This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1889 Excerpt: ...of C corresponding to E2. Now, the power exerted by source A for this position A "Ci A "El will be----, and that exerted by source C will be-----?. The relation of the latter to the former is 2 l FE2 AF' Now, as Ep moves round from Q to R, on the line QE2R, the above relation starts from zero, passes by positive values through a maximum value, and ends again at zero. There will therefore be two positions when the relation of the powers has a fixed value, and one of these will be a position of stability, the other unstable for one or other of the sources. To express the various powers which have been dealt with in terms of the aDgle ABC, which is the angle by which the difference between the phases of the two electromotive forces falls short of two right angles, proceed as follows: Call this angle 6. Also call the angle BAC, $, and the angle BAD, /3. It is first to be observed that the triangles DEA, BCA, if completed, are similar to each other, since each side of the former bears to some one side of the latter the ratio cos /S, or geometrically, AE = ACcos/3 ED = CBcos/3 DA = BAcos#. From the triangle ACB, AC: BC:: sin 6: sin /, AP.sin 6 or Al = / therefore AE =/ e2 cos2 8-ef cos B cos 6-B... TK (). which is in the form desired. Again, from the triangle CEA, AE = AC cos CAE=AC cos 8: . AE2 = AC2cos2& and from the triangle ACB AC2 = BC2 + AB2-2BC. AB cos ABC, =/2 + e2-2 ef cos 6. Hence, AE2 = cos2 8 e2 +f2-2 ef cos 6, AE2 or the power employed to heat the circuit, being r-, is 2K equal to e2 cos2 B+f2 cos2 B-2ef cos 6 cos2 B, a 2R.... P). The difference between these two powers is the power transmitted, and therefore doing work upon source C. Hence the power transmitted is equal to ef cos B cos 6 + B--f2 cos2 B, 2R.... yj. The efficiency of the ...