Question on Ohm's law and motors

  1. I am an electrician, and I recently got into some trouble on an issue. We were trouble shooting a fan motor that was rated at 240v and was only getting 208v. The more experienced electricians I was with said the amperage was going up because there wasn't enough voltage. This makes no sense to me. I get that watts=volts*amps, and that formula makes sense to them as to why the amperage is going up, but I remember in my apprenticeship class when you have a dual voltage motor you can wire the stators in either series or parallel to change the resistance and keep the output equal regardless of the voltage. So, V=IR makes sense to me.

    Basically every electrician I talked to, and I talked to a lot of them, including a few masters and a contractor told me I was wrong. So, I guess my questions are (1)When voltage goes up does amperage go up? and (2)Regarding NEC 430.250, why do the full load amps seem to support their views.
  2. jcsd
  3. vk6kro

    vk6kro 4,058
    Science Advisor

    Your friends are talking about the effects of back EMF on the current drawn by a motor.

    Here is a quote from a similar thread a few days ago:

    If you imagine a motor that is already turning, you have a coil rotating in a magnetic field, just like a generator. There is a voltage generated that opposes the current flowing into the coil from the supply.
    This causes a reduction in supply current as the motor turns.
    If you stall the motor, this generated voltage vanishes and the motor then draws a much larger current.

    This effect is called Back EMF or Counter EMF and is very important in motor behaviour.
    You can read more about it here:

    HOWEVER, the load on a fan depends on how fast it is turning and it is common practice to reduce the voltage on a fan to reduce its speed. This is done without any ill effects, except that the fan runs slower.
    This is because the fan gets less back EMF as it runs slower, but it also gets less load (because the fan has less wind resistance at the slower speed) so the current does not increase as it would with a constant, heavy load.
  4. An electric motor is not a resistor.
  5. If the motor is an induction motor designed for 240 V ac 60 Hz, its quasi-synchronous RPM is ~1740 RPM. If you reduce the voltage to 208 V ac, it still wants to run at 1740 RPM (and deliver the same torque and power). To do this, it must draw more current.

    Bob S
  6. My friends were definitely not talking about back emf. I believe there is a HUGE misconception in the trade about how electricity works and it just baffles me. I just got out of apprenticeship school and people that having been doing this work for 20 or 30 years don't seem to have a clue as to how stuff works. I was told when voltage goes down amperage goes up and that is regardless of the situation. Even something with a static resistance like a water heater element has the same effect.

    for instance, 60w bulb rated for 120v draws .5 amps. They believe if you supply that light with 30v you will increase the amperage to 2 amps. So, I ask, "if you put 400v in the same bulb why does it explode or burn out?" They're answer is usually because the 400 volts blew it up. I try asking why #12awg is rated for 20 amps and not volts, and I just get blank stares. I just can't accept what they're telling me because none of them can explain it to me with any logical sense. That's why I'm here, because I figure I'll either come away with an explanation on why I'm wrong or find that I'm right, which might be worse. Because there are A LOT of people that feel I'm wrong.

    Anyway, thank you for your response.
  7. well, I don't know what you call it, but it has resistance. So wouldn't it be a variable resistor?
  8. What do you mean by variable resistor?
  9. It definitely has resistance, but I think they mean that it is not a constant "linear" resistor. So, as you say it is a variable resistance in a sense. Any dissipation of real power (as opposed to reactive power), whether it is in the form of heat, or sound, or light, or mechanical work, it will show up as an effective resistance on the electrical side.

    In any situation where the system is going to provide nearly constant power, even if the voltage changes a little, you can expect the current to go up when voltage goes down. This may not be always an exact relationship (I mean constant power), unless you have an actual feedback control loop that maintains constant power.
  10. On a dual voltage motor when you wire the motor for 120 volts you get 14amps and when you wire the motor for 240v you get 7 amps. Why do you need put the stators in parallel or series if the process is already done for you? This is the only way you can get the same amount of output from a motor and have less voltage.

    This is the explanation I keep getting, and it just doesn't make sense.
  11. variable: a number that isn't constant and resistor is resistor. The motor will draw like.. 8 times the amount of current when it first starts up then less current. So, depending on the speed of the motor the current will change. And if the current and voltage are changing inproportionately it would lead me to believe so is the resistance.
  12. So, I guess you answered your own question then, haven't you? But, your model is quite imperfect, because you haven't put any quantitative dependence of what this variable resistance actually depends on.
  13. It really depends on the details of the motor, the load and the feedback control. If you run a motor in open loop by simply applying a voltage, it is not likely to maintain constant power if you change the voltage by a factor of two.

    I think the answer is really dependent on the situation. Your way of looking at it has a reasonable basis; however, if you observe a system in which a small change in voltage results in an inverse change in current, you likely have a nearly constant power situation. Mechanical power is torque times rotational speed, so the motor may slow down and have more torque capability at lower voltage. The details matter and the situation is not always simple.
  14. wow, good information. So, you're saying if the voltage only migrates a little the amperage could actually go up? Now, would it, let's say, double if the voltage gets cut in half?
  15. Well, I came here with the answer I thought was right. Everyone at work seems to disagree. I came to an engineering forum because I figured you guys would know for sure.
  16. I think that would only happen if you had a sophisticated feedback control loop. You correctly noted the rewiring of the stator when you convert from 110 V to 220 V, and this is typically done when an induction motor is run open loop by applying AC voltage. If you use a sophisticated vector control loop in a constant power control mode, then you could could maintain constant power over much more than a factor of 2 change in voltage.

    This is definitely true as a rule, but there are always exceptions. An electrician can do work without a detailed understanding of theory. On the flip side, I wouldn't trust most electrical engineers to wire AC in my home, including myself.
  17. If you have a quasi-synchronous induction motor running at 60 Hz, and you reduce the voltage and not the frequency, the RPM change is very small, you have a constant load power situation. I have made measurements on induction motors under load at reduced voltage and observed the current increase.

    [added] When the voltage is reduced slightly, say 10%, the power factor usually increases.

    Bob S
  18. That sounds reasonable to me. I tend to run induction motors in vector control mode with variable frequency, so my 60 Hz experience is limited.
  19. I tried an experiment with a dimmer and an amp probe on my bench grinder. What I found wasn't what I expected. I think dimmers are made a little different today, so I'm not sure if what I found was accurate. I was definitely able to make the motor run slower with less voltage, but the amperage stayed the same once it got up to whatever speed it could. I don't really have the right tools though for that kind of experiment.

    for instance:

    60v 1000rpm 2amps
    90v 1200rpm 2amps
    120v 1400rpm 2amps

    not actual results, but an idea of what I got.
  20. Let's look at a very crude model of an electric motor: the railgun. (scheme attached)


    there are two conducting rails and a cylindrical conductive bar can roll on them freely. There is also a magnetic field perpendicular to the plane of the rails. The load is drawn by the rolling bar.

    There are two governing equations for the motor: the Kirchoff loop equation and Newton's Second Law.

    Let's start with Newton's Second Law for the bar. There are two forces acting on the bar:

    Ampere's force: [itex]F_{a} = B l I[/itex]
    external force (opposite of what the force with which the bar pulls the load) - it is an external parameter [itex]F[/itex]

    The bar has mass [itex]m[/itex]

    The acceleration of the bar is [itex]a = dv/dt[/itex]

    So, Second Newton's Law gives:

    m \, \frac{dv}{dt} = B L I - F

    Let's consider Kirchoff's Law now.
    There are two voltage sources acting in the loop:
    external voltage [itex]V[/itex] which is an external force
    induced EMF: [itex]E_{i} = -B L v[/itex]

    The resistance of the whole circuit is [itex]R[/itex].

    The current through the circuit is [itex]I[/itex].

    Second Kirchoff's Rule gives:

    V - B L v = R I

    The mechanical variable for the system is [itex]v[/itex] - the velocity of the bar (or, in rotary geometries the angular velocity of the rotor). The electric variable of the motor is the current passing through it [itex]I[/itex]. The external parameters are the generated force [itex]F[/itex] (or, generated torque in rotary geometries) and the applied voltage [itex]V[/itex]. The internal parameters are the mass of the bar [itex]m[/itex] (or, the moment of inertia of the rotor in rotary geometries), the length of the bar [itex] L [/itex], the magnetic field [itex]B[/itex] and the total Ohmic resistance of all the conducting elements (rails and bar) [itex]R[/itex].

    If you wish, you can eliminate the mechanical variable [itex]v[/itex] completely and get an equation involving only the electric state variable [itex]I[/itex]. Namely, solve the second equation with respect to [itex]v(t)[/itex]:

    v(t) = \frac{V(t) - R I(t)}{B L}

    differentiate it with respect to time (keeping in mind that the external voltage as well as the current might vary in time)

    \frac{dv}{dt} = \frac{V'(t) - R I'(t)}{B L}

    and substitute this into the first equation:

    \frac{m}{B L} \left(V'(t) - R I'(t)\right) = B L I(t) - F(t)

    That is, we get a first order linear differential equation:

    \frac{d I(t)}{d t} + \frac{(B L)^{2}}{m R} \, I(t) = \frac{1}{R} \left(V'(t) + \frac{B L}{m} F(t)\right)

    How is this equation like Ohm's Law? If anything, it looks like there is a capacitative element:

    v_{\mathrm{ef}} - \frac{Q}{C_{\mathrm{ef}}} = R_{\mathrm{ef}} \, I \\

    I = \frac{d Q}{d t}

    \frac{d I}{d t} + \frac{1}{R_{\mathrm{ef}} C_{\mathrm{ef}}} I = \frac{V'_{\mathrm{ef}}(t)}{R_{\mathrm{ef}}}

    and, taking the effective resistance to be the same as the Ohmic resistance of the true motor (which can be done, by taking [itex]F = 0[/itex] - free running motor), then we see that the effective voltage applied on this circuit is:

    V_{\mathrm{ef}} = V(t) + \frac{B L}{m} \int_{t_{0}}^{t}{F(t') \, dt'}

    i.e. it is time dependent even if the external parameters V(t) and F(t) are constant, and, the effective capacitance is:

    C_{\mathrm{ef}} = \frac{m}{(B L)^{2}}

    Attached Files:

    Last edited: Jun 26, 2010
  21. ok, I get what you're saying, it's not a resistor. Very nice.
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