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Capacitor Charge Discharge Clarification

  1. Jun 12, 2010 #1
    Ok so i am just trying to clarify a few points about capacitor charge, discharge, i was wondering if anyone could help.

    1)In charging, am i right in thinking current decreases exponentially, charge increases exponentially, and voltage increases exponentially.?

    3) Am i write in thinking in discharging, all three quantities decrease exponentially.?

    4) if current decreases exponentially in charging, and then decreases exponentially in discharging how is this possible?.

    5) By removing the power supply and connection to the capacitor, would it discharge, or do you have to discharge through a resistor?

    Thanks in advance
     
  2. jcsd
  3. Jun 12, 2010 #2
    Can no one answer this?
     
  4. Jun 12, 2010 #3
    In the real world there is always some resistance in series with the capacitance.
    EDIT remove text in italics.
    It is the combination of resistance and capacitance that produces the logarithmic or exponential charge/discharge curves.

    You have not quite got the increase decrease or the curves correct so I have sketched some in the attachment.

    note the difference betwen exponential and logarithmic increase
    and the fact the the currents for charge and discharge flow in opposite directions, but the voltage and total charge graphs remain in the first quadrant.

    Consider the experiment in the circuit.

    A timer is set to zero and the switch set to position 1. This charges the capacitor through resistor R from battery voltage E. The current is monitored by meter A and the capacitor voltage by Meter V.
    Current flows into the capacitor and is reckoned positive.

    Graphs of the charging current, voltage and total charge are shown along with their equations. The capacitor commences charging with zero voltage across it, but maximum charge current Ie.

    After a time, t, the switch is set to the centre disconnected position.
    The capacitor now has a total charge Qf and a voltage Vf, both of which are positive. the current Ic has fallen to somewhere near zero.

    The switch is now set the discharge position (3).
    Current now flows out of the capacitor and is therefore reckoned negative.
    At the outset a large current flows as the full capacitor voltage appears across the resistor.

    As the capacitor discharges this voltage diminishes so the current falls.
    Also the charge held in the capacitor diminishes.
    Both the voltage and charge are still positive.

    Graphs and equations for discharge are also shown.
     

    Attached Files:

    Last edited: Jun 13, 2010
  5. Jun 12, 2010 #4

    gabbagabbahey

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    Your logarithmic curves make no sense whatsoever. For a simple RC circuit, the voltage across the capacitor and the current through the resistor are related by Ohm's law and Kirchoff's voltage rule [itex]V_{\text{c}}=IR[/itex] (when charging, the input/battery voltage must be included in the equation: [itex]V_{\text{in}}=IR+V_{\text{c}} [/itex] ). So, if the current is an exponential (and it is!) so is the voltage across the capacitor.
     
  6. Jun 12, 2010 #5

    gabbagabbahey

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    Yes to both questions.

    Only the direction of the current changes.

    It would discharge with or without a resistor. If you somehow managed to build a circuit with no resistance, and just a capacitor (with no internal resistance), the discharge would be almost instantaneous. However, even with superconducting wires, I don't think anyone has managed to build a capacitor with no internal resistance, and since the internal resistance effectively acts in series with the capacitor you always really have an RC circuit (not just a C-circuit).
     
  7. Jun 12, 2010 #6
    Are you asking about the curve f(t) = ket/tau, where k and tau are positive constants?

    Neither the charge on the capacitor, nor the voltage across it, fit this curve.
     
  8. Jun 12, 2010 #7

    gabbagabbahey

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    Of course not, that's an increasing exponential.
     
  9. Jun 12, 2010 #8
    Of course. However, this is how question one reads.
     
  10. Jun 12, 2010 #9

    gabbagabbahey

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    Oh, right.

    @ojsimon, The voltage and charge across the capacitor during charging are increasing, and there is an exponential involved,

    [tex]V_c=V_{\text{in}}\left(1-e^{-\frac{t}{RC}}\right)[/tex]

    but that's not quite the same thing as "increasing exponentially" (Like [tex]e^{\frac{t}{RC}}[/itex] ).
     
  11. Jun 13, 2010 #10
    Ok thanks everyone that has clarified my points i think....
     
  12. Jun 13, 2010 #11
    You are absolutely right Gabbagabbahey. The use 'logarithmic' was inappropriate so I have amended post#2 to reflect this.

    However the graphs were correct.
    Nevertheless I have taken the opportunity to improve the sketch in post#2
     
  13. Jun 13, 2010 #12
    From your new sketches it looks like current increases when it is discharging, doesn't this contradict what was previously said. But thanks again for the sketches they make this point really clear.

    Thanks
     
  14. Jun 13, 2010 #13
    I'm sorry if I haven't managed to make this point clear.

    The discharge current is necessarily in the opposite direction to the charge current.
    When you discharge the capacitor current flows out; when you charge it current flows in.

    So I have shown what you would read on the ammeter. The charge current would read positive and the discharge current would read negative.

    So the discharge current starts from an initially high absolute (negative) value and the absolute value (magnitude) of current decreases towards zero.

    That is why it is plotted on the negative part of the current axis.

    Hope this makes it clear.
     
  15. Jun 13, 2010 #14
    Your drawing and equations look good to me Studiot. For the charge across the capacitor, you're missing a "1" sub-subscript on the IE. But why not replace the RCIE1 term with EC instead, as the asymtotic charge is independent of R?
     
  16. Jun 13, 2010 #15
    You can indeed use Ie = E/R to sunstitute in my expression if you like.

    I just did it that way because each required quantity follows the same pattern.
     
  17. Jun 13, 2010 #16
    Yeah thanks a lot that has clarified this point completely.
    Cheers
     
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