calculate the minimum value of the cross-sectional-area

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noel2507

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Hi this is the first time on the forum so i would like to say hello to everyone. My name is noel I

 
thanks very much for your speedy reply, however i am sorry i have tried this formula and i can't get the answer could you please explain the sum to me thanks very much

 
Added the workings out to my post above.

 
using my question sheet that was the correct answer and you have explained it very clear thanks very much. i have 2 more question i having been trying to answer for 2 days now. if its not to much trouble thanks.

i have supplied the question and the answer but i don't understand how to get this answer could you please explain. this is the last 2 promise.

20 The fault current due to an earth fault of negligible impedance in a 400 V, three phase, four wire circuit

having an earth loop impedance of 0.3 ohms, is

a 1383A

b 767A

c 124.5 A

d 72A.

Answer b 411.4.5

37 A circuit is protected by a 10 A BS 3036 semi-enclosed (rewireable) fuse. The minimum permissible

current rating (Iz) of a conductor protected by this fuse would be

a 7.25A

b 10A

c 13.8A

d 20A.

Answer c appx5 5.1.1

 
20 The fault current due to an earth fault of negligible impedance in a 400 V, three phase, four wire circuit

having an earth loop impedance of 0.3 , is

a 1383A

b 767A

c 124.5 A

d 72A.

Answer b 411.4.5

An Earth Fault means 230V/0.3= 766.67

 
37 A circuit is protected by a 10 A BS 3036 semi-enclosed (rewireable) fuse. The minimum permissible

current rating (Iz) of a conductor protected by this fuse would be

a 7.25A

b 10A

c 13.8A

d 20A.

Answer c appx5 5.1.1

10A x 0.725 (Rating Factor for rewirable fuses) = 13.79A which rounds up to 13.8

HTH cheers

 
Hi

Post 7 is correct but post 8 is sort of :| ....

A 10A rewireable fuse can only protect a cable of maximum 13.8 A as

remember you can overload a wire for 1.45 so 1.45X 13.8 =20A(433 somewhere)

a rewireable has a "fusing factor" of 2 ( showing my age here!) and 2X10=20A (appendix 4)

 
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