
Electrical Engineer
Premium job
 Portchester, Fareham
Responsible for the maintenance and repair of electrical & electronic equipment systems
 Recruiter: Fugro GB Marine Ltd

Engineering Maintenance Training Officer
Premium job
 Oxford and the South West
Responsible for providing a high quality training and support service for a caseload of apprentices.
 Recruiter: JTL

Manager, Network Output Assessment
 Glasgow or London Westminster (moving to Canary Wharf in early 2018)
 £29,200  £35,795 (Glasgow) or £31,500  £38,600 (London)
You’ll need to demonstrate a track record in the analysis of complex data and information
 Recruiter: Ofgem
IET 
search :
help :
home


Latest News:


Topic Title: Adding Currents Topic Summary: Is it right to add currents as values? Created On: 10 January 2013 02:39 PM Status: Read Only 
Linear : Threading : Single : Branch 

Search Topic

Topic Tools

11 January 2013 02:58 PM


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


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


Sorry Im more confused now, but thanks for trying. 



11 January 2013 03:53 PM


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


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  Let the wind blow you, across a big floor. 



11 January 2013 04:08 PM


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) coordinates. Polar coordinates are just a different way of doing this. Consider a point described by the coordinates 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 coordinates. That is your starter for 10 Regards Geoff Blackwell 



11 January 2013 04:33 PM


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


as in current per phase Zeeps
OMS  Let the wind blow you, across a big floor. 



11 January 2013 04:45 PM


If you put it this way then you are absolutely correct. My example was purely hypothetical and theoritical.




11 January 2013 05:16 PM


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


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  Let the wind blow you, across a big floor. 



11 January 2013 06:50 PM


Just to follow on from parsley,
Polar coordinate = magnitude line plus the angle from the reference (usually the hoizontal resistive component, displaying a cosine end result) Rectangular coordinates = 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  http://www.leghrichardson.co.uk deavatared 



11 January 2013 08:46 PM


Thanks for the pointers




12 January 2013 12:48 AM


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/...3598,1136,2193598,15 be nice now 



14 January 2013 09:16 AM


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

Topic Tools

New here?
See Also:
FuseTalk Standard Edition v3.2  © 19992017 FuseTalk Inc. All rights reserved.