| Gallery, Projects and General > How do I?? |
| Measuring DC current >10a? |
| (1/4) > >> |
| loply:
Hi folks, I need to measure a DC current which I know is more than 10a, I'm thinking somewhere between 25 and 30a. My multimeter only has a 10a range and I presume it will either break it, or just not work, if I try. What's normal here? Do I just need a better multimeter? Or use one of those 'clamp' type tools I see sometimes? Thanks for any tips, Rich |
| Bluechip:
IF the circuit allows, put a 0.1 ohm alloy clad 10W or so resistor in series with the load, and measure the voltage across said resistor. EG: http://cpc.farnell.com/welwyn/wh5-0r1-ji/resistor-10w-5-0r1/dp/RE05393 So: 22.5A = 2.25V Yes, get a better meter ... :thumbup: Donkeys years ago I had a 50A shunt for my AVO8, lent it to some miserable snivelling scrote and never got it back. :bang: Dave EDIT BTW a lot of those clamp meters ONLY do AC look at the spec first :thumbup: |
| awemawson:
You use a 'shunt', which is a low value resistance in parallel with your meter. Find something that draws pretty much your 10 amp FSD (full scale deflection). Parallel up low value resistances with your meter until it displays exactly half what it did before, and you now have a 20 amp fsd meter. Make it read 1/3rd and you have a 30 amp fsd meter. As your meter will (or should) have a very low resistance on the 10 amp range, probably a length of iron wire a few inches long will be of the right sort of resistance and you can adjust its value by trimming it's length. |
| philf:
--- Quote from: Bluechip on February 03, 2015, 10:12:33 AM ---IF the circuit allows, put a 0.1 ohm alloy clad 10W or so resistor in series with the load, and measure the voltage across said resistor. EG: http://cpc.farnell.com/welwyn/wh5-0r1-ji/resistor-10w-5-0r1/dp/RE05393 So: 22.5A = 2.25V --- End quote --- A 10W resistor would very soon die. If I remember correctly W=IČR so 22.5 * 22.5 * 0.1 = 57.375 Watts! Edit - Dave's right - the power should be 50.625 Watts! I multiplied 22.5 by 25.5 :( Phil Phil. |
| Bluechip:
Whoops .... :scratch: You're absolutely right Phil. Probably explains why I occasionally re-invent the Direct Arc Furnace .... What I thought of was 0R01, then a delivery arrived and I lost track a bit , so for some reason I decided it was 0r1 and hence screwed it up ... :loco: It should have been one of these: http://uk.farnell.com/welwyn/wh25-0r01ji/resistor-25w-5-0r01/dp/9507159 in which case 22.5A would read as .225V . and the dissipation is 5.06W Sorreeeeeee .... Brain Fart .... BTW I think your 57 .375 watts is adrift :lol: I get 50.625W. Dave Dave |
| Navigation |
| Message Index |
| Next page |