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Let say a motor is rated for 100mA at 3.7V with no load, which means it will draw 100mA whenever it "gets" 3.7V. However, what will happen if my power source is supplying 3.7V with only 10mA current? Does it mean that the motor cannot "take" all of those 3.7V, then where will the rest of the those voltage drop?

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    \$\begingroup\$ that's not how electricity works... \$\endgroup\$ Apr 5 '20 at 11:28
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    \$\begingroup\$ a current-limiting supply will just lower the voltage until the current is as low as specified. You can either defined the voltage or the current at any time, not both. \$\endgroup\$ Apr 5 '20 at 11:28
  • \$\begingroup\$ @MarcusMüller Isn't a power supply (battery) can only output its rated output voltage? A 3.7V battery can output 1.5V at sometimes? \$\endgroup\$
    – mannok
    Apr 5 '20 at 11:41
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    \$\begingroup\$ No, your understanding seems to be fundamentally wrong. Battery current ratings describe how much current the battery can source without damage to the battery, or how much you can draw without incurring more than some (elsewhere specified) voltage drop. \$\endgroup\$ Apr 5 '20 at 11:43
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    \$\begingroup\$ resistors don't "take voltage"; they experience a voltage drop proportional to the current that flows through them: Ohm's law. So, if you add a resistor, both the voltage across your motor and the current through it will drop. Seriously, this site is called "electrical engineering", I'd recommend you revisit Ohm's law, this is really really the minimum basics! \$\endgroup\$ Apr 5 '20 at 12:19
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The starting current of a motor could be around 4-6 times its rated current.

The motor in question won't start but will only act as a dead short on the current-limited power supply.

The power supply voltage would drop sufficiently to sink only 10mA into the dead short.

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