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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.

#### Archy Styrigg

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Electrician's Arms
Trainee Access
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.

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#### Plonker 3

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.

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#### wallyanker

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:

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#### stuartgil

so what your saying is my total load is 22A then this is divided by the voltage?

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#### Plonker 3

No you have you current already, 22 amps. why are you dividing it by the voltage?

#### Archy Styrigg

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No you have you current already, 22 amps. why are you dividing it by the voltage?
Changed my mind!

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#### wallyanker

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

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#### stuartgil

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)?

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#### Plonker 3

But that is not need for calculating volt drop, unless I am being a bit thick :dunce:

#### Archy Styrigg

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But that is not need for calculating volt drop, unless I am being a bit thick :dunce:
I think I'll delete my previous reply!

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#### Plonker 3

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.

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#### stuartgil

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.

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#### wallyanker

But that is not need for calculating volt drop, unless I am being a bit thick :dunce:
Sorry Dill, dont know what you mean?

- - - Updated - - -

ignore that, was a bit late with post

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#### Plonker 3

Sorry Wally, Archy was thinking that the OP was trying to work out the total watts of the circuit.

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#### wallyanker

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

#### Archy Styrigg

Mentor
Electrician's Arms
Trainee Access
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!

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#### Plonker 3

No as the first section will be carrying the full load current, I thought that at first.

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#### wallyanker

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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#### Plonker 3

The load will reduce the further down the cable you go.

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#### wallyanker

The load will reduce the further down the cable you go.
I get it, as the first part of the cable is still carrying current for all loads, then when you go to the next stage, you dont need to include the first load.

The overall voltage drop remains a total of all loads and lengths.

Thanks Dill

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#### stuartgil

cheers Dillb, appreciate your help mate..Looking at what you have worked out is the voltage drop in each section which is great! the question does say the overall Voltage drop so i then went with Wallyankers method. But thanks to you both for your help has eased my stress levels for now lol

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#### Plonker 3

Yes work put the voltdrop for each section, then add them together.

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#### wallyanker

Dont forget to turn the total volt drop into a percentage and mention whether it is within limits or not, extra marks for that

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#### stuartgil

here is my second attempt:

Total current = 10+7+5 = lb 22A

Total metres of cable = 50+30+40 = 120m

mV/A/m value = 4.4

overall voltage drop = 4.4x22x120/1000 = 11.61 volts.

#### yellowvanman

Electrician's Arms
Solar Guru
here is my second attempt:

Total current = 10+7+5 = lb 22A

Total metres of cable = 50+30+40 = 120m

mV/A/m value = 4.4

overall voltage drop = 4.4x22x120/1000 = 11.61 volts.
Not quite right yet:

You've got 3 voltage drops to calculate and then add then together.
1) 1st length is 50m with 22A flowing
2) 2nd length is 30m with 12A flowing
3) 3rd length is 40 with 5A flowing

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#### stuartgil

if i were to calculate voltage drop for each section would it look something like this:

first section ([email protected]) = 4.4x10x50/1000 = 2.2 volts

second section ([email protected]) = 4.4x7x80/1000 = 2.46 volts

third section ([email protected]) = 4.4x5x120/1000 = 2.64 volts

overall voltage = 2.2+2.46+2.64 = 7.3 volts.

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#### Plonker 3

Not quite, section one is 50 meters @ 22 amps. Section 2 30 meters @ 12 amps. Section 40 meters @ 5 amps.

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#### stuartgil

yellowvanman thanks for advice, here i go again:

first section ([email protected]) = 4.4x22x50/1000 = 4.84 volts

second section ([email protected]) = 4.4x12x30/1000 = 1.58 volts

third section ([email protected]) = 4.4x5x40/1000 = 0.88 volts

overall voltage = 4.84+1.58+0.88 = 7.3 volts

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#### Plonker 3

Thats it, do you understand what you have done now, and how you did it?

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#### stuartgil

can i just ask where we get the 12 amps from?

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#### Plonker 3

it is the sum of the current flowing down the second section of cable, you lose 10 amps at the first point. So 22amps-10 amps= 12 amps

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#### stuartgil

so the current halves to the second section and then drops again by the third section and would drop again the longer the run of cable, am i right?

#### trev

##### Regular EF Member
It should be pointed out that in the exam there will be points to be scored along the way in this type of question for the working out, sometimes even for writing out the formula. So show them all every step, you won't lose much time and it could make a difference

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#### Plonker 3

so the current halves to the second section and then drops again by the third section and would drop again the longer the run of cable, am i right?
Not halves no, you just lose the current used at each point along the circuit.

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