Adding up current

Question

In a ring final circuit
(Based on Paul Cooks commentary of the 17th edition)

Very simple curent split using both legs of the ring

Not sure about adding up the currents in each leg
Take leg 2 Than will have the Ia of sockets 4,3 and 2
and the Ib of socket 1
How do these add up so you know the current in this leg?
Hope that makes sense
Thanks

AJJewsbury · Accepted Answer

You need to take the direction of the current (clockwise or anticlockwise) into account. E.g. 5A clockwise plus 3A anticlockwise on a given bit of cable gives a nett value of 2A clockwise. (Mathematicians would pick one direction as positive and so currents in the opporite direction are treated as -ve, so adding up works out). 
 Taking a step back, the whole example is symmetric - same loading and same cable lengths - the left is an exact mirror image of the right, so the currents must also be symmetric - anticlockwise current in L1 must be the same as the clockwise current in L5, likewise anticlockwise L2 must be the same clockwise L4 and so likewise clockwise L3 must be the same as anticlockwise L3 - i.e. L3 must be carrying zero current. If L3 carried any current it wouldn't be symmetric any more. Not true for the general case where loadings aren't so regular, but give you a way of sanity checking your calculations in this particular case. 
 - Andy.

Chris Pearson · Answer

John, I think that it may help if you have different loads at each socket. Try 4, 6, 8, and 10 A. The resistance of the cables is much lower than the loads, so that may be ignored.* 
 It may also help to turn on one socket at a time. So 4 A at socket 1 would be 2 A acw in leg 1 and 2 A cw in legs 2 to 5. Etc. 
 *ETA: my apologies - that is incorrect.

AJJewsbury · Answer

If a clamp meter was placed around the cable? 
 
 If you clamp the whole cable (i.e. both L & N together) you'd likely read zero whatever current the cable was carrying - a the L & N currents should be equal but in opposite directions and so would cancel out the magnetic field the clamp meter reads (actually, if you did get a significant reading, it would suggest that L & N were not balanced - thus indicating a break or high resistance connection in one of the conductors around the ring). 
 If you could clamp just one core (L or N) then yes, you should be able to verify your calculated "nett" values. 
 The bottom line physics is that to get a current to flow along a conductor (which will always have some resistance) you need to have a voltage difference between the ends of that conductor - higher voltage at one end and lower at the other and the current flows in one direction, reverse the voltage difference and the current reverses too. It's not possible for two currents to be flowing in different directions in the same conductor at the same instant. (The currents can flow in different directions at different times of course - which opens the door for a lot of cleverness when it comes to 3-phase work or combining different frequencies) but for d.c. and simple single phase a.c. using r.m.s. values, it can be treated quite simply). 
 AT first glance, your power loss calculation looks right to me (I've not checked your actual numbers though). P=I&sup2;R is a re-arrangement of P=IV where V would be the voltage drop along the conductor - so it all stacks up. 
 - Andy.

Chris Pearson · Answer

JohnE said: In the very first example where each leg was perfectly balanced. the currents add up to zero. This is a mathematical phenomenon of the net current? Im trying to visualise what actually happen say if a If a clamp meter was placed around the cable? It will measure a magnetic field (s) but there is a lot going on in a RFC with clockwise and anti clock wise currents and with AC direction change and unpredictable loads. Its hard for me to visualise... 
 Think of water flowing down pipes. 
 In the first case, let's say 20 l/min flows through the stopcock (the CU). 10 l/min flows anticlockwise to tap 1. 5 l/min comes out of the tap, which leaves 5 l/min to come out of tap 2. Similarly 10 l/min clockwise to tap 4 where 5 l/min comes out leaving 5 l/min for tap 3. All the water has come out of the four taps, so there is nothing left to flow along leg 3. 
 As for a clamp meter, I am afraid that the line and neutral will cancel each other out. You do have two conductors (or pipes) in each cable, but the flow is in opposite directions. 
 To measure the current with a clamp meter, you would have to bring one of the conductors out of the sheath and clamp it on its own, but then you have destroyed your T&E.

Chris Pearson · Answer

AJJewsbury said: AT first glance, your power loss calculation looks right to me (I've not checked your actual numbers though). 
 Agreed. It would be easier to check the numbers if we knew the length of all of the legs!

Chris Pearson · Answer

OK, I have deduced that the lengths of legs 2, 3, and 5 are 6 m, 10 m, and 7 m respectively, in which case the figures seem to be correct save that it is not clear what some of them mean. Calculations attached. I drew it up for 6 sockets, so the last leg and socket have values of zero. 
 (Can we upload e.g. spreadsheets, or just pictures?)

Adding up current

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