FUNDAMENTAL PROBLEMS. 



(a) Current locus with constant resistance and variable reactance. A given 

 electromotive force OE, Fig. 75, acts on a circuit whose resistance R is constant 

 and whose reactance ^"varies. Let the line Of, Fig. 75, represent a possible value 

 of the current. Then RI is the component of parallel to /, and XI is the 

 component of E 90 ahead of I. Therefore the triangle OPE is a right triangle 

 with a constant hypothenuse OE, so that the point P must lie on the circle of 

 which OE is the diameter. That is, the end of the line RI describes a circle as 

 X varies, therefore, since R is constant, the end of the line OI must also describe 

 a circle. The circle described by the end of the line OI is called the current locus 

 under the given conditions, namely, E and R constant, and X variable. The 

 diameter Oa of the circular current locus is equal to EjR and it represents the value 

 of / which corresponds to XO. 



It is especially noteworthy that the current at first decreases imperceptibly in value 

 when, starting from X=o, the value of X is increased. That is to say, the effect 



Fig. 75. 



Fig. 76. 



of a small value of X introduced into a circuit of given resistance is to turn / very 

 perceptibly behind E in phase, but not to produce any appreciable decrease in the 

 value of /. 



(b) Current locus -with constant reactance and variable resistance. A given 

 electromotive force, OE Fig. 76, acts on a circuit whose reactance X is constant, 

 and whose resistance R varies. Let the line OI represent a possible value of the 

 current. Then RI is the component of E parallel to I, and XI is the com- 

 ponent of E 90 ahead of /. Therefore the triangle OPE is a right triangle with 

 a constant hypothenuse OE, so that the point P must lie on the circle of which OE 

 is the diameter. Now, the two angles /? are equal, and XI is proportional to I 

 since X is constant. Therefore the point / lies on a circle ( the required current 

 locus), of which the diameter Oa is at right angles to OE, and equal to the current 

 EjX, which corresponds to R o. 



