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I recently bought a 24V 250W DC motor. In order to test it, I connected it to a DC power supply (rated +-25V / 1A). I set the PSU to 24V and connected the motor (no load) to it. The current immediately reached its cap of 1A and the motor voltage raised and capped at 9.3 V.

I suspect that the current capacity had something to do with it because the motor did not get its ideal starting current (~10A). Is this normal behavior of motors (that the voltage is dependent on the amount of current I can supply to it)? If so, why?

This will help me determine the battery I will use for my motor since batteries have recommended currents.

Any help would be appreciated. Thank you.

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    \$\begingroup\$ Are you serious? \$\endgroup\$ – Harry Svensson Nov 11 '17 at 6:03
  • \$\begingroup\$ @HarrySvensson What do you mean? Yes, I have recently been exposed to motors and I'm having to soak up a lot of knowledge before I start my project! \$\endgroup\$ – user2999870 Nov 11 '17 at 6:14
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    \$\begingroup\$ "Is this normal behavior of motors (that the voltage is dependent on the amount of current I can supply to it)?". Ask this question instead: "Is this normal behavior of power supplies (that the voltage will drop if the load requires more current than the power supply can deliver)?" \$\endgroup\$ – Harry Svensson Nov 11 '17 at 6:20
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    \$\begingroup\$ motors dc resistance is approx. 2.5 ohms here. Now that's like you are asking for 10A from 1A rated supply. 0_0 \$\endgroup\$ – Mitu Raj Nov 11 '17 at 6:41
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    \$\begingroup\$ Something stinks in all the responses here ... if it's really a 250W motor, it doesn't take a 10A starting current and its DC resistance isn't 2.4 ohms. It takes 10A running at full power, and more like 50-100A starting (measure its actual DC resistance, expect somewhere around 0.2-0.4 ohms). The gist of the answer is right, however, a 1A supply won't cut it. \$\endgroup\$ – Brian Drummond Nov 11 '17 at 10:08
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I recently bought a 24V 250W DC motor. ...

From \$ P = \frac {V^2}{R} \$ we get \$ R = \frac {V^2}{P} = 2.3 \ \Omega \$.

In order to test it, I connected it to a DC power supply (rated +-25V / 1A). I set the PSU to 24V and connected the motor (no load) to it. The current immediately reached its cap of 1A and the motor voltage raised and capped at 9.3 V.

This indicates that the PSU has current limiting built in. Since we now know the resistance of the motor we can calculate the current supplied by the PSU as \$ I = \frac {V}{R} = \frac {9.3}{2.3} = 4 \ \mathrm A \$. The 1 A limit on the power supply refers to the maximum current where it can maintain 24 V out.

I suspect that the current capacity had something to do with it because the motor did not get its ideal starting current (~10A).

Correct.

Is this normal behavior of motors (that the voltage is dependent on the amount of current I can supply to it)? If so, why?

It depends on a few factors. Most DC motors would start on reduced voltage if mechanically unloaded. You haven't given enough details to suggest why yours did not start.

This will help me determine the battery I will use for my motor since batteries have recommended currents.

Correct.

Be aware that as the motor speeds up that it generates a back-EMF which opposes the driving voltage. This has the effect of reducing current draw from the supply and at light loads the motor run close to rated speed with low current. You may find my answer to Why does a DC motor spin when there is no load applied to it? helps you understanding a little.

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