538 BELL SYSTEM TECHNICAL JOURNAL 



impressed by the primary, then 



E' = E^ - Ei = 4TzIu (123) 



V = Vs - Vi = 4:TVu (124) 



where 



T = (rsi - Ts2 - Tu + T,,)/4. (125) 



Remembering that /i and Vi are discontinuous at x = 5/2, in accord- 

 ance with equations (117) and (118), it is seen that E' and V are 

 discontinuous at x = s/l in accordance with the following equations." 

 For X > s/2 : 



E' = zLEoT^e-^', Y' = dzEore-T^. (126) 



We are now prepared to formulate the mode-a currents produced in 

 the secondary (3, 4) by the field arising from the primary (1, 2). 

 Since the secondary constitutes a balanced two-wire line in an arbitrary 

 impressed field, it is amenable to the treatment formulated in the 

 subsection following equation (54); thence equations (41), ••• (45) 

 and equations (50), • • • (54) are formally applicable. 



If I'),{^x) and I^^x) denote the mode-a currents at any point x in the 

 secondary wires 3 and 4 respectively, then 



I,{x) = - U{x) = lix), say. (127) 



On referring to the subsection containing equations (41), • • • (45) 

 and applying it to the mode-a effects in the present problem, it will 

 be seen that 7(x) is the sum of the five mode-a constituents j{x), 

 Jo{x), Js{x), Jos{x), Jui.x), corresponding to equations (41), (42), (43), 

 (44), (45) respectively. From the discussion in connection with those 

 equations and from the analysis of the impressed field into two modes, 

 a and c, as described and formulated in the subsection following equa- 

 tion (54), it will be seen that j'(^) is due to the mode-a axial electric 

 force £3(3') — Ei{y) acting at all points y of the secondary, /o(:^) is 

 due to the mode-a impressed voltage ^3(0) — V^iO) acting at the 

 end y = 0, Js{x) is due to the mode-a impressed voltage ^3(5) — ¥4(5) 

 acting at the end y = s, Jos{x) is due to the mode-a impressed voltage 

 ^3(3') ~ Viiy) acting at all points y in the leakage admittances^ between 

 the secondary wires, and /«(x) is due to the discontinuity V'{u — ) 



2^ From mere physical considerations, it is evident that the whole primary field 

 is reversed at x = s/2. 



^^That is, the 'mutual' leakage admittance (equal to the 'direct' leakage ad- 

 mittance between wires plus one half of the ' direct ' leakage admittance from each 

 wire to ground). 



