Decrease font size
Increase font size
Topic Title: Reactive amps
Topic Summary: Calculation
Created On: 24 March 2010 10:47 AM
Status: Read Only
Linear : Threading : Single : Branch
Search Topic Search Topic
Topic Tools Topic Tools
View similar topics View similar topics
View topic in raw text format. Print this topic.
 24 March 2010 10:47 AM
User is offline View Users Profile Print this message


Posts: 14
Joined: 17 April 2009

Hi all

Im using Sin Acos Power factor x full Load current to give me the reactive amps.

PF 0.9
Full Load 92 Amps

So Sin 0.9 = 0.436

0.436 x 92 = 40.1 Reactive amps

Is this correct? If so can someone explain in more detail what the Sin PF is all about?


 24 March 2010 01:02 PM
User is offline View Users Profile Print this message


Posts: 15
Joined: 07 October 2004


It's plain old trigonometry, more use to working with this in powers (i.e. W, VA and VAr) and the power triangle, however.

PF is the cosine of the angle between VA and W. The 'ACos' of this will give you the angle in degrees. Taking the sine of this angle and multiplying it by the hypotenuse ('full load' in your case) will give you the magnitude of the opposite side, or in the case of the power triangle the reactive portion.

Your result looks OK but never really had to work out reactive amps before.

Hope that helps.
 24 March 2010 07:39 PM
User is offline View Users Profile Print this message


Posts: 144
Joined: 29 November 2008

I think you mean Sin(25.84 degrees) = 0.436,

as PF=Cos(?) = 0.9, where ? is the angle. So it is not Sin PF.
IET » Energy » Reactive amps

Topic Tools Topic Tools

See Also:

FuseTalk Standard Edition v3.2 - © 1999-2016 FuseTalk Inc. All rights reserved.