and for the remaining «, 
C' sin 4 nO. 
From this there appears analytically (as could have been seen a 
priori) that the currents which are arranged symmetrically with 
respect to 2, and 7,, are equal in pairs, for the case n— even. For 
the case that n'—odd, the same is true for the circuits arranged 
symmetrically with respect to 7). 
The expressions for « contain also for this case the quantity 6, 
and in order to distinguish the different @’s, we again put for n = even, 
n+ 2/—2 
Oe] = a, = (—1)¥ C cos (4 n — hk) Op = (— IJK C cos (4 n — k) | ————— ] 2 
A0=0) n 
for n' = odd 
3 n' + 2/—2 
Oh = a, = (—1) C sin (4 n'—hk) 6, = (—1)F C sin (4 n'—k) (=5 )e 
G0) 
The currents to be determined being equal in pairs, we find for 
the even case from the initial conditions, only !(m + 2) equations ; 
and for the odd case only 4(m' + 1) equations. Thus in constructing 
the general integrals of the differential equations (9), we only need 
use $(n-+ 2), resp. 3(m'-+ 1) particular integrals of the form (149) 
and (14°), 
So we obtain for the general equations of the currents, 
a Wt n= even 
Pn+2t 
in = C, cos Ann.ert + C,cos$(n+2) 7, et +...4 Crsgcosnm.e ? 
9 
& 
19 Pnt 
wa : 
bp A= | C ,cos(4n —1)a ent 4 Cjcos(42 —1) x .ert+...tCyr4tocos(tn—l1)2r.e ? 
n 
2 
42 Pn+e2t 
n a aii: 
nei, —C, cos (bn — 2)m ent + C, cos (4n — 2) nerd... Cr42cos(in—2)20 e * (15a) 
n ED 
l : Pn+e2t 
i —n—| | s n-- 2 j . at 
i1 — ei ==(—1)? C,cosn.ert + C, cos - Pa! ah Cyrtocos2mw.e * 
—n—1 —n+1 | n Fan 
9 pj 
n [ Pest} 
in = (—1)2 {C, cos 0. est + Ci cos 0. erst + … + Crpocos0.e ? | 
2 / 2 J 
