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I'm confused with the situation below.

Say I have a water pump rated 240 VAC and 240 W. Without a resistor, the drawn current will be 1 A since P = V·I.

By adding a 10 Ω resistor, the voltage drop across the motor will be less than 240 V hence it will need to draw a higher current to maintain the same wattage according to equation below.

Voltage drop across motor pump is 240 - I·R, then power is (240 - I·R)·I.

Using P = V·I, (240·I - I·10)·I = 240 will get I = 1.045 A which is bigger than the connection without the series 10 Ω resistor.

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    \$\begingroup\$ Why would your water pump need to draw the same power all the time (irrespective of mechanical load)? \$\endgroup\$
    – Andy aka
    Commented Aug 9, 2022 at 6:37
  • \$\begingroup\$ I get this question from my friend asking is it possible for a water pump motor to experience higher rated amp if the wiring diameter is too small. Unfortunately, I do not get the information regarding the motor. This is the reason why I assume for this to happen, the motor need to maintain the same power for it to happen but I'm not sure in practical, does motor working in such a way. \$\endgroup\$
    – chuackt
    Commented Aug 9, 2022 at 9:38
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    \$\begingroup\$ That's not how motors work (constant power) unless they have built-in electronic controllers. But if you reduce voltage enough to stall the motor, it'll sit there drawing high current all day delivering no power - at least until it burns out. \$\endgroup\$
    – user16324
    Commented Aug 9, 2022 at 10:47

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hence it will need to draw a higher current to maintain the same wattage according to equation below.

That is an unwarranted assumption. The equation can just as easily be satisfied if the current is less than 1A and the power is less than 240 W. Typical motors do not contain some sort of internal regulation system that causes them to consume a fixed amount of power regardless of voltage. In reality, as the voltage drops across the resistor, the motor will feel less voltage, too, and so it will consume less power, not more. Eventually the circuit will likely settle at some new equilibrium state where I is less than 1A, and P in the motor is less than 240 W.

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    \$\begingroup\$ This is of course correct. But it's worth mentioning that there are devices (such as power supplies for computers) that will draw more current if the input voltage is lower. \$\endgroup\$
    – psmears
    Commented Aug 9, 2022 at 14:58
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    \$\begingroup\$ @psmears I debated mentioning that. But in the end I decided not to. I left the door open just a bit by saying "typical motors do not contain some sort of internal regulation system". So an Atypical motor COULD contain such a system. And devices other than motors could contain such a system. \$\endgroup\$
    – user57037
    Commented Aug 9, 2022 at 15:52
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    \$\begingroup\$ "Typical motors do not contain some sort of internal regulation system" - in a lot of cases wrong. A typical case - a synchronous motor that has more or less constant speed. A constant load (e.g. a water pump at the same water levels) will get you the same mechanical power. What will change is the current and the current vs voltage phase angle. \$\endgroup\$
    – fraxinus
    Commented Aug 9, 2022 at 15:54
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    \$\begingroup\$ @fraxinus synchronous machines are becoming more common. Typically they have drive electronics since they cannot run directly on DC, and they typically cannot start up automatically from AC line voltage at 50 or 60 Hz. So I am still considering synchronous motors to be atypical. A lot of standby generators use synchronous machines for generation, though. And those are common. But they are not being used as motors and don't have any type of constant power property. \$\endgroup\$
    – user57037
    Commented Aug 9, 2022 at 16:06
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    \$\begingroup\$ @mkeith It is almost the same with asynchronous AC motors. They are pretty much common and even more common in water pumps. \$\endgroup\$
    – fraxinus
    Commented Aug 9, 2022 at 16:39
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You can help us by adding information about your water pump. What model is it?

Depends on the pump. If it is Current-controlled in any way aka tries to draw 1A of current constantly, then It will keep trying to draw 1A of current, no matter the resistor you put in series to the AC line. If you put such a large resistor that drawing 1A of current is no more possible, it will start drawing less current or not work at all.

But I doubt someone made the internal electronics of a water pump so sophisticated. Probably it works like a floor fan (lets call it "Voltage-driven") where it just consists of primary and secondary coils, which will draw less current if you put a resistor in series with the AC line.

If the pump is "voltage-driven", using the WATTS here would not help you, as the 240W is a nominal value, for normal operation of the water pump. If you start putting in-series resistors to the AC line, it is not meant to work this way and the Watts information will not help.

If it is current-driven, with internal electronics trying to draw constantly 1A, then it would maintain the 1A current draw (until you put a large enough resistor that the pump would no more could draw 1A). It would not draw more current.

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    \$\begingroup\$ Actually, most circulation pumps in central heating systems are "smart" nowadays and can try to maintain a constant pressure or flow rate as the load changes (e.g. as individual radiators are turned on and off by their thermostatic valves). But 240W is high for that type of pump unless it's in a commercial setting. \$\endgroup\$
    – Finbarr
    Commented Aug 9, 2022 at 7:47
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Generally, the current drawn by a motor is proportional to torque, and the voltage is proportional to speed. If the pump is regulated by pressure, as for a well pump, it will probably try to draw 2-3 times rated current when starting (locked rotor current), so your 240W 240VAC pump may draw 3 amps as it starts up, and the 10 ohm resistor will then drop the voltage to 210V.

Once the motor comes up to speed (assuming it does not stall because of low voltage), it will likely draw less than the rated 1 amp, but as the pressure builds, it may draw 1 amp if that corresponds to the torque required, but usually a motor would not be specified to work at full rated output.

What will probably happen is that the motor will run slower because of the lower voltage, and it may take longer to reach full pressure where the controller will turn off. But if it runs too slow, the "slip" may be excessive, and it might stall, where it will overheat and a thermal cutout may trip.

This is based on the assumption that this is an induction motor, PSC or maybe capacitor start. There would never be a good reason to purposely wire a resistor in series with a pump motor, but this question may be to see what may happen if there is a lot of resistance in the wiring to the pump.

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