25 ohm at 10 ohms, 4 ohms at 25 watts. according to bobwind if i am measuring a 50 amp source i will need atleast a 600 ohm resistor, is this correct anyone. tia.
ps, i have considered getting an inductive pickup but thought there may be a less expensive method. again tia
2006-08-09 05:27:08 · 5 answers · asked by billfoxaccessories 1 in Cars & Transportation ➔ Other - Cars & Transportation
I had to read your other questions to know what you are talking about. I assume you want to measure current over 10 amps in a 12V circuit, right? The original answer that suggested a 10 ohm resistor is way off. You will never get 10 amps through a 10 ohm resistor on 12V. Some have suggested a clamp on ammeter. This will not work for DC.
I don't have a good practical solution. If you can find the right resistor in series, measuring the voltage will tell you the current. This is limited by the precision of the resistor. If you had a 0.1 ohm resistor, your 10A will cause a voltage drop of 1V. If that isn't acceptable, you'll need a smaller resistor. That 0.1 ohm resistor would be dissipating 10W.
A more basic question for you. What do power windows have to do with the ignition circuit of your car?
2006-08-09 09:04:48 · answer #1 · answered by An electrical engineer 5 · 0⤊ 0⤋
25 ohms at 10 ohms? That's totally incorrect. 4 ohms at 25 watts is correct. 4 ohms is the resistance of the component, and 25 watts is the amount of power it can safely dissipate (operate at) I don't know who or what bobwind is, and without knowing what it is you want to measure at your 50 amp source, and whether or not the source is AC or DC, I have no idea why a 600 ohm resistor was recommended.
I'm a retired electrical engineer with 2 PhD's. If you'd care to email to me exactly what it is you're trying to accomplish, I might be able to help you. dickn2000@yahoo.com
2006-08-09 05:39:36 · answer #2 · answered by Anonymous · 0⤊ 0⤋
good question. The wattage is the max capability the resistor can tolerate without burning. The formulation for capability is going: P = V x I, or P = v^2 / R, or P = I^2 x R. Given: max P = .5 watt, and R = 2400 ohms use formulation v^2 = P x R then v^2 = .5 x 2400 = 1200 then take sq. root v = 34.641 volts
2016-12-11 05:46:22 · answer #3 · answered by ? 4 · 0⤊ 0⤋
I think your question needs editing.
25 ohm at 10 ohm ???
Get a variable resistor and adjust to suit your application.
2006-08-09 05:35:20 · answer #4 · answered by echiasso 3 · 0⤊ 0⤋
Depend on the wattage of the resistor. And a source simply has a maximum amperage. The voltage across the resistance determines the amperage flow. E=IR P=IE Just do the calc....
Actually the internal resistance, is the real determiner, but that's another calculation.
2006-08-09 05:37:44 · answer #5 · answered by blackfangz 4 · 0⤊ 0⤋