Self-induction of Wires, 19 



The solution is a special case of the second of (1626), 

 Part IV. , which we may quote. In it take 



S" = K + Sp, R"=B'+I/p, ... (36) 



p meaning cl/dt so far. Also put ? 2 = 0, ^2 = ^0 sm nt j and 



-m*=F 2 ==(K + Sp)(R' + L>), . . . (U) 



and put the equation referred to in the exponential form. 

 Thus, 



(F/S" + ZQe«'-> + (F/W-ZQe-W-* . ■ 



^(F/S^ + ZO (F/S' / -Z )-e- F HW-Z 1 )(F/S'' + Z ) v o sm ™- ^> 



This is the differential equation of C in the line. Now in F, 

 S", Z , and Z 1; let d 2 /dt 2 = — n 2 . It is then reducible to 



F + Q'j t (A'F+ B'Q'n 2 ) + (A'Q'-B'P')| 



:—— r V sinn^=- A* + B*n» V oB inn*, (6*) 



giving the amplitude and phase-difference anywhere ; and 

 the amplitude is 



C =Y (A' 2 + B'% 2 )-*(F 2 + Q'V)^ ... (76) 



A' and B' are functions of z, whilst P' and Q' are constants. 

 Put 



F =P + Qi, ) where z = ( — 11 f > 



Z 1 =R\ + I/ 1 m, p^ p re » i- l > . . (Sb) 



.} wl 





The values of P and Q are 



P= $)*{(&* + 1/ VJ*(k" + 8W)* + (KB'- L'Sn 2 ) pi rQ ^ 

 Q=.. { - . . . • pj' w 



possessing the following properties, to be used later, 



P2 + Q2 = (K 2 + S 2 n 2 )*(R' 2 + L'*n 8 )*, ) 



p2_Q2 = KR'-L'Sn 2 , > ■ (106) 



2PQ = (R'S + KL')n. J 



The expressions of R ', R/, L r , L/ can only be stated when 

 the terminal conditions are fully given. Their structure will 

 be considered later. P and Q depend only upon the line. 



Let 

 A = R' - Sn 2 (E ' IV + BJ LqO + K (R ' E/ - L ' L/n 2 ) ; "> 



B = I/n + Sn(B 'R 1 , -L 'L 1 'M a ) + Kn(R 'L 1 ' + R l 'L 3 1 



a=P(R ' + R 1 ')-Qn(L ' + L 1 '); 



b = Q (B ' + R/) + Pn (L ' + Lj') . J 



C2 



