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Topic Title: Adding Currents
Topic Summary: Is it right to add currents as values?
Created On: 10 January 2013 02:39 PM
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 11 January 2013 02:58 PM
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zeeper

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r√(x^2+y^2 ) = √(14.1^2+13.9^2 )= 19.79 Amps


Its the 14.1 and 13.9 I cant find, I dont know j notation.

when I draw a phasor with an angle of 45 I get 14 as the result. how ever maybe its because I am treating the circuits as one.
 11 January 2013 02:59 PM
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Apostolos1983

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Oh my god, I do not know where should I begin!!
How can you strip a current from the phase angle?
Imagine two currents:
Ia = 10@0 degrees
Ib = 10@90 degrees
The sum is:
Ia+Ib= 10*sqrt(2) @ 45 degrees

So the magnitute of the emerging current is 14.1 and not 20 A as it would be from a simple add.
 11 January 2013 03:29 PM
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zeeper

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So the magnitute of the emerging current is 14.1 and not 20 A as it would be from a simple add.


Sorry Im more confused now, but thanks for trying.
 11 January 2013 03:53 PM
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Apostolos1983

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No, no, no this was not an answer to you!!!
It was an example for my friend OMG.
Sorry for the misunderstanding!
 11 January 2013 04:03 PM
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OMS

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When was the last time you had two motor loads with unity PF and effectively the other with total inductance and no resistance.

I accept in theory that if we want to add the currents you need to accoiunt for the power factors - as geoff demonstated above.

you aren't talking theory however,you were asking about motors - and the motors wouldn't be varying much between 0.6 and 0.8 lag - which Geoff also demonstrated would be almost inconsequential in terms of deriving the ampacity of the input feed to the motor MCC or switchboard.

I thought you were asking for the simple explanation as to why your engineer has suggested you just use the phase currents in order to initially size the feeder ?

regards

OMS

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Failure is always an option
 11 January 2013 04:08 PM
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GeoffBlackwell

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zeeper
This is not easy without diagrams.

The explanation needs to be tailored to the level of electrical principles that you have studied. I suspect that you have been introduced to power triangles [S (kVA); P (kW); Q (kVAr)] and the impedance triangle (Z; R ;X) were X might be Xl for inductance or Xc for capacitance.

Have you been shown how to add two currents that are at different phase angles ?

Let's try and open the lid on this (just a bit). I may be trying to tell you things you already know.

I assume that you will be familiar with representing points on a graph in rectangular form. This is when the position of the point is described by (x,y) co-ordinates. Polar co-ordinates are just a different way of doing this.

Consider a point described by the co-ordinates x = 7.07 and y = 7.07.

Plot this point on a graph.

Next draw a line from this point to the intersection between the x and y axis at zero. Now measure the length of this line and then measure the angle between the line and the x axis. You should get 10 for the length and 45 degrees for the angle.

This point then can be represented by a line of length 10 at an angle of 45 degrees to the x axis. So 10 @ 45 degrees are the Polar co-ordinates.

That is your starter for 10


Regards

Geoff Blackwell
 11 January 2013 04:33 PM
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zeeper

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phase currents in order to initially size the feeder


line currents .

Thanks Geoff

We did a lot of phasors at college but I forget.

So I can now add 10 + 10 at 90 degrees.

which is (10x10) + (10x10) = square route = 14.1

Some thing like that
 11 January 2013 04:38 PM
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OMS

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as in current per phase Zeeps

OMS

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Failure is always an option
 11 January 2013 04:45 PM
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Apostolos1983

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If you put it this way then you are absolutely correct. My example was purely hypothetical and theoritical.
 11 January 2013 05:16 PM
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zeeper

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sorry OMS

I was think motors where the phase current would be different depending on star or delta.

Last time I was at college everything was line current . except the star ,delta of transformers or motors which is phase current.
 11 January 2013 05:26 PM
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OMS

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For sure Zeeps - probably bad terminology but the current per phase in distribution will be the line current in effect.

do a google for Converting Polar to Rectangular coordinates and back As a starting point - the horizontal (X) axis is the real numbers and the vertical (y) axis is the J or imaginary numbers

There's a simple starter here

regards

OMS

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Failure is always an option
 11 January 2013 06:50 PM
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Legh

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Just to follow on from parsley,

Polar co-ordinate = magnitude line plus the angle from the reference (usually the hoizontal resistive component, displaying a cosine end result)
Rectangular co-ordinates = resistive and reactive components added by the use of pythagoras, the resultant line becomes the magnetude and the angle is measured between that line and the horizontal component (answer now given in polar notation)

Also at level 3. You are usually only shown phasor diagrams for series circuits which means the current stays constant and the voltage across components varies.

You need to start looking at parallel circuits where the voltage stays constant and the current through the components varies.
The best book for this level is Electrical Principles by John Whitfield then follow the link that OMS has given

Legh

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http://www.leghrichardson.co.uk

"Science has overcome time and space. Well, Harvey has overcome not only time and space - but any objections."
 11 January 2013 08:46 PM
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zeeper

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Thanks for the pointers
 12 January 2013 12:48 AM
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Frankmul

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I don't know if this will confuse things even more or if you will be able to down load this.
Apoligies for any errors in the excel file

http://www.wuala.com/frankmul/...3598,11-36,2193598,15

be nice now
 14 January 2013 09:16 AM
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zeeper

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Ok had a revelation in the shower last night. And managed to get me head around the numbers this morning. It help alot when I remember how to use the cos and sin buttons on me calculator lol. J notation is pretty straight forward once the concept is understood even for a level 3 sparkie LOL. I find it really annoying when I got a problem I cant work out. So thanks again.
IET » Wiring and the regulations » Adding Currents

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