[Elecraft] KPA-500 110 or 220?

[Adrian]

E=IR in a circuit. To determine volt drop in a circuit it is a simple
circuit resistance multiplied by the current. Simple ohms law and nothing
squared.
Bob refers to IR loss in the circuit feeding the amp as per above line and
not meter efficiency equations.

Circuit;  1 ohm x 20 amp = 20v volt drop  (120v)

To halve current we doubled the voltage and reduced load by half with the
same circuit(tap change etc).

Circuit;  1 ohm x 10 amp = 10v volt drop  (240v)

Looks like a 50% loss(read reduction in volt drop) with current halved not
75%

As far as power improvement(You see I removed the voltdrop at load end
equation);

100 x 20  = 2000w

230 x 10 = 2300w

So a 300w or 15% power improvement, and a much higher IMD improvement no
doubt.

-----Original Message-----
From: elecraft-bounces at mailman.qth.net
[mailto:elecraft-bounces at mailman.qth.net] On Behalf Of Monty Shultes
Sent: Tuesday, 7 August 2012 5:00 AM
To: elecraft at mailman.qth.net
Subject: Re: [Elecraft] KPA-500 110 or 220?

Actually it's "I-squared R drop", and a 1/2 reduction in current results in
a 75% loss reduction.
Monty K2DLJ

On Aug 6, 2012, at 1:27 PM, Bob <WB4SON at gmail.com> wrote:

> Electric meters measure current consumption on both 120v legs when 
> calculating total watt hour consumption.  So it matters not one bit if 
> the loads are balanced or unbalanced from a billing point of view.  
> Rumors to the contrary are simply not correct.
> 
> The reason for running the amplifier at a higher voltage is that it 
> will reduce the IR voltage drop.  Because the operating voltage has 
> doubled, the required current has been cut in half.  Therefore the IR 
> drop (loss in the
> wiring) has been cut in half.
> 
> 73, Bob, WB4SON
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