How to calculate Voltage Drop in Single/Three Phrase Circuit?

Question

Assuming there are 10nos of 2KW lights at a distance of 300m away. (Single phrase - 240Vac, Three phrase - 400Vac)
How do we calculate the voltage drop in percentage for single phrase using 3cores 16mmsq armour cables?
How do we calculate the voltage drop in percentage for three phrase using 4cores 16mmsq armour cables?

Answer 1

Too early for me.
The link may help, it a .PDF file and you will need Adobe Reader or similar to view.
http://www.tycothermal.com/assets/NorthAmerica/English/Documents/Miscellaneous/Products/216/H57611_AppendixB.pdf

Answer 2

First you get modern-day. 5.685V / one thousand = 0.005685A the present will consistently be a similar with the aid of all aspects in a collection circuit. Now which you already know modern-day you could calculate the voltage and resistance for R2 & R4 R2 = 5V/0.005685A = 879.5 ohms V4 = 560 ohms * 0.005685A = 3.1836V all of the load voltages ought to upload as much as the source voltage. Vs = V1+V2+V3+V4 V3 = Vs - (V1+ V2 + V4) = 21.21 - (5.685 + 5 + 3.1836) V3 = 7.3414 V The sequence modern-day continues to be a similar, yet now you already know the voltage in the time of R3. R3 = 7.3414V / 0.005685A = 1291.36 ohms it is it...

Answer 3

OK, you ask "...how do we calculate..." so I'm not doing it for you....
IT all comes back to good old Ohm's Law.
You know you have 10 X 2,000 Watt lights, and you know your voltages. So you can calculate the current.
Now you go to
http://www.powerstream.com/Wire_Size.htm
and look what resistance your wire has, over a distance of 300meter (remember, those Ohm values are for one wire, but you have a pair!).
Then take Ohm's Law yet again, and calculate how much the voltage will drop along the wire's resistance, at the current you have calculated above.
Then do it again for the 3-phase scenario...
Don't know Ohm's Law? See:
http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.intro.html