496 Mr. 0. Heaviside on Resistance 



this result may be roughly stated as about 60 kilometres if it 

 be a single copper wire of 6 ohms per kilometre, 4 metres 

 high, with return through the ground ; but it varies con- 

 siderably, of course.) 



But if leakage be now added, it will increase the relative 

 importance of the magnetic energy, so that the length of the 

 circuit requires to be increased to produce a balance. This 

 goes on until K reaches the value RS/L, when, as an exami- 

 nation of (36) will show, the length of the circuit needs to be 

 infinitely great. The same formula also shows that if K be 

 still greater, L cannot be made to vanish at all, being then 

 always positive. 



11. Now let the circuit be infinitely long. Equation (35) 

 reduces to the irrational form 



Z=±(R + Lp)*(K + Sp)-*, . . . (38) 



with ambiguity of sign. Of course the positive sign must be 

 taken. The negative appears to refer to disturbances coming 

 from an infinite distance, which are out of the question in our 

 problem, as there can be no reflexion from an infinite distance. 

 But equation (38) may be obtained directly in a way which 

 is very instructive as regards the structure of resistance- 

 operators. Since the circuit is infinitely long, Z cannot be 

 altered by cutting-off from the beginning, or joining on, any 

 length. Now first add a coil of resistance R t and inductance 

 Lj in sequence, and a condenser of conductance K x and per- 

 mittance S,, in bridge, at A, the beginning of the circuit. 

 The effect is to increase Z to Z 2 , where 



Z^Ki + S^+^ + L^ + Z)- 1 }- 1 ; . . (39) 



i. e. the reciprocal of the new Z 2 , or the new conductance- 

 operator, equals the sum of the conductance-operators of the 

 two branches in parallel, one the conducting condenser, the 

 other the coil and circuit in sequence. (39) gives the qua- 

 dratic 



Z 2 2 +(R 1 + L ljt >)Z 2 =(E 1 + L lJt >)(K 1 + S lj p)- 1 . . (40) 



Now choose R 1; L 1? K 1? S x in exact proportion to R, L, K, and 

 S, and then make the former set infinitely small. The result 

 is that we have added to the original circuit a small piece of 

 the same type, so that Z 2 and Z are identical, and that the 

 coefficient of the first powder of Z 2 in (40) vanishes. Therefore 

 (40) becomes 



Z = (R + Lp)*(K + Sp)-* (41) 



This fully serves to find the sinusoidal solution. Differen- 



