If you couldn't guess, I'm new to electronics.

So, I've just received my 3V air pump, it didn't come with any specs, so I'm trying to figure everything out. I hooked it up to a DC variable power supply (at 3V) and a multimeter.

I switched the multimeter to 10A non-fused and placed the lead there as well, pinned a wire from the power supply ground to the pump, and used the multimeter as the bridge between the positive terminal of the supply and the other side of the pump.

I got a reading of 0.17 under normal conditions and 0.21 under full load.

Now, here's the question, 0.17-0.21, is that amps? Does that mean I am pulling 170-210 milliamps? When I toss these numbers into Ohm's law, I get 17.64706 Ohms resistance. Does this seem reasonable?

Since I'm just learning this, I just want to make sure this is all correct. If I am wrong, I would love to understand more!

  • \$\begingroup\$ By "variable power generator" do you mean "variable power supply?" I'm guessing it's not actually a gas-powered generator of some kind. \$\endgroup\$
    – JYelton
    Commented May 9, 2014 at 17:17
  • \$\begingroup\$ Good point, sorry about that! Yes a Variable Power Supply \$\endgroup\$
    – ntgCleaner
    Commented May 9, 2014 at 17:22
  • \$\begingroup\$ And remember you're dealing with the real world, that 17.64706 has way too many figures ;) \$\endgroup\$ Commented May 9, 2014 at 19:14
  • \$\begingroup\$ haha, good call. What is acceptable? Should I have any points after the decimal? \$\endgroup\$
    – ntgCleaner
    Commented May 9, 2014 at 22:42

1 Answer 1


On the 10A range, your multimeter probably won't have great accuracy with low current, but your measurement sounds reasonable. 0.17 to 0.21 A, or 170 to 210 mA.

Presumably your air pump is a small DC motor, which will have different current measurements at startup, and during running conditions with different loads.

To calculate resistance, you use Ohm's law:

$$R = E / I$$

If the voltage (E) is 3, and the current (I) is 0.17 A, then:

$$\frac{3}{0.17} = 17.6\Omega$$

Therefore, your resistance calculation seems correct. Just keep in mind that a DC motor will measure different resistances (currents) depending on what it is doing, due to induction/reactance.

  • \$\begingroup\$ Thank you for this great answer. I know it probably seems simple to most, but for someone just starting out, it's a spider web of information. And yes, that motor runs at a constant .17A when there is nothing connected to it and it runs at .21A when it's completely plugged, so I figure that's a good range to look at \$\endgroup\$
    – ntgCleaner
    Commented May 9, 2014 at 17:28
  • \$\begingroup\$ Starting out can be daunting and complicated! So you ask good questions, and that's how you learn. Glad to have helped. \$\endgroup\$
    – JYelton
    Commented May 9, 2014 at 17:40
  • \$\begingroup\$ Quick question - if I wanted to step down the voltage supplied to the motor, Would I use the higher range Amperage? Should I do my math on the .21A or the .17A? I plan on having a 6V power source, but I will need to step it down 3V, which means 3V difference, I would need to use a 15ohm resistor (at .21A)? (Or is there a better way?) \$\endgroup\$
    – ntgCleaner
    Commented May 9, 2014 at 18:44
  • \$\begingroup\$ You should use a voltage regulator, a resistor might not suit you because when the motor starts it draws a lot of current so the voltage might drop too much and it won't start at all. \$\endgroup\$ Commented May 9, 2014 at 19:13
  • \$\begingroup\$ A voltage regulator like this for example m.linear.com/product/LT1763 with the 3V output of course. \$\endgroup\$
    – RawBean
    Commented May 9, 2014 at 20:43

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.