1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A circuit with changing EMF

  1. Aug 2, 2008 #1

    mtr

    User Avatar

    1. The problem statement, all variables and given/known data
    There is a circuit which consists of a capacitor with capacity C, a diode and a source of changing EMF such that during time [tex]\frac{1}{2}T[/tex] EMF equals E and during another half of T it is -E (current flows in the opposite direction). What will be the voltage on the capacitor after a long time?


    2. Relevant equations
    [tex]C=\frac{Q}{U}[/tex]


    3. The attempt at a solution
    First of all, I don't understand what is the role of the diode in this circuit. Does it only influence on the direction of the current? And in fact I do not have any reasonable idea how to solve this whole problem.
     
  2. jcsd
  3. Aug 2, 2008 #2
    A diode is a device which allows current in one direction, but not the other. Thus, the capacitor can be charged, but not discharged.
     
  4. Aug 2, 2008 #3

    mtr

    User Avatar

    So after a long time the capacitor is charged completely, but still how can I find the voltage? Does it equal EMF? Or maybe I should find Q somehow and use [tex]C=\frac{Q}{U}[/tex]?
     
  5. Aug 2, 2008 #4

    Defennder

    User Avatar
    Homework Helper

    You didn't provide the angular frequency of the voltage source. What 1/2 T here? The period of the sinusoidal AC voltage source? Is the source sinusoidal in the first place, or is it a square wave?

    Note that the voltage across the capacitor and the source is always the same at all times because it is connected across the capacitor.
     
  6. Aug 3, 2008 #5

    mtr

    User Avatar

    Actually, the EMF is produced by a pendulum in a magnetic field, with period 2T.

    Isn't there any drop of voltage that is connected with the diode?
     
  7. Aug 3, 2008 #6
    In an ideal diode there is no voltage drop. In a realistic case you should know the internal resistance of the diode, but it shouldn't be much greater than that of connecting wires, etc.
     
  8. Aug 3, 2008 #7

    mtr

    User Avatar

    So it seems, that I have to find the maximum value of the EMF and it will be the voltage on the capacitor. Am I right?
    On the other hand what is the capacity C for? It is among task's data, but it hasn't been used yet.
     
    Last edited: Aug 3, 2008
  9. Aug 3, 2008 #8
    I came to the same conclusion myself.
     
  10. Aug 3, 2008 #9

    mtr

    User Avatar

    What about C?
     
  11. Aug 3, 2008 #10
    Strange problem... Anyway, you can use dimensional analysis arguments to show that you can't use C (or, more likely, [tex]1/\omega C[/tex]) and E to obtain a voltage, because capacity introduces another dimension of ohms, which doesn't cancel with anything. Probably, if you had a resistor somewhere, you could and should incorporate C somehow.
     
  12. Aug 3, 2008 #11

    mtr

    User Avatar

    I don't exactly understand what you mean. How can we obtain a voltage using C? And is E the EMF or the stored energy in the capacitor?
     
  13. Aug 3, 2008 #12

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    No circuit diagram was provided, but I assume the diode and capacitor are connected in series. The EMF is applied to that series combination, not directly to the capacitor. This is the typical way to introduce the concept of a rectifier.

    The forward voltage drop of a real diode is typically 0.6 to 0.8 V in practice. It's not clear if they are assuming an ideal (Vf = 0) or a real diode though.
     
  14. Aug 3, 2008 #13

    mtr

    User Avatar

    That's right, the capacitor and the diode are connected in series. I suppose they are considering here and ideal diode, otherwise there should be something about its properties.
     
  15. Aug 3, 2008 #14

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Yes. :smile:

    At least that's right for an "ideal diode" that has no voltage drop across it.

    For a real diode, the capacitor voltage would be about 0.6 to 0.8V less than E.

    E = Vdiode + Vcap
    Vcap = E - Vdiode

    Ideal diode:
    Vcap = E - 0 = E

    Real diode:
    Vcap = E - (about 0.6 to 0.8 V)
     
  16. Aug 3, 2008 #15

    mtr

    User Avatar

    That's true, but still there is the capacity which wasn't used. I think that there might be something about impedance, as Irid suggested.
     
  17. Aug 3, 2008 #16
    That's the point, we can't. Look at it this way. You have some physical quantities (E, C, ...) which are measured in some units (V, F, ...). Now you must combine these quantities in some way to obtain the voltage across capacitor - U. How can you do it? It's trivial to see that in this problem the only possible way is to make

    [tex]U=E[/tex]

    up to a dimensionless constant. You can't plug in C anywhere here, because it will ruin the dimensions. Volts must equal Volts, and unless the Farads have some other Farads to cancel with, they have no place in the answer.
    If you still have doubts, search the web for dimensional analysis.
     
  18. Aug 3, 2008 #17

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    What this circuit does is, given an AC voltage source, a DC voltage is created which can be used to run DC electronic devices.

    For this homework assignment, it only matters that C > 0.

    The value of C comes into play once you connect something which will draw current. Drawing a current will reduced the charge, and hence the voltage (=Q/C), during the half-cycle where the EMF is negative. It's part of electronic design to choose a large enough C so that the drop in voltage is acceptably small.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A circuit with changing EMF
  1. EMF and circuits (Replies: 2)

  2. EMF and Circuits (Replies: 6)

Loading...