[Waverley ARS] Dummy Load calculations

[Adam Carmichael]

Hello List,

I'm reading into how to calculate the components for a 50 Ohm dummy load 
- and so I decided to test myself, unfortuneately I cannot grade myself! 
I was wondering if somebody would be kind enough to check what I'm doing 
is correct.

The question I gave myself:

I want to build a 50 Ohm, 1kW dummy load. What resistors are required?

So I set out to find out. 1W resistors would mean I would need 1000 
resistors, which is too much soldering! 5W resistors would mean a lot of 
soldering, but 200 resistors is still manageable; and so I continue.

Total Resistance required: 50 Ohms
Number of resistors: 200
Power dissipation per resistor: 5W

Remembering the equation for parallel resistances of equal resistors:

R(total) = R1 / (number of resistors)
50 Ohms = R1 / 200

Or, rearranged:

50 * 200 = R1
          = 10 000 Ohms

Therefore: 200 * 10K Ohm 5W resistors are required to build a 1kW dummy 
load.



Thanks in advance,

Adam

-- 
Adam "carneeki" Carmichael - VK2FNRD
p: +61415371990
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