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I have a power supply from a broken 3d printer. I want to test if it is still okay. It is specified as 24V, 15A (360 W). It is giving me perfect 24V at idle, but I wanted to test if it is still performing well under (almost) full load.

But what could I use to dissipate 360W at 24V? And: my multimeter is only specified for 10 A, so I will have to use something as a shunt resistor...

I'd appreciate any suggestions avoiding expensive equipment.

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To test if your 24-V power supply can deliver 15 A of current, you need to get or make a massive load resistor with value of 1.6 Ohm. Better make it 1.5 Ohms, since every power supply must have some margin.

You can make it out of piece high-resistance wire, or maybe out of a spool of household iron wire. Or you can get 10 pieces of massive 50-W 15-Ohms resistors in parallel (you need your load to hold about 340 W). Many variants exist. Connect the load with thick copper wires (12-14 AWG), and see if your PSU still delivers 24 V.

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  • \$\begingroup\$ Sounds a little frightening. Picture myself as Dr. Frankenstein. For a moment I thought about just connecting it to the radiator ;-) But yeah, that's the problem, at 24 V there is almost no existing load device other than plain wire, that is able to heat 360W. \$\endgroup\$ – oliver Jul 14 '18 at 17:53
  • \$\begingroup\$ @oliver, not really frightening. Some heater element for a kitchen stove might work as well. You just need to make sure that your DMM is accurate enough to measure 1.5 Ohms, which is not usual. Better make 15 wires of 15-Ohms each, and connect them in parallel. \$\endgroup\$ – Ale..chenski Jul 14 '18 at 18:00
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    \$\begingroup\$ @oliver, and if you value your time, a 15-0hm 50W 1% wirewound resistor costs about $4, so $40 will do your load. Or a single 1.5-Ohm for about the same price. Or 3 of those from a surplus store, mpja.com/1-Ohm-120W-Power-Resistor/productinfo/17785+RS \$\endgroup\$ – Ale..chenski Jul 14 '18 at 18:25
  • \$\begingroup\$ Ah, yes, I also found some power resistors on ebay for an even cheaper price, and packaged in an aluminum heat sink. I think I will buy those. Thank you! \$\endgroup\$ – oliver Jul 14 '18 at 18:45
  • \$\begingroup\$ For $40 you can purchase a new 15A 24V Supply \$\endgroup\$ – crasic Jul 15 '18 at 0:44
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I see this question often. Here is my common answer.

Use a 5 gal bucket (load dependent) and add salt or baking soda to the water till you get the current you want.

For smaller loads such as your 360W or higher voltages, you can just hang 2 wires in the bucket. For bigger loads, I use 1/2in copper pipe and attach my wires to it with electrical tape.

Note:

  1. Your water will undergo electrolysis, and generate oxygen and hydrogen (not an issue). The salt will create Cl and Na, this can be more of an issue for high power loads in a small space.

  2. The surface area of your conductor in the water and the distance apart will affect current. So You may want to use wood etc. to fix your conductors in place. 2x4s with holes in them on the top and bottom of the bucket work well.

  3. Ensure you have a safe distance between you and the bucket with high voltages. This ensures any spilt water etc. does not connect you to the circuit.

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    \$\begingroup\$ Very interesting idea. Did you do a rough estimation whether at saturation concentration of NaCl in water it is possible to reach a conductivity that allows 15 Amps? \$\endgroup\$ – oliver Jul 15 '18 at 5:07
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    \$\begingroup\$ I have elaborated this a little further. en.wikipedia.org/wiki/Conductivity_(electrolytic) suggests a conductivity of 5 S/m for sea water. With electrodes of area 10cm x 10cm, at 10cm distance apart, I am actually getting R=2 Ohm, so it might work with metal bars instead of wires. Warning to anyone who might try this: be prepared to get large amounts of chlorine and Knallgas. \$\endgroup\$ – oliver Jul 15 '18 at 5:23
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    \$\begingroup\$ So I have done this for a 120v 1500W load. It worked well. I ended up using baking soad instead of salt. \$\endgroup\$ – MadHatter Jul 15 '18 at 5:24
  • \$\begingroup\$ Amazing! I love mad science. :-) \$\endgroup\$ – oliver Jul 15 '18 at 5:25

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