Ring Circuit Calculations.

[C&GStudent]

Apologies for the photo first of all!

I'm trying to calculate how much current each leg in the hypothetical ring circuit is carrying.

I'm sure I've seen an example on how to do it here some where.

Can someone point me in the right direction, or run through the calculation for me please.

You don't have to tell me the answer, just give me the the theory.

The boss at work is always giving us this "homework!" lol.

Thanks again.

 

[Lectrician]

9Kw? No current flow. Fuse has blown.

 

[C&GStudent]

Thanks for your input Lectrician!

I'm just interested in the theory of calculating how much current each leg of the ring may be carrying, if the loads and lengths of the legs are known?

 

[Adam_151]

The volts at a socket terminal will be lower than the volts at the DB due to volt drop, happy so far?

So you have two lengths of cable, with the same volt drop (cos they are connected at both ends) but different resistances, the current that will flow in each one will obey ohms law, so I=V/R will give you the current, and as R is propertional to length, if you just want the ratio of the currents you can just divide the currents into the radio of the reciprocals of the cable lengths

So if you had leg lengths of 2 and 8 metres, and a current of 10A

2^(-1) = 0.5
8^(-1) = 0.125

for 2M: (10/(0.5+0.125))*0.5 = 8A
for 8M: (10/(0.5+0.125))*0.125 = 2A

 

[ebee]

It all depends on how much of a stickler he is.
example say the top RH load of 2.0KW is fed from 3 + 3 + 4 = 10 m off the left leg and 4 + 5 = 9m off the right leg .
The two lengths total 19M
So the 9m leg will draw 10/19s and the 10m leg will draw 9/19s of the total for that load.
you divide 2.0 KW by 230v (or 240v in old money) to get total current for that 2.0KW load.
You can then do this with all the loads to get the LH & RH figures.
Hopefully that is all he requires.

Or he might prefer the current thru each indiviaul length of cable which you could also equally calculate by addition.

However if he wants to calculate volt drop and resistance of each bit of cable then you would have to recalculate several times (I really do not think he will want you to go that far though in a million years)

 

[C&GStudent]

Thanks guys. I've got to calculate how much current is flowing through each individual leg.

 

[plugwash]

pretty simple.

We assume that volt drop will have a negligable effect on current draw and therefore each load can be modeled as a current source. This lets us calculate everything by the principle of superposition

Assuming a supply of 250V (i'm being lazy ) the 3.0KW load will be drawing 12A. The clockwise route is 16m the anticlockwise one is 3m. The anticlockwise route is about 5 times the length (actually 5 and a third but as I said i'm being lazy) of the clockwise one and therefore has 5 times the resistance. Therefore 10A flows down the shorter leg and 2A down the longer one.

obviously its a good idea to do theese calculations more accurately if you want it for real, i just CBA pulling out a calculator

repeat this for each load to get the total in each leg.

once you know the total current flow in each leg its trivial to work out the current flow in inividual segements.

 

[pdcelec]

C+G, I have a fully worked example from my ONC if you would like a copy. I could send it to your email

 

[pdcelec]

just found how to post a pic

 

[C&GStudent]

just found how to post a pic.....

Thank-you pdcelec. I employed your method.