parabolae. 
The descending branch is PB, and would be continued along PB’. 
Its maximum is to be found at J’, So the curve, really passed through, 
is APB. The experimental values agree perfectly with the calculated 
values, now that the figure is made on this scale; only those found 
for E=0 and H=4 1,02 (those with?) do not agree (as is 
indicated in the figure by the sign x). 
38* 
