798 Dr. H. H. Poole on Vector Methods for 



Evidently 



c iri = c 2 r 2 1 



K^o) K 2 &) J 



2 n 



In the case of an imperfect condenser some power is 

 absorbed and the angle OAB is obtuse. This waste of 

 power may be due either to actual leakage or to dielectric 

 hysteresis, or both. In any case, the current will be nearly 

 identical with that which would pass through a perfect con- 

 denser K and a resistance R in series [or alternatively a 

 condenser K' and a resistance R' in parallel]. K and R 

 will probably depend somewhat on the P.D. and frequency, 

 and the current through the actual condenser will probably 

 not be perfectly sinusoidal, so the representation is only 

 approximate. 



If we obtain a balance with two such condensers, the con- 

 ditions must be represented by the lower vector diagram. 

 Evidently, as before, 



n = Ri = _ K, 



r 2 R 2 Ki * 



We shall almost certainly have to insert a resistance in series 

 with one condenser in order to obtain a balance. 



If we have a perfect condenser K x [say, an air condenser] 

 we can use this method to measure the effective value of R 2 

 for another condenser K 2 , and hence find its power-loss for 

 any given frequency and voltage. We cannot hope for per- 

 fect silence, as a sinusoidal P.D. wave would probably not 

 give rise to a sinusoidal current through an imperfect 

 condenser. 



Comparison of a Self-Inductance with a Capacity 

 [Anderson's Method] . 



The branch OAjC is similar to OA x B in fig. 1, and is 

 similarly represented on the vector diagram. The current c B 



through the condenser (assumed perfect) is ^ in advance of 



the P.D. (OA) ; hence, as AB represents the fall in pressure 

 due to this current passing through the resistance ?* 3 , the 

 angle OAB is right. The vector OB must represent the fall 

 in pressure in the arm OB. The lengths of these vectors are 



