# How to calculate suitable dummy load for a power-supply?

A beginner question - feel free to shoot me down

I'm looking to modify a retired AT SMPS to drive my ham-radio. This is to be achieved by tying all the +12V wires together so the maximum amperage may be drawn from it.

The SMPS lay unused for nearly 10 years now. Reading on the internet indicates the SMPS may burn itself out if powered up without a load. To test whether the SMPS is alive, I plan to put a dummy 10ohm 10W load across one molex connector.

$12V \times 8A = 96W$ so I guess it should be good for about 80-85W

For $10 \Omega$ $10W$ load:

$10W = I^2 \times 10 \Omega$

$\dfrac{10W}{10\Omega}=I^2$

$I^2 = 1 \frac{W}{\Omega}$

$I = \sqrt{1 \frac{W}{\Omega}} = 1A$

So the $10 \Omega$ resistor can handle $1A$ current.

In reality though

$V = I \times R$

$12V = I \times 10\Omega$

$I = \dfrac{12V}{10\Omega} = 1.2A$

$P = I^2 \times R = (1.2A)^2 \times 10\Omega = 14.4W$, or about 44% overload

What I need to know here

• Are my calculations above in order?
• If yes, should I get a higher rated load, say 20W (if such a power-rating is available)?

EDIT: Just a short note to convey I finally got around to verification using a halogen car bulb rated at 110W (+: Thanks a tonne, all!

• 12V @ 10 or 20W is called a (halogen) light bulb. Don't make it more difficult than it is and enjoy the useful 'free' desk light. – jippie Apr 18 '12 at 19:12

# re: Halogen light bulb

Since I'd like to illustrate with a picture, I'm writing this as an answer rather than comment. This is a response @jippie to his comment to the O.P., which suggests using a light bulb instead of a resistor.

Here's a caveat to be aware of. Resistance of the filament of the light bulb will go up with temperature. By design, the filament will get quite hot. This non-linearity is illustrated by this plot:

For more discussion on light bulb resistance, see pp.21-25 here.

# re: O.P.

Calculations in the O.P. are correct.