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Question on Ohm's law and motors

  1. Jun 26, 2010 #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. Jun 26, 2010 #2

    vk6kro

    User Avatar
    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:
    http://en.wikipedia.org/wiki/Counter_emf


    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. Jun 26, 2010 #3
    An electric motor is not a resistor.
     
  5. Jun 26, 2010 #4
    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. Jun 26, 2010 #5
    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. Jun 26, 2010 #6
    well, I don't know what you call it, but it has resistance. So wouldn't it be a variable resistor?
     
  8. Jun 26, 2010 #7
    What do you mean by variable resistor?
     
  9. Jun 26, 2010 #8
    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. Jun 26, 2010 #9
    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. Jun 26, 2010 #10
    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. Jun 26, 2010 #11
    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. Jun 26, 2010 #12
    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. Jun 26, 2010 #13
    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. Jun 26, 2010 #14
    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. Jun 26, 2010 #15
    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. Jun 26, 2010 #16
    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. Jun 26, 2010 #17
    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. Jun 26, 2010 #18
    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. Jun 26, 2010 #19
    Let's look at a very crude model of an electric motor: the railgun. (scheme attached)

    attachment.php?attachmentid=26695&stc=1&d=1277611731.png

    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:

    [tex]
    m \, \frac{dv}{dt} = B L I - F
    [/tex]

    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:

    [tex]
    V - B L v = R I
    [/tex]

    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]:

    [tex]
    v(t) = \frac{V(t) - R I(t)}{B L}
    [/tex]

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

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

    and substitute this into the first equation:

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

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

    [tex]
    \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)
    [/tex]

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

    [tex]
    \begin{array}{l}
    v_{\mathrm{ef}} - \frac{Q}{C_{\mathrm{ef}}} = R_{\mathrm{ef}} \, I \\

    I = \frac{d Q}{d t}
    \end{array}
    [/tex]

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

    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:

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

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

    [tex]
    C_{\mathrm{ef}} = \frac{m}{(B L)^{2}}
    [/tex]
     

    Attached Files:

    Last edited: Jun 26, 2010
  21. Jun 26, 2010 #20
    ok, I get what you're saying, it's not a resistor. Very nice.
     
  22. Jun 26, 2010 #21
    A rail gun is not a good model to use for an induction motor; however I understand the motivation to use a simple illustration to make a point.

    Generally, we shouldn't expect to get an Ohm's law relation. However, it is a basic fact that real work must show up as a resistive term on the electrical side. Reactive power can be inductive or capacitive loading. But real work looks resistive.

    A simple model for an induction motor, at least when it is stopped, is a transformer. A locked motor has a small resistance of the windings, but since no work is being done, the main load is inductive. When an induction motor is spinning rapidly in steady state and doing real mechanical work, then the electrical load looks mostly resistive. This resistance is frequency dependent and doesn't behave like a typical resistor. So in a sense the resistance has frequency dependence similar to reactance; however, there is no phase shift like a reactance and the load is mostly resistive.
     
  23. Jun 27, 2010 #22

    vk6kro

    User Avatar
    Science Advisor

    I did some tests on a 250 volt 50 Hz desk fan using a variac to provide variable input voltage.

    Volts 115 V current 0.20 A Power factor 0.91 (stalled.)
    Volts 156 V current 0.26 A Power factor 0.91
    Volts 189 V current 0.26 A Power factor 0.93
    Volts 207 V current 0.27 A Power factor 0.93
    Volts 222 V current 0.28 A Power factor 0.92
    Volts 240 V current 0.28 A Power factor 0.91

    At no stage did the current increase with decreasing voltage, and the range of speed available was from 1500 RPM to zero. The power factor was basically unaffected.

    This is probably a shaded pole motor. Industrial fans using squirrel cage motors would behave differently.

    Notice that, by calculation, you can see that the impedance of the motor increased from 575 ohms at 115 volts to 857 ohms at 240 volts, but this does not translate into increased current at lower voltages because of the reduced supply voltage.
     
  24. Jun 27, 2010 #23
    Out of curiosity, what was the fan doing incorrectly for you to be troubleshooting. If it wasn't moving, a motor that doesn't get enough voltage to overcome starting friction will stall and act as a transformer with the secondary shorted, and appear to draw more current than it should.

    You also said no-one could tell you why wires are rated by current carrying capacity. Thinner wires have a higher resistance per length, Power lost in the conductor = I^2*R which produces heat. The current rating is the point where the heat generated is within acceptable limits, more will result in insulator degradation and fires.
     
  25. Jun 27, 2010 #24
    omg, it was a nightmare... We were replacing some fans at a paint facility for a nuclear power plant. The original fans were 240/480 3phase, and we showed up with two 240v split phase fans. Well, rather than just reordering the right fans we decided to just put these fans in and take power off the 120/208v 3phase panel that was right next to the 480 panel. So, we get the fans installed and run some new wire to them. We turn the first fan on, and it's running great. Move to the next fan flip the switch and... pow! The breaker trips. The fan starts up and won't come below 80 amps (14amps full load). We check the fan to make sure it's moving freely everything seems good. We leave and come back with more material, and now we can't get the first fan to turn on. It's doing the same thing. It won't hold all the time. It holds some of the time and runs normal, but not all the time. So, we change the breakers, the motor controls, the wire gets re-pulled with heavier gauge to prevent voltage drop. We do everything new, everything oversized. We can't find anything wrong other than the motor is a 240v motor on a 208v line. We even stuck a buck/boost transformer on the one fan to try and get it to hold. My boss is getting ready to order two new 200v motors, but before that we ask the utility company to have one of their electricians check their transformer to see if there are any loose wires that might be fluctuating the voltage. Well, it turns out that the transformer supplying that panel was supposed to be changed 2 years ago and has been sitting in the paint shop's corner since then. They've been having problems like this the whole time. It took us 3 weeks with 2 guys to install the 2 fans. Needless to say, we lost money on that job.
     
  26. Jun 27, 2010 #25
    What I meant was.. I was trying to use logic in a round-a-bout way to get them to see my point. They know that every gauge with a specific insulation is rated for a specific amperage. My point was to explain that voltage doesn't create heat, amperage creates heat. If higher voltage drops amperage than putting 1000v into a small motor should make it run with less heat. It doesn't though, it will melt it because the amperage will shoot through the roof.
     
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