of the Compound Dynamo. AT7 
which corresponds to » must be less than that which corre- 
spond to m. Hence 
meee 
Atm B+m Puy 
le Q Py 
Solving for P and Q, we get 
pa $i bom—9(B+m) . 
l+q (A—B)(u—m) Gees in) 
ee ea—m—qA+m) Be 
Sg CABG ny OTM, 
Now since A—B is negative, we must, if P and Q are 
positive, have 
q>(u—m)/(A+m), 
and, a fortiori, 
> (u—m)/(B+m). 
Further, since P>Q, 
{q(B+m)—(u—m)}(A+p)(A +m) > 
{q(A +m) —(%—m)}(B+p)(B+m); 
“. gQA+m)(B+m)(B—A) <(#—m){B?—A? 
+(u+m)(B—A)}; 
“. g(A+m)(B+m) <(w—m){A+B+p.+4+m)}. 
Cases V. and VI—A/B> /P//Q>1. 
Taking next the case in which the values of ¢ corre- 
sponding to w and m are equal, we deduce from (18) 
p—_2 (A+e)(At+m) 
l+p A—B ” 
® (B+p)(B+m) 
1+p A-B 
Substituting these values in the equation 
b= (/ P—/ Q)?/(A—B), 
& 
we get 
(A—BY(1+p)={/ (A+ p)(A4m)—V/ (B+ p)(B+m)}3 
Phil. Mag. 8. 5. Vol. 19. No. 121. June 1885. 2K 
