418 SUBMARINE CABLE LAYING AND REPAIRING. 



opposite in direction to the testing current it will act like a 

 resistance added to tlie cable and the balance will be too high. 

 The correction to apply to an ordinary CR bridge test will 

 therefore be subtractive from the bridge result. In Schaefer's 

 test the correction for a like condition is additive. This may 

 be a little difficult to understand at first, but it is the natural 

 result of the mathematical treatment. We start first of all 

 by setting out the earth current correction as subtractive 

 as it properly should be when opposing the testing current : — 



a; + F = A--. (1) 



and ^. + ^=B-- (2) 



The earth current correction is the last term on the right- 

 hand side of each of these equations. 

 Expressing equation (1) in the form 



a; + F = A-w(— ) 

 \%cl 



and subtracting (2) from it, we obtain 



Y~^^^ = {k-^)-~{n-\). 

 ^ t/n, - 1 ^ ' nc^ ' 



Let the factor by which F is divided = P. Then 



F=(A-B)P-^^(^-l)P (3) 



This represents the total resistance of the break. Now, by 

 equation (1) we have 



a; = A-F--. 

 c 



Substituting the above value for F we obtain 



^,=A-[(A-B)P-^.(«-l)p]-^. ... (4) 



From this we see that the break resistance (the quantity 

 within the large brackets) has a component term depending 

 upon the earth current. When there is no earth current this 

 term is nil and the resistance of the break is simply (A - B)P. 

 With a negative earth current which we are now considering 

 the term (A - B)P representing the break resistance is too high 



