[Boatanchors] Ohms Law

[jmfranke]

No, your first analysis was correct.  Something else is wrong.  The first 
thing to do is measure the voltage across the 15 Ohm resistor.

John  WA4WDL
--------------------------------------------------
From: "Barrie Smith" <barrie at centric.net>
Sent: Sunday, April 18, 2010 6:40 PM
To: "Boatanchors Mail List" <boatanchors at mailman.qth.net>
Subject: [Boatanchors] Ohms Law

> I've been juggling 5 volt transformers all day, trying to find one that 
> would actually produce close to 5 volts for a pair of tubes.
>
> They were all either too high in voltage, or too low.  One did give me 
> almost exactly 5 volts, but it ran way too hot.
>
> So I've selected one with plenty of current capacity and am playing with 
> resistance in the primary to get the secondary voltage down to 5 volts.
>
> Using a 25 ohm, 150 watt rheostat, I find that 11 ohms gets me real close. 
> The rheostat runs quite warm, which leads me to my question.
>
> The voltage drop across the rheostat is a measured 13 volts.  Knowing that 
> and the resistance, Ohm's law tells me that I have 1.18 amps running 
> through, which equals 15 watts.
>
> I don't want to install the rheostat in the transmitter because it is 
> quite large, so I thought I'd substitute a fixed resistor.
>
> I tried a 15 ohm, 30 watt fixed resistor.  It got very, very hot real 
> quick!
>
> I must be making a mistake.  I thought I should use the "dropped voltage" 
> in Ohm's law.  Perhaps I must use the full 115 volts, instead.
>
> The amount of heat I'm producing indicates that is true.
>
> Thoughts?
>
> 73, Barrie, W7ALW
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