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Capacitor - working

  1. Jun 2, 2006 #1
    How can we explain the blocking of a dc current by a capacitor and passing through of ac, physically, not mathematically. In case of dc we know that the capacitor will initially act as a short circuit till it gets charged fully and then naturally acts as an open circuit since it cannot be further charged and hence no current can flow through it. Similarly, how can we explain the same in case of ac?o:)
  2. jcsd
  3. Jun 2, 2006 #2
    Usually I just recall the impedance of the cap.
    [tex]Z= \frac {1}{j \omega C}[/tex], so at low frequency [itex]\omega = 0[/itex], then Z is big.
  4. Jun 2, 2006 #3
    Thanks corneo, but it is the mathematical explanation. I want to understand it physically as i said earlier.

  5. Jun 2, 2006 #4
    ***(be careful to not search for answers that are so fundamental or "physical" that you end up ignoring valid explanations. I only say this because I myself often over analyze things and dig so deep that I ignore the treasure I already dug up. In fact I recently did this about a topic related to the one you are asking about)***

    Now for an attempted answer....


    It takes time for a capacitor to develop a voltage across it because the voltage is produced by a displacement of charge on the respective plates...the capacitor "looks" like a short initially at DC because there is no electric field built up between the plates to resist current flow.

    As charges flow onto one plate and off on the other one an electric field is established to resist current flow. Eventually this electric field (or voltage) is large enough to prevent any more charge from moving to or from the plates....so at this point the capacitor "looks" like an open circuit.


    Here we have a kind of tug of war between the charges.....the charges are constantly added and removed from the plates as the current direction changes..so you could think about it like the capacitor never really has the chance to accumulate charge on any one plate because you keep taking it off as soon as it gets there.

    Another thing to realize is that the plates voltage levels are related to one another through the electric field between them ....so if you rapidly change the voltage on one plate(like grounding one plate suddenly) you have essentially created an oscillation in the voltage level. This creates an oscillation in the electric field which translates into an oscillation in the voltage level of the other plate. You can not instantaneously force the voltage across the plates to change, because if you try to do so without taking the time to transport charge the electric field will adjust itself to keep the voltage across the plates the same.

    Hopefully that helps.

  6. Jun 2, 2006 #5


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    The complete understanding comes from looking at just the DC case a little carefully. Why exactly is DC "blocked"? What does that really mean?

    Let's concern ourselves with the positive terminal of a power supply (or battery) which is connected to one plate (or electrode) of the capacitor. The supply terminal is always at some fixed positive voltage +V (for a DC supply). Initially, the capacitor plates are at zero voltage. When the power is turned on, or a switch is thrown, a current flows between the battery terminal and the capacitor plate, because there is a voltage difference between them (Ohm's law). The current is nothing but the rate of flow of charge - so charge travels from the battery to the capacitor plate. Having nowhere else to go (since the one capacitor plate is not electrically connected to anything else), the charge starts building up on this plate. As the charge builds up on the capacitor, the voltage between its plates also increases at the same rate (V=Q/C). After a while, the capacitor plate has reached a voltage that's essentially equal to +V. Now, there's no voltage difference between the battery terminal and the capacitor plate. Hence, no current will flow between them. Thus, with a DC supply, there will be practically no current flow after a very short while. This can be pictured as the the capacitor trying to reach the battery voltage and force the current to zero.

    If you had an AC supply, the positive terminal doesn't stay at a constant voltage anymore - it keeps changing. So, by the time the capacitor voltage reaches what was once the terminal voltage, the actual terminal voltage has changed. The capacitor now has a new target to chase. This game goes on endlessly as long as the terminal voltage keeps changing. So, since there's always a difference in voltage between the battery and the capacitor, there's always a current flowing between them.
    Last edited: Jun 2, 2006
  7. Jun 4, 2006 #6
    Thank you

    Thank you all for ur responses. It is now clear to me. These were some good and solid answers.
  8. Nov 22, 2007 #7


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    Hi all,

    I am also crazy enough to think on those lines.

    I went through the above explanations and found it useful.

    But one thing still is not clear to me. If we see that it requires time for charge to move across and produce a drop in voltage, why is that a capacitor with lower farads ( lets say pf) has higher cut off frequency than a cap with few more farads (lets say nf). Here i assume that permitivity and the area of plates is same. distance is the only factor in consideration. I believe that having larger distance would reduce the cap and hence would require more time for charges to move across.

    Let me know if I am wrong...

  9. Nov 22, 2007 #8
    That sounds like a resonable explanation to me. I would think of a simple RC circuit with the time constant tau = RC. If C is smaller, then the time constant is smaller, which would relate to a higher cutoff frequency...
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