34 Mr. T. H. Blakesley on a Geometrical Determination 



7. Suppose, then, that AB, BC are the revolving represen- 

 tatives of two electromotive forces. Then AC is their resultant ; 



CAE is an angle whose tangent is equal to 7^7, as explained ; 



CE, BF are perpendiculars upon AE. Then AE is the final 

 resultant or effective electromotive force, merely requiring 

 division by the resistance to give the current. 



A 



The power derived from the source of AB is AF 



AE 



AE 



2r 



the 



power transferred to the source of BC is FE "~ ; and the 



AF AF 2 

 power heating the circuit is AE — — =— ; — 

 1 6 <2r 2r 



As regards the projection of BC, viz. FE, since (as here 

 drawn) FE is in a contrary direction to AE, there is a 

 transfer of power to its source. Had F been situated nearer 

 to A than E is, the source of BC would do work and assist 

 in heating the circuit. This obviously depends upon whether 

 BC, AE are inclined to one another at an angle greater or 

 less than a right angle. 



If we denominate these three powers as the power of the 

 active source, the pow r er of the recipient source, and the 

 heating-power, they will be to each other in the proportion 



AF : FE : AE ; 



FE 



and the efficiency of transmission will be -r^, the ratio of 



, , . AB A * 



waste being -r^y 

 AF 



8. Since AFB is a right angle, F always lies upon the 



circle described upon AB as diameter. Describe this circle. 



We may call it briefly the F circle. 



At A set off the angle BAD in the negative direction 



