Re: AHF efficiency



 2.5KW will be the loss in AHF,but to generate 50A @440V the energy drawn will be same?

ie output power(KVA) + AHF loss(KW) = Input power (KVA) so that KVA demand will increase.Is that correct assumption?

AHF draws fundamental current from system and produce harmonic current to remove harmonics that means for generating 50A harmonic current a 50A funda.current + (current for AHF loss) has to be drawn by AHF which will increase KVA consumption.

  My assumption is – to remove 50A harmonic current from the system AHF must draw more than 50A fundamental current.

Is it correct?