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What makes the bulb lose resistance?

  1. Oct 4, 2012 #1
    I have included a diagram of this simple circuit I have constructed. When hooked up to the battery, the motor starts out slow, and then speeds up to a ~constant speed.

    The bulb, however, starts out bright but immediately gets dimmer, almost to the point of no light being produced.

    If hooked up in series the current flows through the motor and then through the bulb. I know once a motor gets started it takes less current to keep it moving.

    the circuit, when reversed, seem to produce the same result. I know if these components were hooked up in parallel everything would work as if it were connected individually . However, I am just so curious as to why the light bulb gets dimmer when in series.

    I have so many thoughts about why it is happening I simply need someone else to verify the reason.

    So tell me why, please.
     

    Attached Files:

  2. jcsd
  3. Oct 4, 2012 #2

    vk6kro

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    The motor draws less current when it starts turning because it also acts as a generator and generates an internal voltage which opposes the flow of current into the motor from the battery.

    So, this is like putting another battery in the circuit with opposite polarity to the existing battery. The combined EMF is less if you do this.

    There is less EMF driving current around the circuit, so there is less current and the bulb does not glow as brightly as before the motor started to turn.

    If you like to look at it another way, you could say the motor seems like a higher resistance when it is turning and this higher resistance in series with the lamp causes it to glow less brightly.
     
  4. Oct 5, 2012 #3
    Thanks for the reply. I understand now. Although, I thought that motors produce a high initial current, and then taper off for a constant A draw? I noticed this when initiating the circuit.

    My meter shows an initial current of ~.2A, once the motor increases speed the bulb dims, and the current is down to ~.16A.

    The motor was adjusting its speed to make Kirchhoff's law work.

    And to confirm Kirchhoff's law , I checked, and the current flow out of everything was .17A

    So due to Ohms law, the motor increases speed until the current through everything is the same, yes?

    Amazing fundamental.
     
    Last edited: Oct 5, 2012
  5. Oct 5, 2012 #4

    vk6kro

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    If you just put the motor across the battery and measured the current, the effect would be similar.

    The motor just draws less current when it rotates and the lamp is indicating this effect by dimming.

    Incidentally, Kirchoffs Laws apply all the time and they are correct at any instant. The current in a series circuit can vary if one of the components changes, but the current is always the same at every part of the series circuit.

    The resistance of lamps increases with increased current, so you might like to work out what effect this would have on your experiment.
     
  6. Oct 6, 2012 #5
    ( Buzzer sound )

    The motor alone pulled .25A, the bulb by itself pulled .2A., and together they pull .17A

    I confirm the resistance theory. But the law that states the current is the same throughout the circuit has me convinced that the motor is adjusting to keep the current the same through each component. The EMF is supposed to drop across each component, but that is hard to calculate considering both components can change resistance.
     
  7. Oct 6, 2012 #6

    vk6kro

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    The current in a series circuit just depends on the total resistance at that time, but it is always the same at all points in the circuit at any instant.

    Motor behaviour is complex, though.
    A motor which will run with a certain resistor in series with it may not start to turn with that same resistor in series.
     
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