0
\$\begingroup\$

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?

\$\endgroup\$
7
  • 4
    \$\begingroup\$ 100kOhm is small ? \$\endgroup\$
    – Eugene Sh.
    Commented Aug 25, 2017 at 16:12
  • 2
    \$\begingroup\$ If you use a ONE milli-ohm resistor you can measure 15mV at 15A at just under 1/4W. \$\endgroup\$
    – Trevor_G
    Commented Aug 25, 2017 at 16:19
  • 1
    \$\begingroup\$ Your problem isn't that the resistor is too small size wise, but that the resistor is too large resistance value wise. As Trevor said, you want a 0.001 ohm resistor max if you want to measure 15A without going over the wattage rating. Remember your other calculation V=I*R. Using that equation you should quickly see why you don't want to use 100,000 ohms in a circuit that can pull 15A. \$\endgroup\$
    – I. Wolfe
    Commented Aug 25, 2017 at 16:25
  • \$\begingroup\$ @EugeneSh. by "small" I am referring to the size of the resistor. I read that it is the size of the element that determines its ability to dissipate power. \$\endgroup\$
    – user3646493
    Commented Aug 25, 2017 at 16:31
  • 1
    \$\begingroup\$ NO... The only way you can force a resistor to carry more current is to increase the voltage across it.. which in turn makes it dissipate more power and get hotter. For your application you need a resistor with a very small resistance so as to a) not reduce the voltage to the motor appreciably, and b) stay cool enough that it's resistance remains close to the indicated value.. \$\endgroup\$
    – Trevor_G
    Commented Aug 25, 2017 at 16:52

1 Answer 1

Reset to default
2
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.