Self-induction of Wires. 427 



none elsewhere, and let (X + ^ Y)A and (X + ^ 1 Y)A 1 be the 

 currents on the left (nearest 2=0) and right sides of the seat 

 of impressed force. We have to find q , q ly A , and A x . 

 The condition Y = Z C at 2 = gives us, by (153), (154), 



-(X; + g Y ')^S "=Z (X + $0 Y ); 

 therefore 



2o =-(X ' + S %X )-f-(Y ' + S "Z„Y ). . . (159) 

 Similarly, Y = Z 1 C at z = l, gives us 



^-(X.' + Sx'^XO-W + S/'Z^). . . (160) 



Here the numbers and a mean that the values of X, &c. and 

 S" at 2 = and at z = l are to be taken. 



Now, at the place z — z 2 the current is continuous, whilst 

 the Y rises by the amount <? 2 suddenly in passing through it. 

 These two conditions give us 



(X 2 + ^ Y 2 )A = (X 2 + g 1 Y 2 )A 1 , 

 - S 2 "* 2 + (X 2 ' + q Y 2 ') A = (X 2 ' + qi Y 2 ') A D 



where the 2 means that the values at z=z 2 are to be taken. 

 These determine A and Ai to be 



a nr a - ( x 2 + ?i¥ 2 > 2 or (X 2 + q Y 2 )e 2 ( . 



A or Al - ( S 2 ")-XX 2 Y/-Y 2 X 2 ')teo-^)- ' * l •' 



Now use (155), making the denominator in (161) to be 

 h(q — qi). We have then, if C and Oi are the currents on 

 the left and right sides of the seat of impressed force, 



(X + y Y)(X 2 + ?1 Y,) 



• k^t-"- j.. . . (m 



c== (X+ gi Y)-fX 1 +yoY a ) 



1 Hqo-qi) 



These are, when the p is throughout treated as d/dt, the 

 ordinary differential equations of C and Cx arising out of the 

 partial differential equation of C by subjecting it to the ter- 

 minal conditions and to the impressed force discontinuity. 

 Now make use of the algebraical expansion 



iM = t _Ae) . . . . (163) 



1 



