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Re: Wire-wound resistors as dummy test load



Original poster: "Edward Wingate by way of Terry Fritz <twftesla-at-qwest-dot-net>" <ewing7-at-rochester.rr-dot-com>

Tesla list wrote:
> 
> Original poster: "David Speck by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<dave-at-davidspeckmd-dot-org>
> 
> Ed,
> I think we are speaking different languages, but saying the same thing.
> 
> If the 4 resistors are in series, they will total 800 ohms.  Passing 1
> amp through the string will dissipate a total of 800 watts, 200 watts in
> each resistor, with a voltage drop of 200 volts across each resistor.
>  Power (watts) = (I ^ 2) * R. = V * I
> 
> If the four resistors are in parallel, they will have a combined
> resistance of 50 ohms.  Putting 200 volts at 4 amps across the parallel
> set will still dissipate 200 watts in each resistor, or a total, again,
> of 800 watts.
> 
> When you series resistors of differing wattages, the current of the
> chain is determined by the power dissipation ability of the smallest
> wattage rated resistor in the chain.  Whatever level of current makes
> the maximum dissipation in the smallest rated resistor will be the
> maximum current the chain can sustain without burning out the smallest
> resistor.
> 
> Dave

Dave,

You are correct!

I sure didn't state the case very well and should have been more
specfic.

Ed Wingate RATCB