IET
 Topic Title: Reactive amps Topic Summary: Calculation Created On: 24 March 2010 10:47 AM Status: Read Only Linear : Threading : Single : Branch
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 24 March 2010 10:47 AM deleted_1_Bikerz 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? Cheers Sheldon 24 March 2010 01:02 PM Robo Posts: 15 Joined: 07 October 2004 Sheldon, 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 eswnl 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.
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