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Resistence in DC Generator

  1. May 17, 2008 #1
    Hi, this question might seem really basic but I can't seem to wrap my head around it. For a DC generator (like the hand cranked ones used in science experiments) the generator gets harder to crank when connected to a resistor since it is harder to move the electrons through the wire/circuit. The generator is easy to crank when there is no resistor and electrons can flow easily through the circuit. But why is the generator also easy to crank when the wires are disconnected and the circuit is incomplete? Shouldn't it be really hard to turn the generator when the wires are disconnected since the electrons would theoretically have to move through the air to go from the positive to the negative wire?

    I hope I explained my questions good enough.

    Thanks
     
  2. jcsd
  3. May 18, 2008 #2
    "Shouldn't it be really hard to turn the generator when the wires are disconnected since the electrons would theoretically have to move through the air to go from the positive to the negative wire?"

    This would be true only if the air gap was small enough to otherwise provide a current path with a high enough voltage to "break" the gap.
    So, generally, the answer to your question is no.
     
  4. May 18, 2008 #3

    pam

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    Since there is I^2 R loss in the resistor, conservation of energy says that you have to do more work in turning the rotor. The torque that you have to overcome comes from the magnet acting on the current in the windings of the rotor.
     
  5. May 18, 2008 #4
    The generator converts the mechanical energy of spinning a coil within a magnetic field to electromagnetic energy. No energy conversion occurs without a closed circuit.

    The generator charges reactive loads. It can increase magnetism in a superconductor, or increase voltage's energy in a capacitor.

    A resistor can convert voltage's energy (or magnetic energy) to thermal energy.
     
  6. May 18, 2008 #5
    I think I'm still a little confused. If the rotating magnets in the generator move electrons through the wires as they pass by it gets harder to move the electrons and hence turn the magnets and turn the crank if a resistor or load is connected into the circuit. But why in a broken circuit does it suddenly become easy to turn the generator. Should'nt the magnets still be trying to move electrons?

    pzlded, you explained "No energy conversion occurs without a closed circuit."
    I'm confused about why this is.
     
  7. May 18, 2008 #6
    An open circuit coil cannot convert kinetic energy to magnetism because the open circuit prevents the existence of magnetism in the coil.

    A magnet can only add centripetal force to a point charge. Centripetal force cannot add kinetic energy or any other energy to the point charge, w/ or w/o a closed circuit.

    Similarly, a changing magnetic field in a transformer primary cannot add kinetic energy to secondary coils. Changing point charge kinetic energy requires force and distance in a direction that is not perpendicular to point charge velocity.

    The rate a magnetic field changes does not determine coil voltage. When a superconductive coil moves a certain distance toward a magnet, the coil acquires a fixed amount of magnetic energy, regardless of the speed the coil travels along that distance.

    Force x distance relationships do not depend on the time it takes to travel the distance.
     
  8. May 18, 2008 #7
    Maybe it will help to think of it in terms of power. The electric power consumed in the circuit has to be matched by the mechanical power put into the generator. If you have an open circuit no power is consumed so none need be produced.
     
  9. May 19, 2008 #8

    Redbelly98

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    Power generated = voltage x current

    If either voltage = 0 (short circuit) or current = 0 (open circuit), no power is generated so turning will be easy.

    It should be hardest to turn when you use the resistance that maximizes v*i.
     
  10. May 19, 2008 #9
    Whenever induction comes up, these questions arise. Here is an abridged answer. A peer-reviewed university text will elaborate, and should be referred to as a reference.

    When open circuited, the generator is easy to turn. With minimum output current, the magnetic field due to the output current is also minimum. When a load is added, assuming constant voltage output, the usual case, the output current has a magnetic field which opposes the inducing magnetic field in accordance with the law of Lenz. This counter mmf (magnetomotive force) results in a counter torque. Hence, the shaft will be harder to turn because there is a counter-torque acting in opposition to the input torque. Heavier loading (more output current) results in increased counter-mmf and counter-torque, making the shaft harder to drive. In order to maintain constant voltage output, the speed must be held constant. To do so requires more input torque to overcome the increased counter torque.

    As far as a dead short is concerned, for constant current output, this is the no load condition. To obtain constant current, the drive at the input must be constant torque. Torque is related to current, and speed to voltage. Driving a generator at constant speed results in a constant voltage output. Driving at constant torque produces constant current. With a short circuit load, the input torque is immediately countered by the counter torque. As a result the speed is very low, as well as the output voltage. The power is minimized with a short circuit load and constant current output. As the load resistance is increased, less counter torque is produced, and the speed increases, as does the output voltage and power.

    But if the shaft is driven at constant speed with the intent of producing constant voltage output, then lowering the load resistance results in an increase in power.

    I hope this helps. BR.

    Claude
     
    Last edited: May 19, 2008
  11. May 22, 2008 #10
    I see the source of your confusion. When the wires are disconnected that's a really BIG resistance. It's so big no power goes out. Keeping that in mind, adding any value of resistor actually makes the resistance less.

    When it's shorted is the worst. There's still some resistance from the windings. If you go back and read what these other folks talked about power and current 'n stuff, it could begin to make sense.
     
    Last edited: May 22, 2008
  12. May 22, 2008 #11
    there is something called the maximum power transfer theorem. it says (which can be easily derived) that the power transferred from a source to a load is maximum when the internal resistance of it is equal to the source's internal resistance. the curve for the power transferred thus hits a peak when Rsource = Rload. it falls of to zero on either side of it. so if your load resistance is too low(~zero) or if it is too high then the power transferred is very low.
    hence...
    you can look it up in any decent book for networks analysis.
     
  13. May 23, 2008 #12
    But --- *transferred* is one thing. Total power generated and dissipated is another. A 5 volt constant voltage source with a 50 ohm internal resistance is loaded with a larger resistance, say 450 ohms. The total power generated is 50 mW, of which 45 mW is transferred to the load.

    Now, if the load resistance is 50 ohms, total power is 250 mW, and 125 mW, or half is transferred to the load. This condition results in the largest amount of power reaching the load.

    Now if the load is 12.5 ohms, the power transferred is just 80 mW, but the total power is 400 mW, larger than the case with matched load and internal resistance.

    Total power is one thing, but transferred power is another.

    With a generator, the "no load" condition is open, but only for *constant voltage* operation. For constant current operation, "no load" is a short circuit.

    But, if a generator is forced to spin at constant speed, the output is constant voltage, and reducing load resistance results in greater power generated. The assumption that low load resistance results in low power is based upon constant current mode of operation.

    Regardless of CC or CV mode of operation, the generator shaft is always harder to turn with a lower value of load resistance. Have I cleared things up? BR.

    Claude
     
  14. May 24, 2008 #13
    the power generated has to be transferred to the load.. when you crank a generator the idea is to convert the mechanical energy you supply into electrical energy which is then transferred to the load.
     
  15. May 24, 2008 #14

    Redbelly98

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    Claude, I've been thinking about this statement, which is at the heart of your argument, and disagree with it.

    For low enough resistance, the induced current is large enough to create a magnetic field which mostly cancels the change in the applied magnetic field (i.e. the one due to the magnets moving relative to the coils).

    Actual, net magnetic field is the sum of the applied field (from the moving magnets) and the induced field (from the induced current). And induced emf is the result of this net field. If total resistance (load + internal) is small enough, the induced field nearly cancels the external field change, so induced emf is smaller than it is for a larger resistance load.

    Regards,

    Mark
     
  16. May 24, 2008 #15
    What induced emf? For a generator w/superconductive windings and a short (superconductive load), ALL VOLTAGE IS ZERO. Cranking the generator just adds more magnetic energy to the inductive load. Electric generators produce magnetism. EMF (volts) only occur if magnetic energy converts to volts.

    Think of the generator's coil as an inductor. If (before cranking) voltage (a charged capacitor) is connected to a superconductive generator's coil; superconductivity will prevent voltage's energy from existing within the coil and there will be no voltage drop along the coil's wire. The applied voltage that initially appears across the coil is a measurement of the voltage of the voltage source. After the coil begins converting voltage's energy to magnetic energy, this magnetic energy is applied to the circuit's resistances.

    Eventually voltage applied to the coil will be virtually zero, because enough magnetic energy converts to voltage's energy in the resistances to make resistor voltage equal applied voltage. If the generator is now cranked, the magnetic energy from cranking will convert to additional voltage's energy.

    Now isn't that simple?

    Yes? Ok, the above is over-simplified. Consider that if a circuit becomes an open circuit, applied voltage will appear across the open. Similarly, applied voltage initially appears across an inductor (the generator's coil). Current travels through the resistor in two directions. Initially, voltage's energy from the charged capacitor load passes through the resistor w/o increasing resistor voltage (coil voltage = applied (capacitor) voltage). After the resulting magnetic energy (within the closed circuit) is applied to the resistor, resistor voltage increases.

    In other words, initially the resistor acts like a capacitor to short voltage to the coil. Then magnetic energy converts to voltage's energy within the resistor (similar to charging a capacitor). Voltage's energy within a resistor is unstable, thus converts to thermal energy. (If indeed a resistor converts magnetic energy instead of voltage's energy to thermal energy, then a resistor would retain voltage (similar to a capacitor) upon occurrence of an open circuit.)

    In other words, the existence of voltage requires a capacitor to store voltage's energy. A resistor is a capacitor, albeit voltage's energy converts to thermal energy within the resistor.
     
  17. May 25, 2008 #16
    Good grief! Read my last half dozen posts in other threads and I've been saying the same thing. The problem with your position is that you are unwittingly assuming constant *torque* not constant speed. Of course the induced mmf/emf cancels the generated, reducing the output. But please remember that the result of the large mmf/emf due to the low output resistance is a counter *torque*. This counter torque opposes the input shaft torque and if the input torque is constant the speed will decrease and the voltage decreases. Your argument is correct for constant input torque.

    But I stipulated constant shaft *speed*. When the counter mmf/emf due to the low load resistance cancels the generated mmf/emf reducing speed and voltage, the input torque can then be *increased* to cancel the counter torque. Thus the generated mmf/emf increases. Of course so will the counter mmf/emf increase, requiring more input torque. Eventually the original constant output voltage value is obtained, as long as the resistance of the load is greater than the output loop reactance. Suppose that the load induced mmf/emf results in 90% cancellation of the generated mmf/emf. The net output voltage is, of course, only 10% of no load, for constant torque. If we increase the input torque so that we generate 1000% of the no load mmf/emf, then 90% of 1000%, which is 900%, of the no load generated mmf/emf is cancelled. Thus the output voltage is 100% of the no load. It works out mathematically. Forcing constant speed results in constant voltage. Note that for a given torque and speed, we get a V and an I. Dropping the load R to 1/10th results in constant V with I increasing 10X. To obtain constant speed, we must increase the input torque 10X. Power is torque*speed mechanically, and I*V electrically. Conservation of energy is upheld.

    If the resistance is lowered while the speed is held constant, the output will be constant voltage. The power dissipated in the resistance is (V^2)/R. The problem with lowering the resistance down to zero is as follows. There is an inductive reactance associated with the loop formed by the winding and the termination. When the resistance drops below this reactance, the induced mmf/emf is no longer in phase with the generated mmf/emf. The total impedance is the phasor sum of the resistance and the reactance, in quadrature (90 degree phase). At this load value, the reactive power exceeds the real or dissipative power. Reducing R further will then reduce the dissipated power, as the voltage divider action between the reactance and resistance results in a smaller voltage reaching the resistor. For the reactive component of the current, the counter torque is displaced by 90 degrees.

    But, although the generated mmf/emf can be forced to remain constant by forcing constant speed, the voltage divider reduces the terminal output voltage, For a zero ohm load, i.e. superconducting, all the voltage drops across the inductance and none across the superconductor. Thus no dissipation takes place.

    The source of confusion is in assuming real or resistive loading all the way down to zero ohms, which cannot happen. Although superconducting loops have no R, they always have L. We can't get around that.
     
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