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Homework Help: Power dissipation in RC circuit

  1. Aug 4, 2010 #1
    When a capacitor is charged by a battery half the power is dissipated as heat and half is stored as potential energy within the field of the capacitor. Similarly a resistor dissipates all power that the battery supplies.
    Now here's my question: In an RC circuit where is the power dissipated?
    I can show that the heat the resistor dissipates before the arrival of steady state is equal to final potential energy stored in the capacitor with the sum of both the heat and potential energy summing to give the work done by the battery.
    The only plausible conclusion I can draw is that heat is dissipated only by the resistor.If this is indeed correct then why is the capacitor not losing heat? My teacher tells me that the capacitor loses heat because of fringing of its field. Does it somehow disappear in an RC circuit?That would be strange.
  2. jcsd
  3. Aug 4, 2010 #2
    Every real circuit things do have resistance, so in practice, capacitor does lose heat. But you can look at the real capacitor in this way: it has 2 parts, one of which is a theoretical capacitor (which has zero resistance and capacitance C only) and the other is a resistor r. Let's say, the real resistor you put in the circuit is R. Then the theoretical resistance of the circuit is r+R, and the theoretical circuit that you always work with in problems is (r+R)-C circuit.

    About field leakage, I don't think that's the main cause. Strictly speaking, yes, energy is also lost in that way, but that's a negligible amount.
  4. Aug 4, 2010 #3
    In general sense, capacitors store energy in them. Even if fringing might cause some energy to be lost, we consider it to be negligible in comparison to the amount stored in them, because in general, parallel plate capacitors have their cross sectional area quite large in comparison to the distance between. (It is the same approximation as that of a frictionless pulley! It never exists in reality but we can neglect its friction if its a smooth one).
    In an RC circuit when a capacitor has been charged, and a steady state is reached, the amount of energy stored in the capacitor is half of the amount supplied by the battery, so naturally, it is true that the rest of the energy is dissipated as heat through the resistor.
    In that particular derivation that you are talking about, fringing of electric field is neglected, and hence this result is obtained.
    Hope its clear, now!!
  5. Aug 4, 2010 #4
    What you mean to say is that fringing is not the reason why capacitors lose heat when they are charged by a battery(not in an RC circuit)?
  6. Aug 4, 2010 #5
    Fringing (or more exactly, electromagnetic radiation emission) is one of the reason why they lose ENERGY. The other reason is heat dissipation. This applies to both cases (charging and discharging).
  7. Aug 4, 2010 #6
    Is there a difference b/w losing heat and losing energy?
  8. Aug 4, 2010 #7
    I'm not sure what people mean by losing heat, but strictly speaking, energy is the right word. Heat is a way of energy transferring, just like work.
  9. Aug 4, 2010 #8
    Well you are right on spot to say that but I meant heat dissipation in the sense most texts use.
  10. Aug 4, 2010 #9
    Then fringing doesn't have anything to do with losing heat ;)
  11. Aug 5, 2010 #10
    I am confused and I think you should tell me what you really mean. By my understanding the reason for energy dissipation in a capacitor was fringing or electromagnetic radiation emission. If it were negligible as I can gather you telling me, what causes the capacitor to lose energy when charged by a battery? And what causes it not to lose energy when we have a resistor in the circuit too?
    Please I do not have any background knowledge of fringing so pardon me if I am messing up heat and energy.
  12. Aug 5, 2010 #11
    Okay, I'll explain a bit more in details.

    When the charge or discharge takes place, there is current ---> there is change in charge on capacitor ---> there is change in E-field inside capacitor ---> there is induced B-field by that change of E-field ---> the B-field changes with time ---> there is induced E-field changing with time ---> there is B-field and so on. This, as a whole, means there is electromagnetic wave. The wave propagates into the surrounding and carries an amount of energy away. This is one way that energy of the circuit is lost. (1)

    The real capacitor and real circuit always have resistance. Therefore, when there is current ---> there is heat dissipation. The heat goes into the surrounding. This is another way that energy of the circuit is lost. (2)

    So how much energy lost due to (1) compared to (2)? When there is resistance, energy lost in (1) is negligible; (2) is the main reason. However, in the extreme case where resistance = 0 approximately, (1) must be taken into account.

    In conclusion, by the energy conservation law:
    _ Charging: Work done by battery = energy stored in capacitor + energy lost due to both (1) & (2).
    _ Discharging: Initial energy stored in capacitor = energy lost due to (1) + (2).

    So back to your post #10:
    When charged, the capacitor doesn't lose energy. It stores energy gradually. Only battery loses its energy. The energy from the battery = work done by battery = energy stored in capacitor + energy lost.
    When discharging, the capacitor loses energy. It lets its energy out, and the energy goes to (1) and (2).
    Last edited: Aug 5, 2010
  13. Aug 5, 2010 #12
    Actually I did mean the battery and not the capacitor losing its energy to the surroundings.

    Anyways thanks a lot!
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