Could someone check my voltage drop calculation?

The voltage drop on a circuit supplied from a 230v source by 16mm two core copper cable 23m long, clipped direct and carrying a design current 33Amp.

Actual Voltage Drop = MV/A/M x Ib x Length then divided 1000. Therefore. 2.8 x 33 x 23 divided by 1000 = 2.12 volts (2.8 from table 4D2B page 277 BS7671)

The answer given in the book I am using is written this way..

Vc = mV x Ib x L divided 1000. Therefore. 28 x 33 x 23 divided 1000 = 21.25 volts


So where does the author get 28 for his calculation? I'm reading 2.8 in the IEE Regulations.

What is the answer? Is it 2.12 volts which is my understanding or the books author 21.25 volts?

Best Regards
Martin

You will get more replies if you post it in Wiring and the regulations.

By the way 21.25 volt drop does not look like the right answer.

Regards
Chris Chew

Martin,

Your result looks OK to me. Looks like a typo in the book you have.

Just to confirm, voltage drop figures are given in mV/A/m, or in the case of 16mm sq = 2.8x10-3/A/m. So no need for the divide by a 1000 if you start off in mV.

Regards

Robo

Thank you for your input "sfchew" I posted it into "Wiring and the regulations" with success.

Robo, thank you for your knowledge.

Regards
Martin

hello thre. i`m newbies here... anyone from you have a some picture for earthing installation for transformer.

thanks

Mohdyazel,

You must start a new thread and provide more information regarding the transformer.

Thanks