THE LOCALISATION OF BREAKS AND FAULTS. 457 



the two observations. Then In the second pair the end pre- 

 viously lower tests the higher of the two. The added resis- 

 tance is, therefore, too high, and is then reduced by rather 

 more (20 or 25% more) than the difference between the 

 readings, so that it is too low, and the third pair of balances 

 consequently comes out unequal. This gives the two outside 

 values and the respective differences in the balances for 

 each, from which, by the above formula, the correct value of R 

 is found. 



For example, preliminary balances to false zero from both ends 

 on a faulty cable give 



A = 2,015 from the near end 

 and B=4,785 ,, distant end. 



Resistance was therefore added to the near end. 

 With 3,500 ohms added (Rj) 



A = 5,530, B = 5,065, difference {d{) = 4:65 



As A exceeds B the added resistance is too high. 

 With 2,500 ohms added (R.^) 



A=4,530, B=5,027, difference (4) = 497 



Here the added resistance is too low, as B exceeds A. 



3,500 ohms being too high, and 2,500 too low, the right 

 amount to add was 



This was very near, as for equal balances of 5,050 from both 

 ends, the added resistance was 3,020 ohms, The CR of the line 

 being 5,460 ohms, the distance of fault was — 



5,460-3,020 



^ = 1,220 ohms from the A end. 



The N.R.F. correction should be applied if the cable is 

 known to have a low insulation at any particular place. For 

 instance, if in the above example the N.R.F. is 0-6 megohms, 

 situated 500 ohms from B end, the readings at that end would 

 be too low, and applying the Rule given on page 446, the 

 correction to add to the reading would be : — 



(5,050 - 500)2 20-7 



0-6 ~ 0-6 



=34 ohms. 



