For a project I need to switch on a simple handheld vacuum cleaner with an Arduino. I am thinking of using a relay or mosfet. I opened the case and a 7.4 V 2200 mAh Li-Ion battery powers the motor, through a regular switch.

On the manufacturer's website, the power of this vacuum cleaner is 120 W, so according to my understanding:

I = P/V

I = 120 W / 7.4 V = 16.2 A

So according Ohm's law, the resistance of the motor has to be 0.46 Ohm right? Isn't this too low?


  • \$\begingroup\$ Isn't this too low? Too low for what? \$\endgroup\$
    – Andy aka
    Commented Apr 30, 2019 at 12:12
  • \$\begingroup\$ We need to take into consideration the inductance of the motor also. \$\endgroup\$
    – 19aksh
    Commented Apr 30, 2019 at 12:13
  • \$\begingroup\$ A brushed DC motor will have a much lower resistance at rest than you would expect from the current it draws when running. This is due to the back-EMF. That's why they draw a big surge when you first apply power. \$\endgroup\$ Commented Apr 30, 2019 at 12:17
  • \$\begingroup\$ @SpehroPefhany What do you mean by "drawing"? Does the motor lower it's resistance under load, so that a higher current can flow? \$\endgroup\$
    – Verman
    Commented Apr 30, 2019 at 12:22
  • \$\begingroup\$ @Verjan Yes, higher load (higher torque) requires more current (power), so the apparent resistance of a motor (due to back-EMF) changes with varying load. \$\endgroup\$
    – JimmyB
    Commented Apr 30, 2019 at 12:26

1 Answer 1


The calculations you did are right. But, the value is the peak power. Your motor will draw power according to the motor controller setting and load. Keep in mind when the motor is not moving, its resistance is equal to the wire resistance which can be less than 0.1 ohms; essentially a short circuit. That is the reason for the initial "spike". When it starts rotating the back emf comes into play and changes the effective resistance which decreases the power draw.


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