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Homework Help: A few questions

  1. Mar 29, 2007 #1

    Zeus12345

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    Over the past few weeks in my HS physics classes I've come up with several questions that my teacher can't answer (because they are a maths teacher not a physics teacher :uhh: ). I've tried to find the answers on the internet but can get no straight answers, I hope you guys can help me.

    1. When a current flows through a conductor is ALL the potential energy released to the conductor, ie. do the charges return to the opposite terminal without any energy. If a charge moved between two points of varying electrical potential in empty space, how and to what is the potential energy transferred?

    2. A current carrying conductor has a magnetic and electric field around it, do these fields require energy in order to exist. Is energy lost in a conductor creating these fields? Why is it that a charged particle has to be moving before it can have a magnetic field?

    3. A solenoid with AC running through it acts just like any other device, yes? It is converting electrical energy into heat due to resistance in the conductor. When a second coil is brought near you have a simple transformer, what effect, if any, does the second coil have on the current in the first coil? Now that electrical energy is being induced in the second coil does that mean that less energy is being lost as heat etc in the first? If not then how is energy being conserved? Will a transformer consume the same amount of power when the second coil is present (or simply a complete circuit) as to when the second coil is not present (or is an incomplete circuit)?

    I had a couple more questions but I will save them and see if I can understand your answers to these ones first. Any help is much appreciated :biggrin:
     
    Zeus12345, Mar 29, 2007
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  3. Mar 29, 2007 #2

    haiha

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    I think now i can answer your first question
    1. when charges from one to the other terminal, they release energy. But electric current must always be a close loop, so inside the battery (or any other power supply) there is exactly the same current going through and part of the potential energy is released here. If a charge moves in empty space, the energy becomes kinetic energy and be released when the charge hits the opposite terminal.
     
    haiha, Mar 29, 2007
  4. Mar 29, 2007 #3

    berkeman

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    Staff: Mentor

    Thread moved to Homework Help, Intro Physics.

    Welcome to the PF, Zeus12345. Homework and coursework questions must be posted in the appropriate Homework Help forum, not in the general forums. And when you post homework or coursework questions, you need to show some of your own work and thoughts, in order for us to help you the best. Also, when you start a new thread in the Homework Help forums, you will be provided a Homework Help Template, which you should use to help us provide you with assistance.

    Now on your present questions, I can provide a little feedback to see if it helps.

    -1- Remember that electrons are what actually flow in the circuit, but we refer to positive "current" flowing out the + terminal of the battery and around the external circuit. But since negatively charged electrons are physically what is moving, they actually flow out the - terminal of the voltage source and around the external circuit, and return to the + terminal of the voltage source. But normally we just refer to + current, so if there were + charges moving, then yes, they would have zero potential energy at the - terminal of the voltage source, their potential energy gets pumped up as they move through the voltage source to the + termainal, and then all that potential energy is given up as heat (via collisions with the atoms as the charge moves through the wires, resistors, etc. of the external circuit) as the charge makes its way back to the - terminal, where the whole process happens again.

    -2- Energy is required to create the current. Energy is lost to heat when a DC current flows in a wire. Energy may also be lost to EM radiation when an AC current flows in a wire, if the AC frequency is high enough for the wire to start acting like an antenna.

    -3- What can you tell us about how transformers work? What are the equations for voltage and current and impedance transfer ratios? What are self inductance and mutual inductance?
     
    berkeman, Mar 29, 2007
  5. Mar 30, 2007 #4

    Zeus12345

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    Thanks for the great replies, very much appreciated. Sorry about posting in the wrong place :blushing:
    So this would be a form of power loss in electrical transmission lines? My course only mentioned loss due to resistance and induction of eddy currents in the pylons. I take it that a DC current carrying conductor does not expend energy through the formation of electrical and magnetic fields (not alternating EM).
    Well I know the formula for the ratio of input voltage to output voltage depending on the number of turns in each coil. I understand that a force is applied to the electrons in the second coil resulting in the polarity of the conductor and hence current. I also get that this current flow induces a magnetic field that opposes the one that created it, creating 'back emf' in the primary coil. I only get lost when it comes to the transfers of energy occurring. From what I've read I believe that the primary coil will be transforming the same quantity of energy irrespective of whether the second coil is present. However the presence of the second coil and the 'back emf' would suggest that it's presence would result in less current flowing through the primary coil. So I get lost with the concept of how the primary coil 'transfers' energy to the secondary coil despite the primary magnetic field not requiring energy for its formation.

    Please understand that I'm not lazy and have earnestly tried to find the answers to these questions but am still at a loss.
    Again, any help us much appreciated.
     
    Zeus12345, Mar 30, 2007
  6. Mar 30, 2007 #5

    berkeman

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    Staff: Mentor

    It's quite apparent that you're not lazy, and you are asking good questions in your learning. With coupled inductors (like a transformer), there is self-inductance and mutual-inductance. The mutual inductance (or coupling coefficient) represents how tightly the two windings are coupled magnetically. If all of the flux from one coil goes through the 2nd coil, then the coupling coefficient is 1.0, which would be an ideal transformer. Real transformers with cores made of an appropriate magnetic material (for the frequencies involved) come pretty close to having a coupling coefficient of 1.0.

    When the coupling coefficient is high, and when there is a load on the secondary, then you are correct that the output current induces a back-flux in the core, which gives you a back emf in the primary. This effectively lowers the input impedance of the primary, and requires more primary current to support the input voltage. Since you know how the turns ratio affects the voltage and current ratios, what can you say about the impedance ratio? That is, if you have an output secondary load of R, and primary:secondary turns ratio of n, what is the effective load seen when looking into the primary?

    So now if we put the two concepts of reflected load impedance and coupling coefficient together, I think we have the answer to your question. If the coupling coefficient is high (close to 1), then the primary and secondary are tightly coupled, and the primary current generates a flux that generates a secondary current through the load impedance, and this current generates a back-flux that requires more primary current to support the primary and secondary voltages. The closer the coupling coefficient is to 1.0, then the more this system acts like an ideal transformer. But as the coupling coefficient decreases (like the adjacent coils you mentioned without a common core to confine the flux to the two coils), then there is less flux linkage between the two coils, and there is less current induced in the secondary coil by the primary flux. So less current induced in the secondary coil means less back flux to generate the back emf, and the transformed load impedance is higher than it would be with tightly coupled coils.

    Does that make sense? Good questions!
     
    berkeman, Mar 30, 2007
  7. Mar 30, 2007 #6

    Zeus12345

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    Thanks so much, that makes lots of sense. When discussing it in class and reading about it, it never really brought in the electrical impedance aspect and it appears that is the missing link :biggrin:
    Thanks to you I think I finally have the main concepts under my belt,
    Cheers!
     
    Zeus12345, Mar 30, 2007
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