TRANSIENT SOLUTIONS OF ELECTRICAL NETWORKS 123 



to an applied voltage £o cos (co/ + d). This can be obtained from 

 equation (38) by taking the real part of the equation and rejecting 

 the imaginary part. The process of doing this is simple and the 

 result obtained is 



I'm, 



E_ 

 2R 



cos {wt-\- e - 2m <p) - 2e-<'/^'-^« 



/ _2t_ \ '"-I 

 {m - 1) ! 



cos (6— if) cos<^ 



/ 2/ Y'-2 



\ -ylLC , 



+ ~7—. KTT [2 cos^ ^cos (6 — 2(p) — mcos ^cos {9 ~ ^)] 



+ 



(w - 2) ! 



/ _2^ Y~^ 



{m - 3) ! 



4cos^ v'cos (d — 3(p) — 2m cos- ^cos {6 — 2(p) 



m(m — 1) .. ."] . 



+ — ^-yj ^cos ^ cos (d - cp) \ + 



(39) 



where tan v' = w-^LC. 



For example, the solutions for one and two section networks take 

 the form 



2R 

 E 



^i = 2R f^^°^ (co/ + - 2^) - 2e-('/^^^> cos ^ cos (^ - ^)] 



«2 = 2D COS (co/ + ^ - 4<^) - 2e-('/^^« 



2/ 



2cos2^cos(^ - 2(^) (40) 



2 cos ^ cos {Q — ip) -\- ■ jYj^cos {6 — <p) cos <^ 



]]■ 



It appeared desirable to obtain some numerical calculations on the 

 building up of current in this type of network. This calculation has 

 been carried through for two section, four section, and six section 

 networks. The current has been calculated for a constant voltage, 

 for an alternating voltage whose frequency is the resonant frequency 

 of the network, and for one whose frequency is twice the resonant 

 frequency of the network. The current building up for a constant 

 applied voltage is shown for the three networks on Fig. 5. The 

 current building up in a two section network, in which an alternating 

 voltage whose frequency equals the resonant frequency of the net- 

 work, is shown on Fig. 6. The steady state and the transient terms 



