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stuartgil

i'm in my first year and i have a bit of math home work, i just need to know that i'm on the right tracks with this equation. Am looking if one of you could please help me in some way?

Here i go:

A 10mm, Two core armoured 70 degree Pvc Insulated cable (copper conductors) is clipped direct to a surface and forms a two wire radial distributor loaded at three points. From the supply end at 230 volts, the first load is 10A and is at a distance of 50m, the second point is 7A and is a further 30m and the third is a 5A and is 40m away from the second load point.

* Find the current in each section
* Calculate the overall voltage drop using IET Wiring regs.

So i have, first section (10A @ 50M)

Current lb= 230/10 = 23A

From the IET Wiring Regs i have found the mV/A/m value of 4.4

calculation of voltage drop = 4.4x23x50/1000 = 5.06 volts.

Please can one of you more experienced electricians see if i'm on the right track? Would really appreciate your help, thanks stuart.
 
Firstly, the current in the 1st section is the sum total of all the loads on the cct.



Draw the circuit with all the information you've been given on it.
It will help you visualise what is going on and where.
 
Last edited by a moderator:
i'm in my first year and i have a bit of math home work, i just need to know that i'm on the right tracks with this equation. Am looking if one of you could please help me in some way?

Here i go:

A 10mm, Two core armoured 70 degree Pvc Insulated cable (copper conductors) is clipped direct to a surface and forms a two wire radial distributor loaded at three points. From the supply end at 230 volts, the first load is 10A and is at a distance of 50m, the second point is 7A and is a further 30m and the third is a 5A and is 40m away from the second load point.

* Find the current in each section
* Calculate the overall voltage drop using IET Wiring regs.

So i have, first section (10A @ 50M)

Current lb= 230/10 = 23A

From the IET Wiring Regs i have found the mV/A/m value of 4.4

calculation of voltage drop = 4.4x23x50/1000 = 5.06 volts.

Please can one of you more experienced electricians see if i'm on the right track? Would really appreciate your help, thanks stuart.

As Archy says, the total load current is the 3 figures highlighted in red.

Secondly, just want to say thankyou for explaining where abouts you are in your training, and giving us your take on what the answer might be. there are quite a few people who just simply ask the question without even attempting to answer it.
 
I might be way off here but I dont think you have typed the question correctly.

* Find the current in each section?? The current at each point is in the question, are you sure it should not have read "find the volt drop at each section?"

Also, the second question is asking for the overall volt drop so you must calculate this using the end point of the radial circuit (120M) and the design current will be all sections added together.

Thats how I read it, I may be wrong though. Some else thoughts will confirm. :redface:
 
Last edited by a moderator:
No you have you current already, 22 amps. why are you dividing it by the voltage?
 
to work out total voltage drop, you add all total loads up for your design current Ib

Add all lengths up, which in your case is 120m, then put the figures into the equation.

So....

mv/a/m X L X Ib/1000 becomes
4.4 (using your figure in first post) X 120 X 22 = 11616 / 1000 = 11.61v


Long time since I did the calculations, hope the above it right
 
hi wallyanker, it says find the current in each section and the second question is calculate the overall voltage drop?? i am reading it the way you are explaining it as well and it doesnt make much sense as i thought the 20A,10A,15A were the current at each point. so like you said for the second question i calculate the voltage drop on the total meterage (100m)?
 
There will be three sections of volt drop
1. 50 Meters 22 Amps
2. 80 Meters 12 Amps
3. 120 meters 5 Amps

That is how I am reading the question.
 
sorry ment 120m lol thanks for all your help, this is making sense at last!! A massive THANK YOU TO ALL OF YOU. At least i know there are ppl willing to help on this site if i ever mis-interprit something again. THANKS once again.
 
Sorry Wally, Archy was thinking that the OP was trying to work out the total watts of the circuit.
 
There will be three sections of volt drop
1. 50 Meters 22 Amps
2. 80 Meters 12 Amps
3. 120 meters 5 Amps

That is how I am reading the question.

There will be three sections of volt drop, also second question is asking for overall volt drop, so all sections added together.

Think your above amps to lengths are in the wrong order to, 50m section has a load of 10a
 
Sorry Wally, Archy was thinking that the OP was trying to work out the total watts of the circuit.

Yeah, I thought he might have been thinking along those lines with him making the incorrect division.

Stuart, post your new answers, I'm sure someone will 'mark' it for you! :)
 
I thought if you calculate the volt drop at the 50m mark, your use the first load. Calculate at the second mark, you use both 1st and 2nd lengths and loads, etc..
 

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