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RLC circuit voltages

  1. Aug 11, 2011 #1

    vrc

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    oké, I'm ghaving a hard time with this one, bt it should be really simple acutally:

    picture shows RLC circuit:

    RLC_series_circuit.jpg


    by kirchhoff law we can stat that:

    +V-R*i(t)-(1/C)*integral(i*dt)-L*(di/dt)=0

    I do understand V, R*i(t) and )-(1/C)*integral(i*dt), because those voltage physically are there with that polarity ( they have given energie by the work done by the source

    However, the voltage across the coil, i don't get, L*(di/dt), this is the formula for de induced voltage by the coil itself, because i(t) is function of time and there fore by de law of Lenz must be in induced currect to oppose the increase of induction (given by di(t)).

    So why they don't write down: Ecoil=E0-L*(di/dt), because the voltage E0 wich was already there has a different polarity than the induced voltage (induced voltage is producd by the coil...)

    Am I totally confusing funamentalistic issues here ?

    thank !
    grtz
     
  2. jcsd
  3. Aug 11, 2011 #2
    Before the circuit is connected there is no Ecoil or E0 so you cannot say that the voltage on the coil was already there.

    Connect the circuit to the voltage V, and now you have 3 things happening.

    The voltage on the resistor is equal to i(t)*R - that is easy.

    The voltage on the capacitor starts out as zero and gradually increases as the capacitor charges or Vc = 1/C * integral ( i*dt) . You can also say the current going into the capacitor is C*dv/dt. The current starts out at some value I.

    For the inductor, the current starts out as zero, but the voltage starts out depending on the other values of R at some value Ecoil. What is Ecoil.
    Well the voltage on the capacitor is zero as it is just beginning to charge.
    The inductor opposes any change in current, so the starting current is zero.
    Thus the voltage across the resistor is zero.

    So your equation reduces to V = Ecoil.
    Or, V=Ecoil=Ldi/dt.

    If as you suggest what they write down, you would have
    v = Ecoil=E0-Ldi/dt.
    which doesn't make much sense now. ( ie V-E0 = 0 = Ldi/dt )

    I agree that it should be simple but for some reason it takes more brain power to figure out inductors than capacitors.
     
  4. Aug 12, 2011 #3

    vrc

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    hey,

    first thank you for you're respons, grateful

    suppose in my picture the start voltage of de ac source at t= 0 is has a + pole above, so the start currect goes clockwise

    as I understood:
    * t=0 the currect is only dependt on the value of R , because the capacitor must charge first
    and de inductor will not react because there wille be no change in currect

    * because of the V/R , a start currect will flow, so the coil willl have a voltage across it( it's also just a wire = resistance) ,
    called Ecoil; because we know that a difference in current will produce a difference in magnetic field (produced by the coil) and
    this will produce a current

    * the direction of the current will be that it opposes the the difference in magnetic field (produced by the first difference in current, V/R), where the current flows to , we call the positive pole

    * so calling Ecoil=L*di/dt, the first voltage on the inductor Ecoil must bequal to the induced voltage, I do understand that becaause an intial current di will give an initial voltage dVcoil, so the opposed voltage dVL will be equal to it

    so in the circuit the call VL=-L*di/dt, because of de induced voltage has a oppossed voltage

    and what if there was already a voltage across the inductor ?

    sorry for all my question, but I do want to get my head clear of this, important stuff...

    thank you !
     
    Last edited: Aug 12, 2011
  5. Aug 12, 2011 #4

    vrc

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    it's hard to explain my confusing, so I did a little drawing,
    I wrote most of the way I see it, the blew + pole at the inductor is because the di' will flow to th resistor, so create (it's ofcource created voltage) an electric field

    Only do I not understand why in the formula for the inductor voltage they used the notation
    -L*di/dt...


    RL.jpg

    I hope this is a correct intertretation !

    thank you

    grtz
     
  6. Aug 14, 2011 #5
    You need to use Faraday's law to properly understand inductors.

    What is often misunderstood about inductors is that there is no voltage drop across an ideal inductor, it imposes an EMF in the opposite direction. Kirchhoff's law does not hold for inductors.

    What does hold is:
    [tex]\oint_C \vec{E} \cdot \vec{dr} = -\frac{\partial \phi_B}{\partial t}[/tex]

    In the case of an inductor
    [tex]-\frac{\partial \phi_B}{\partial t} = -L \frac{\partial i}{\partial t}[/tex]

    Hopefully this makes sense, just remember that Kirchhoff's law does not hold for inductors.
     
  7. Aug 15, 2011 #6

    vrc

    User Avatar

    my course state that for a dc source whit the switch closed at t=0:

    E-L*(di/dt)-Ri=0

    Mau I also assume that an ideal conductor has no resistantce,otherwise a voltage would be acrros it!
    There I didn't saw my fault, because I was thanking about an inductor as a comonent with a resistance. Ofcourse the ideal conductor does not exist because there always will be some resistance.

    thank you vm!
     
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