1 8 ALTERNATING CURRENTS 



we may conveniently lay off this current vector along the horizontal 

 axis, as in Fig. 12 (5). 



Let 0i, 2 be the phase differences corresponding to the two given 

 impedances, and let Zi, Z 2 be their numerical values. 



In the vector diagram, assume the current to have a value of 

 unity. Then the p.d. across AB is given by a vector OVi, Fig. 12 (&), 

 of length Zi, making an angle 0i with the current vector. Similarly, 

 the p.d. across BO is given by a vector OV 2 .of length Z 2 , making an 

 angle 2 with the current vector. From this it follows that the p.d. 

 across AC is given by the diagonal 0V of the parallelogram con- 

 structed on OVi and OV 2 as sides, and the angle made by this 

 diagonal with the current vector is the angle by which the current 

 lags behind the p.d. across AC. 



Since, however, we have assumed the current to be unity, it 

 follows that the length of 0V will correspond to the numerical value 

 of the total impedance, and the angle which OV makes with the 

 current vector will correspond to the angle of phase difference for 

 the total impedance. Thus the required impedance is completely 

 determined. 



It is now evident that the rule for the composition of two 

 impedances in series with each other is identical with the rule for 

 the composition of two forces acting at a point we have simply to 

 apply the parallelogram law. 



The same rule may be extended to any number of impedances 

 connected in series, and by applying the polygon law we can easily 

 find, by a purely graphical method, both the magnitude of, and the 

 angle of phase difference corresponding to, the total impedance. 



Although a purely graphical method enables us to deal with this 

 problem, yet where accuracy is required it may be preferable to have 

 recourse to calculation. The most convenient method is then as 

 follows. 



Let Zi, Z 2 , Z 3 . . . be the given impedances, and 0i, 2 , 3 . . . 

 the angles of phase difference, or the phase angles* as we may briefly 

 term them, between the current and the p.d. across each impedance. 

 Eesolve each impedance into two components, one of which lies 

 along the horizontal, and the other along the vertical axis. Find 

 the sum of all the horizontal and the sum of all the vertical com- 

 ponents, square the two sums, add them, and extract the square root ; 

 this gives the total impedance. Thus 



total impedance = 

 \/{Zi 



* The phase angle is to be reckoned positive if the current lags behind the p.d. , 

 and negative if it leads. 



