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B Where does the current go

  1. Oct 16, 2016 #1

    pug

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    If I have a power source (4V and 50 mAmps) going into a motor that is 4V with a running current of e.g. 200mAmps. What will happen? e.g. will the current from the power source just heat the motor? Or if I had a capacitor attached to the motor that was e.g. was 4V and 200mAmp...would a charge build in the capacitor and then discharge sporadically into the motor?

    when I think of volts I think of an electrons "height" and when I think of current I think of how many electrons are flowing. so my thought is that the height (volt) is sufficient and the current will build on the capacitor (dam) until it bursts/discharges.
     
  2. jcsd
  3. Oct 16, 2016 #2
    Have you learned about electromagnetism yet?
     
  4. Oct 16, 2016 #3
    Are you asking what would happen if you connect a source rated for 50mA to a load that draws 200mA? The current would flow at 200mA until the source blew up.

    There may be some voltage droop. Real sources have some output resistance. When you drive a current across a resistance the voltage is lower on the output side of that resistance. As you can probably guess the output resistance of a source is an important parameter for determining the current rating for that source.

    That's more or less true. Capacitors have voltage ratings. If you apply too much voltage the capacitor will arc internally.
     
  5. Oct 17, 2016 #4

    i like this analogy *thumbs up*
     
  6. Oct 17, 2016 #5

    CWatters

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    What sort of power source?

    If it says 50mA on the label that's usually the maximum you should draw. If you try and draw 200mA then that's an overload and the power source will do something that depends on the quality of it's design...

    It might try and deliver 200mA, overheat and switch itself off until it cools down. Switch back on, overheat, repeat.
    It might blow a fuse (switch off until fuse is replaced - not always possible).
    It might limit the current to 50mA by reducing the output voltage below 4V.
    It might just deliver 200mA for awhile then fail after a few mins, days, or weeks.
    Some combination of the above.

    What happens depends on what the power supply does when subjected to the overload. If the power supply limits the current by reducing it's output voltage then the motor might slow down and stall/stop depending on the mechanical load on the motor. In that case the current would just heat the motor.

    No. The capacitor will most likely make no difference to what happens. It might be possible to design a circuit that stored energy in a capacitor and used it to run the motor sporadically. This is similar to how some photographic flash guns work although they bump up the voltage rather than the current.
     
  7. Oct 17, 2016 #6
    Have you tried this?... If so...what happened?
     
  8. Oct 17, 2016 #7
    I always thought of it like this. Voltage is like how many electrons there are and amperage is how fast those electrons are moving. So many electrons moving slowly would be like high voltage and low current.
     
  9. Oct 17, 2016 #8

    davenn

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    This is very incorrect

    please do some reading on voltage, current, amperes and charge, their meanings and how they relate

    electrons travel very slowly in a circuit ... google drift velocity and also read my comments below in post #9


    Dave
     
    Last edited: Oct 17, 2016
  10. Oct 17, 2016 #9

    davenn

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    unfortunately it isn't totally correct and will lead to misunderstandings

    basically right .... when referring to the potential difference between the power supply terminals

    current is just the flow of electrons. to go a step further to understand that .... you need to refer to the Coloumb

    https://en.wikipedia.org/wiki/Coulomb

    The coulomb (unit symbol: C) is the International System of Units (SI) unit of electric charge. It is the charge (symbol: Q or q) transported by a constant current of one ampere in one second:


    also not quite right....
    As some one else mentioned, capacitors have voltage ratings. This means as long as you don't exceed that voltage you can have a voltage across a capacitor for eternity and it will NOT "burst like a dam"
    The capacitor will just energise to that applied voltage and stop there

    Dave
     
  11. Oct 20, 2016 #10
  12. Oct 20, 2016 #11

    David Lewis

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    If the motor runs continuously, usually the power supply gets hot or burns out. If the motor runs intermittently, and 200mA represents short term current spikes, you might get away with it.
     
  13. Oct 21, 2016 #12
    Why is voltage called potential difference? Does this have to do with the difference in potential energy between two regions of opposite net charges? Why? In mechanics as I learned potential energy is the energy an object has due to its shape or position (change in PE= Change in H * G*M), what does this mean in a circuit?
     
  14. Oct 21, 2016 #13
    Yes, it is exactly the same thing only per unit of charge. Moving 1 coulomb of charge through the circuit across 1 volt PD takes (or releases, depending on sign) 1 joule of energy.
     
  15. Oct 21, 2016 #14
    The question should be "why in some countries the potential difference is called voltage?".
    "Potential difference" is the basic name of the physical quantity described (in various languages of course).
    Just because the unit of measure in SI is volt, some people use a shortcut and called it voltage.

    Like they called the length of a piece of cloth something like "metrage" in my home country.
     
  16. Oct 23, 2016 #15
    Voltage is esseintially the electric field on a charge mutliplied into the distance moved as a net effect of that force. Q1 is the origin of the electric field, Q2 a much smaller charge without any effect on Q1.

    Electric field = Force / magnitude of charge.

    Force between two charges is found through Coulumbs law, which we substitute.

    Electric field on a charge or a point:[ Q1*Q2*K/r^2] / Q2 = Q1*K/ r^2

    Voltage = work done on a charge by an electric field / magnitude of the charge.

    V = F * D / Q2

    Q1*Q2*K/ r^2 * D / Q2 simplifies to Q1*K/ r^2 * D. And Q1*K/r^2 is the electric field at that point. I think this is correct?
     
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