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Mr. W. G. Rhodes on 



and E, e as adjacent sides, but, since E is the E.M.F. of the 

 generator, we take that which gives the component of E along 



Fig. 3. 



Oc in the same sense as the current. The other parallelogram 

 would make e the generator. We may notice that the possi- 

 bility of constructing these two parallelograms affords a proof 

 of the fact that, in general, either of two alternate current 

 machines may be driven as a motor by the other, irrespective 

 of their relative E.M.F. 's. An analytical proof of this is 

 given by the energy equation 



remembering that 



p + c 2 R = cE cos yjr, 

 p=zce cos (£, 



and that (f> and ty are independent. 



The condition that E represents the generator E.M.F. 

 limits our choice of the two parallelograms to OET0 (fig. 3). 



We then have 



angle cOE =^, 



angle cOT x = 0, 

 and angle cOe =<j>. 



15. Now through T draw PSTQ parallel to the line of 

 current, and draw PM, SO, TR, and QN through e, 0, T, 

 and Q respectively, at right angles to the line of current. 



