Here's the small resistor
Why: I need to measure current drawn by a large motor (from electric lawnmower), but my multimeter has a max rating of 10Amps. Workaround is to attach a known resistor in series and measure the voltage drop across the resistor, then use ohms law to solve for current in the circuit.
This resistor has a power rating of 250mW. Based on other lawnmower motors, I am assuming this is approx a 350W motor, and with a 24V source, the current will be around 15Amps (using P=IV).
Question: If I place such a resistor into the circuit, will the resistor blow up, or will it SAFELY dissipate whatever small amount of energy it is capable of absorbing and let the motor take the rest?
Best Answer
Based on Ohms law the maximum current going through a 100 kOhm resistor at 24 V is: I = U / R = 24 V / 100 kOhm = 0,24 mA.
Your motor will never run if you put that huge resistor in there.
And a resistor has no knowledge on how much power it is able to dissipate and intelligently let more or less current pass. So it is the duty of the one designing the circuit to make sure it will be able to handle the typical power dissipation expected from the design. And probably include some safety margin for stuff the designer did not expect.
Current sense resistors should be chosen in a way, that the influence on the circuit is small and have a power rating so they won't burn down.
Now to sense 15 A, and don't influence the circuit much, the voltage drop should be lower than 1 V, probably a lot lower, like 150 mV or so.
So the resistor would have R = U / I = 150 mV / 15 A = 10 mOhm.
The power dissipation of that resistor would be P = U*I = 150 mV * 15 A = 2.25 W.
Measuring motor current might have some difficulties with induced noise and stuff, but I don't have any experience in that field. The phase currents might also be higher than the average you calculated there - but again not sure on that.