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Power dissipated from a capacitor?

Discussion in 'Electronics & Electrics' started by RSDXzec, Oct 7, 2012.

  1. RSDXzec

    RSDXzec Member

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    It is my understanding that capacitors don't dissipate power, they just store energy and release it back. I need to calculate power dissipated across a circuit for an assignment and when I asked my lecturer about the caps he said to treat them as impedance and multiply by the current through it much like you would with a resistor. Just wanted to know what others here thought, resistors constantly dissipate power when current flows through them but a cap just stores and releases energy so i'm not sure what to do here...

    cheers.
     
  2. ChoppedLiver

    ChoppedLiver Member

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    Home work ???

    ALL devices dissipate power.
    Think about what a capacitor will allow 'through' it.
    Think about what a capacitor won't allow 'through' it.

    Maybe look at some capacitor electrical specifications for a better understanding.
     
  3. OP
    OP
    RSDXzec

    RSDXzec Member

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    I wouldn't consider it homework since it's an extra bit of work that I've decided to do for my investigation... nor am I in highschool... so I don't get "homework".

    I know as the cap charges it allows less current to flow through it which would be the impedance I should calculate, so I'll just go with that.
     
  4. P.YO

    P.YO Member

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  5. RobRoySyd

    RobRoySyd Member

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    The impedance doesn't cause the capacitor to get hot as the V and I are not in phase. Only the parasitic series and parrallel resistances will cause work to be done.
     
  6. 2xCPU

    2xCPU Member

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    How exactly is that not homework?

    You seem to be having trouble differentiating between ideal (perfect) components and the real world.

    Pick a random brand, search the data sheets for a random capacitor and search the characteristics to get your answer.
    If the terms aren't self explanatory, you haven't been paying attention in lectures.

    2.
     
  7. OP
    OP
    RSDXzec

    RSDXzec Member

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    it's not homework because it's not work I have to hand in
    this stuff was not taught in this subject and I'm doing extra purely out of interest.
    I don't appreciate being talked to like a child

    thanks to everyone else though for helping out, I've already done all the calculations I need so it's all sorted.

    Cheers.
     
  8. 2xCPU

    2xCPU Member

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    Am I the only one who sees a conflict in these statements?
    It seemed a fairly basic question. Sounded like you wanted a "hold your hand" type answer.

    For interest, what level are you studying?

    2.
     
    Last edited: Oct 7, 2012
  9. The Wolf

    The Wolf Member

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    Yes you are.
    He was required to calculate power dissipated across a circuit. Subsiquent to that he questioned the lecturer as to how he would extent the learning objectives of the task to other components. Mainly a capacitor


    The power dissipated across a capacitor:
    P=I(squared)R
    I = the rms current through the capacitor
    R = The Series Resistance of the capacitor.

    I= 2 (pie) FcV,
    F=The frequency in Hertz
    c=The capacitance in Farads
    V=The rms voltage
     
  10. 2xCPU

    2xCPU Member

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    (spelling corrected)
    That's not what is in the first post.
    Clearly "the caps" are part of the assignment circuit and not some extracurricular learning the OP has decided is worthwhile.

    2.
     
    Last edited: Oct 7, 2012
  11. stupot

    stupot Member

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    I think it is reasonable to provide the OP with some helpful advice. In the end, he will need to understand and be able to solve the question himself, whether he recieves advice here or somewhere else.

    My tip:
    consider that the power dissipated may not be 'real'. Also, as noted above, there is always some parasitic resistance in electrical componentry.

    Good luck.
     
  12. dakiller

    dakiller (Oscillating & Impeding)

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    Ideal capacitors don't dissipate any power. In circuit analysis, if the ESR or leakage of the capacitor is important, it should be added to the model as separate elements and be properly defined otherwise you just have pure conjecture.

    If it was homework, and no ESR or leakage was mentioned or given, I would always assume that the capacitor was ideal, otherwise you are never going to get anywhere if you have to deal with all real world parasitics, as you can never account for them all.
     
  13. paulvk

    paulvk Member

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    Real power is disipated in a capacitor due to resistance in internal wiring, stress in the dialectric material and resistance of the material that forms the plates of the capacitor.
     
  14. rainwulf

    rainwulf Member

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    Impedance is not resistance.
    However, impedance can be replaced by resistance in calculations to produce a wattage or power.
    Impedance however is directly related to the frequency of the signal across the capacitor.

    For example, putting DC across a capacitor will create a short current flow as the cap charges, but then nothing, hence, impedance is infinite (apart from leakage resistance which depends on a whole bunch of factors)

    However, ac across a capacitor will cause a small amount flow, with an associated PD across the terminals. Then you can use P=VI.
    "I" however is the resultant current through the cap, which is directly related to the frequency.

    All THAT however is not at all related to actual LOSSES through a capacitor, since there is no such thing as a perfect capacitor. You have resistive, capacitive and inductive losses.


    I am pretty sure this is all correct, but my ac and capacitor theory was a long time ago in terms of beers and years.
     
    Last edited: Nov 5, 2012
  15. mtma

    mtma Member

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    In terms of theoretical impedance, only resistance can actually dissipate power.

    Inductance and capacitance hold energy and give it back one way or another.

    So in a practical component model such as one of a capacitor all of the losses will occur in the resistances of the model. The parasitic inductances and capacitances may contribute to the loss but only if they cause current to flow through a resistance.

    This is also not to say that a practical component model includes all of the applicable losses - the parasitic effects, which give back the energy can give it back as external waveforms which then dissipate in other components external to the capacitor.
     
  16. zero_velocity

    zero_velocity Member

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    It all depends if the circuit is using AC or DC sources.

    If DC, you must consider the caps as open circuits (or infinite impedance)

    If AC, then you must calculate the impedance of each cap in accordance with the frequency used.

    If the caps are expressed in farads, then you must do conversion, otherwise if they are given in an ohmic form, you can treat them as resistors, however it is worth noting for these sorts of problems that for DC circuits the caps wont 'dissipate' any power.
     
  17. paulvk

    paulvk Member

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