dumy load question

[gcb]

as you will get from this question, i'm not knoleadgeble in eletronics, but i'm trying to test a few 17V battery and need to attach a 2A load to them

I = V/R, 2A = 17V/R, R=8.5

if i add a 8.5Ohm resistor to the battery, will that generate the load I want? will the resistor/battery burn/explode/kill my first born? I have to run it for several minutes...

i only have cheap hobist resistors for china btw. I can salvage components from a highend receiver power supply that i have somewhere.

 

[(*steve*)]

17V battery sounds like an unusual voltage, but hey...

Yes, an 8.5 ohm resistor will cause a current of 2A to flow from a 17V source.

the power dissipated will be V^2/R = 17*17/8.5 = 34 Watts.

I would use a 50W resistor.

Now 8.5 ohms is not a standard value (but 8.2 ohms is, and that's probably close enough.

Let's assume you have 1/4W resistors, and that you have many 8.2 ohm 1/4W resistors.

To dissipate 50W, you need 200 of them.

If you connect 14 8.2 ohm 1/4W resistors in series, and make 14 of these strings, and then connect them all in parallel, you will have a resistor of 8.2 ohms capable of dissipating close to 50W

Beware, because all those resistors floating around can short out easily.

However, I bet you don't have 196 8.2ohm 1/4W resistors.

 

[gcb]

17V battery sounds like an unusual voltage, but hey...

Yes, an 8.5 ohm resistor will cause a current of 2A to flow from a 17V source.

the power dissipated will be V^2/R = 17*17/8.5 = 34 Watts.

I would use a 50W resistor.

Now 8.5 ohms is not a standard value (but 8.2 ohms is, and that's probably close enough.

Let's assume you have 1/4W resistors, and that you have many 8.2 ohm 1/4W resistors.

To dissipate 50W, you need 200 of them.

If you connect 14 8.2 ohm 1/4W resistors in series, and make 14 of these strings, and then connect them all in parallel, you will have a resistor of 8.2 ohms capable of dissipating close to 50W

Beware, because all those resistors floating around can short out easily.

However, I bet you don't have 196 8.2ohm 1/4W resistors.

ah, the power dissipation rating. i knew i was forgeting something

Is that indicated anywhere on the resistor? i have a few bulky ones on some component bin... they only have 4 bands. and a green (teal?) body instead of the yellowish (ocre) one. but are way thicker then the 1/4W ones...

 

[Brianj_92505]

Another cheap alternative would be an old electric heater that you don’t mind scrapping. The ni-chrome wire that the heating coil is made from usually handles 10 to 15 amps just cut off 2 ohms worth or uses a heavy jumper wire with an alligator clip to make it a variable resistor. Hook up a meter and adjust it to 2amps. It will get hot as would any component dissipating that much power, don’t go burning yourself. I used to like to place ¼ watt 1 ohm resistors across a 9v battery and watch them smoke and burn. It fascinated me because shorting it with a wire (or tongue) had no effect.

 

[(*steve*)]

The rated maximum power dissipation is gnerally determined by:
a) the material
b) the surface area
c) the allowable temperature rise

In general, (a) and (c) go together, so within a type of resistor, the surface area is a reasonable predictor of power rating.

If you can calculate the ratio of length and circumference then (assuming they're the same composition) you can have a pretty good guess at the rated power dissipation.

But, a 50W resistor is likely to be a large wirewound resistor in a metal jacket, and will likely have tabs to bolt it on to a heatsink.

 

[gcb]

Another cheap alternative would be an old electric heater that you don’t mind scrapping. The ni-chrome wire that the heating coil is made from usually handles 10 to 15 amps just cut off 2 ohms worth or uses a heavy jumper wire with an alligator clip to make it a variable resistor. Hook up a meter and adjust it to 2amps. It will get hot as would any component dissipating that much power, don’t go burning yourself. I used to like to place ¼ watt 1 ohm resistors across a 9v battery and watch them smoke and burn. It fascinated me because shorting it with a wire (or tongue) had no effect.

i think the heating resistence is a very good idea. can i cut two 16 Ohm pieces and put in parallel for even better heat dissipation and still get the 2A?


Also i tried to find the bolt-on huge resistors Steve talk about, just to see if buying one would be a better idea, but i can only find 10ohm upwards for 25W+... price is OK... <$10.

 

[(*steve*)]

i think the heating resistence is a very good idea. can i cut two 16 Ohm pieces and put in parallel for even better heat dissipation and still get the 2A?

Sure.

It's the total resistance that's important (assuming you have the power handling capacity)

If you place n identical resistors in parallel the resulting resistance is 1/n of the resistance of each resistor and the power handling capacity is n times the individual power rating.

For safety it's generally a good idea to be generous with power ratings.

 

[BobK]

12V auto lamps make pretty good power resistors. Use at least 2 in series for a 17V source.

Bob

 

[gcb]

12V auto lamps make pretty good power resistors. Use at least 2 in series for a 17V source.

Bob

nice. i will try that, as it seems safer than hooking jumper wires on portions of a waffle machine heating element.

thanks again everyone!

 

[gcb]

just a closure to the thread. I used a waffle iron... it was perfect for what i was doing (seeing how long a 14v battery would drop bellow 12v.

the waffle iron circuit is AC plug = a thermistor thing that probably does nothing in low temps = 12 Ohm resistence

and it has a battery operated kitchen timer on the base.

i got a sweet .9999A when connected to my battery, and the timer was incredible handy

only bad news is that the battery was dead. got to 13V in under 5min. 12.0 V under 10min. and it was rated 3Ah

thanks again for the knowledge and suggestions.