410 Currents in a Primary and a Secondary Coil. 



at the point, p, of contact of a tangent to the curve drawn 

 from the point where the right line AQ meets Ox. 

 Construct also the right line OT whose equation is 



k 2 V = Wx; ....'.. <9) 



then, taking any value, ON, of x, draw the ordinate NP, and 

 we have 



t/ = VN; */ 2 =B'N; y=PN;7? = YN. 



Moreover, it is obvious from the previous values that 



'1=1, :.y^\ ...'.. (10) 



which shows that the point P on the primary hyperbola is 

 deduced from the point P' on the secondary by the simple 

 construction or calculation of a fourth proportional. (Though 

 not belonging to the physical problem, it may be noted that 

 one asymptote of the primary hyperbola is the parallel to Oy 

 at the point where the secondary line, A' I/, cuts Ox, the 

 other asymptote making with Ox the angle whose tangent 



. AM 

 v k 2 N 2 ' J 



Finally, as regards the phase-angles, take ^ first. We 



N *S x 



have tan y= ; 



A r 2 



,2 „,_ -*-2 _ | 1/2 _ $h 



sec X=ZT2 = 



r 2 2 r 2 2 OA" 



;> w\/l§ ^ 



Hence, describing a circle on NB' as diameter, and drawing 

 A'E parallel to Ox, meeting the circle in E, we have 



%= ENE'i 

 ,\ 0= re-entrant angle ONE. 

 Again, we have 



Sm ^ a 2 + b 2 ~ U'Wy'x ~~ Wny' " &%' ' 



B*x 

 But from (6), h = 



k 4 v ' 



/. sin 2 ^r=— , ...... (12) 



y 



which shows that if we construct a circle on NP' as diameter, 

 and take the point, D, in which this circle is cut by the 

 tangent HD at H to the secondary hyperbola, we shall have 



^ = DN« (13) 



