2391-10 Written Exam - 01/12/11 | Page 6 | on ElectriciansForums

Discuss 2391-10 Written Exam - 01/12/11 in the Electrician Courses : Electrical Quals area at ElectriciansForums.net

C

cr0ft

Just sat this tonight. What does everyone that sat it tonight think of the paper?
 
Votage drop calculation formula= R1 + Rn x 1.2 x Ib
So in this case = 0.38 x 1.2 x 30
Voltage drop = 13.68v = 5.95 % DOES NOT COMPLY!!

1.2 is the multiplying factor to convert the resistance of a copper conductor at a temp' of 20 degrees C to a temp' of 70 degrees C.

The question gave resistances of R1 and Rn but did not say at what temperature. So why are you using 1.2 ??
 
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1.2 is used because you have to take maximum operating temp of the cable when making this calculation. Otherwise when the circuit is in use and the temp of the cable rises the calculations and current carrying capacity will change. Worst case Scenario. Otherwise it would comply when not in use but possibly not when in use.
 
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1.2 is used because you have to take maximum operating temp of the cable when making this calculation. Otherwise when the circuit is in use and the temp of the cable rises the calculations and current carrying capacity will change. Worst case Scenario. Otherwise it would comply when not in use but possibly not when in use.

1.2 is used to convert a 20 degree resistance reading to a 70 degree resistance reading. See p105 of GN3:

For examplea reading taken at 20 degrees C could be converted to an equivalent reading at 70 degrees by determining the correction factor obtained from p105:
(1 + 0.004 (amb temp - 20))

= 1 + 0.004 (70 - 20)

= 1 + 0.004 x 50

= 1.2 So this is the multiplying factor to convert 20 degrees C to 70 degrees C.

BUT THE IMPORTANT THING IS THE QUESTION DID NOT SAYWHAT TEMP THE READINGS OF R1 & Rn WERE TAKEN AT.
 
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