I could be wrong, but I don't think a 1% resistor necessarily means the value is within +/- 1%.  I think there are some statistics involved which might say, for example, that the standard deviation is 1%, meaning there are about 40% of the resistors in a particular lot that would fall outside the 1% value and a few percent even falling outside the 2% value.

And, of course, David is right that the error (or at least some of the error) could be coming from the measuring device itself.  As they say, if you have one clock, you know what time it is.  If you have multiple clocks, you have no idea...

72,
John AA0VE

On 05/11/2015 02:20 PM, David Martin davemrtn@... [4sqrp] wrote:

 

Yes but, as you can see from his photo, he paid for a 1% resistor !!

For $11 (+ shipping), I would be calling the vendor.

OR MAYBE, the error is in the device he is using to measure the resistance ??

On 5/11/2015 2:05 PM, Wayne Dillon wayne.dillon@... [4sqrp] wrote:

 

Hi Darryl

Should be 2K2, this equates to 50.0355 Ohms total effective resistance. Not sure on power as the 51.2 is only about 3% of the 2K2. Anyway why worry about the error? the error is less that 3% and that's better than most commercial dummy loads.

Just my $0.02 worth.

Be blessed

Wayne - KC0PMH

On Mon, May 11, 2015 at 1:49 PM, kk5ib01@... [4sqrp] <4sqrp@...> wrote:

 

This is my new low cost dummy load, made with a Caddock thick film non-inductive resistor. This one is 50 ohms at 1%, and rated at 100 watts, and costs about $11.00 from Mouser or Digikey. I realize the current heat sink will not be adequate for 100 watts, but it will suffice for the 5 and 50 watt calibrations of my K3, it does get warm to the touch. Only problem is resistance measures 51.2 ohms. What value of a parallel resistor would bring that down to 50 ohms? I calculate 1234.56 ohms or 1.2 K. Is that right? Also what would the power rating of the resistor need to be? Thanks.

Darryl, KK5IB

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"It's only "work" if you'd rather be doing something else" - Dean Kamen