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I want to simulate a motor with a resistor and inductor. It means I want to connect a resistor and inductor to 1-phase or 3-phase instead of a motor.

For achieving this goal I have to measure both active and reactive power and calculate the value of resistor and inductor. Is it possible? I mean is it a right way to consume active and reactive power with resistor and inductor?

And if it is right, how is the connection of resistor and inductor, parallel or in series?

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  • \$\begingroup\$ What about back EMF? \$\endgroup\$
    – user57037
    Commented Feb 17, 2019 at 5:55
  • \$\begingroup\$ I just want to consume active and reactive power \$\endgroup\$
    – Hamed P
    Commented Feb 17, 2019 at 6:00
  • \$\begingroup\$ The resistor should consume some power. But it is not clear what your goal is. Maybe experimenting with a simulator would help you accomplish your goals. \$\endgroup\$
    – user57037
    Commented Feb 17, 2019 at 6:04
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    \$\begingroup\$ You can put the components in either series or parallel, and calculate how much actual and reactive power they will consume, in order to test your meter. Then having gained confidence in your meter, you can use it to measure your motor. Is that what you want to do. \$\endgroup\$
    – Neil_UK
    Commented Feb 17, 2019 at 6:15
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    \$\begingroup\$ As Neil said I want to test my meter and I wanted to know if it measures the reactive power right or not. And I don't have any motor or other devices that consume reactive power and I thought of using inductor instead of a motor \$\endgroup\$
    – Hamed P
    Commented Feb 17, 2019 at 6:23

2 Answers 2

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You are assuming that the motor is equivalent to a resistor and inductor. That is not a given. If you use a three-phase synchronous motor without load, there is an excitation of the rotor where the motor will consume only real power. With underexcitation, the consumption will have an inductive component. With overexcitation, the consumption will have a capacitive component. Large idling synchronous machines are traditionally used for power phase compensation purposes in either direction. The temporary storage of energy necessary for reactive power consumption other than the natural inductivity is done with rotational inertia, so you may need flywheels and good bearings for best results in that capacity.

Rotational inertia makes for a lot of the electrical behavior of motors, so a replacement by static components is only a reasonable approximation for stationary behavior and even then will not be a good model of the harmonics that may occur.

Starting and behavior under changing mechanical loads is not really easy to model with passive circuitry.

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Since I can not comment, I will post my comment as an answer. I concluded that you do not have a lot of experience since you posted this: "If I connect a 100 μH, 10 A inductor in series with a 500 W resistor is it possible".

From your initial question I deduced that you want to connect the resistor and inductor to the mains voltage, do not do this without knowing the resulting current.

Preferably, do not do this with the mains voltage at all; get a signal generator. Depending on the frequency of the signal you are working with calculate the inductor impendance \$ Z=\omega\cdot L \$, and make sure the inductor is within its limits. Next choose a resistor with approximately the same resistance in ohms as the inductor. Connect them in series or parallel and work with that.

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  • \$\begingroup\$ Thanks for your reply. As you said I don't have enough experience in this field. Is it important that the resistor and inductor be in the same value? \$\endgroup\$
    – Hamed P
    Commented Feb 19, 2019 at 17:37
  • \$\begingroup\$ The resistance of the resistor should be about the same value as the impendance of the inductor at the given frequency. This assures that the voltage across each of them is about the same when they are connected in series, and your measurements will therefore be more accurate. \$\endgroup\$
    – Ziggy
    Commented Feb 20, 2019 at 9:09
  • \$\begingroup\$ Thanks for the useful information. I want to test it with 3-phase and it seems it's a little dangerous! what is your advice for me? \$\endgroup\$
    – Hamed P
    Commented Feb 20, 2019 at 9:50
  • \$\begingroup\$ Get an ac power source with current limiting, or use somthing like this: link, just make sure you dimension the components(the resistor and inductor) so that the resulting current the source provides does not exceed its limit, and the current trough the individual components does not exceed the components limit. \$\endgroup\$
    – Ziggy
    Commented Feb 20, 2019 at 10:53
  • \$\begingroup\$ Thank you, Ziggy. I will test it and share the results. Can I have your email or something else to get in touch with you? This is my Gmail: [email protected] \$\endgroup\$
    – Hamed P
    Commented Feb 23, 2019 at 7:39

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