Memoirs presented to the Cambridge philosophical society on the occasion of the jubilee of Sir George Gabriel Stokes, bart., Hon. LL. D., Hon. SC. D., Lucasian professor.

DISCHARGE OF AN AIR CONDENSER, WITH A DETERMINATION OF "v." 157
In the first instance neglect the terms depending on the resistances, then
S'2L'- 2 = Q = 2 cos 0,
S'2L' = 2 (1 + cos 6)= 4 cos.
Now if the whole of the capacity S, were concentrated at one part of the circuit
connected by a wire of inductance L we should have
1 I
XSL 1n (n - 1) S'L'
In the most important of the cases with which we have to deal a large part of the
inductance is so concentrated in the capacity S. We shall suppose therefore in solving
the equation that X2S'L' is a small quantity of the order 1/n2. So that 2 cos - is of the
order 1/n, and 0 is not much different from wr. Put 0 = r- q: ultimately (b will be
treated as small, though for the present the solution is general: then
S'X2L'= 2(1 + cos 0)= 2 (1 - cos).
Whence substituting
{S'X2L'- 1} {(S'2L'- 1) sin (n - 1) g + sin (n - 2) b}
+ (S'X2L'- 1) sin (n - 2) ( + sin (n - 3)
- {(S'X2L' - l)sin (n- 1) + sin(n- 2) } =0.
Thus 2S'X2L' {sin (n - 2) - sin (n - 1),p}
+ sin (n - 1) ( + sin (n - 3) q - 2 sin (n- 2)
-S sin (n - 2) b - sin (n - 1) + S'X2L' sin (n - 1) b}
+ S'2X4L'4 sin (n- 1) = 0,.'. 2S'X2L' {sin (n - 2)- sin (n - 1) /}
+ 2 sin (n-2) ( (cos - 1)
S- {sin (n - 2) - sin (n - 1) ~ + S'x2L' sin (n - 1) /}
+ S'2X4L' sin (n- 1) = 0.
But 2(cos -1) = - S' L'....................................(A).. L'{sin (n - 2) b - 2 sin (n- 1) ~}
- I {sin (n - 2) q - sin (n - 1) + S';2L' sin (n - 1) b}
+ S'À4L'2 sin (n - 1) =...................................................(B).