waveforms capacitor voltage / current

Discussion in 'General Electronics Chat' started by kk57, Oct 24, 2007.

  1. kk57

    Thread Starter New Member

    Oct 24, 2007
    1
    0
    I am looking at a simple ac source in series with a capacitor. In the book I have it says that "initially when the capacitors plates are uncharged,they will not oppose or react against the changing current and therefore maximum current will flow and reactance will be low. As the capacitor charges it will oppose or react against the charge current , which will decresase, so the reactance will increase." i look at the capacitor voltage and current waveform and that seems right.Then the book says The discharge current is also hiest at the start of discharge as the voltage of the charged capacitor is also high. the confusion start there. When I look at the waveforms just agter the positive peak of cap voltage the current is minimum not max.Can you straighten me out. Also have had no calculus but can see when cap first starts charging although current is maximum the rate of change of current is minimum and when cap voltage is max current is minimum but its rate of change is maximum. Can you eplain that to me? Thankyou
     
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Is it actually minimum - zero amps, or is it maximum in the reverse direction?
     
  3. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    567
    12
    Consider a switch that goes to +5V and a capacitor. When you close the switch, the capacitor will appear as a short and draw high current. As the capacitor charges, and approaches the supply voltage, it draws less current. You saw this this.

    Now open the above switch, and consider a switch placed across the charged cap. When you close the switch, the capacitor will discharge with high current. As the cap discharges, the available current is less because the voltage drops.

    The discharge current is the current available to the load, it is highest when the cap is fully charged. If the load is constant, say 1K, then 5V over 1K is more current than a partially discharged cap of 3V over 1K.

    BTW -- if you connect a constant-current source to the cap, then you will create a linear ramp based upon:

    dt = C*(dv/i)

    where:
    C = capacitance
    dt = change in time
    dv = change in voltage
    i = current

    For example, plug "1mA", "+5V", and "1000pF" into the above equation. What do you get?


    Answer: a 5us ramp, from 0V to +5V.
     
  4. Murod

    Active Member

    Dec 24, 2005
    30
    0
    I think you observe the voltage and the current of a cap being applied to AC source, and trying to compare it with textbook's transient voltage analysis of cap applied to DC source, and off course it's different. Am I right?
     
  5. techroomt

    Senior Member

    May 19, 2004
    198
    1
    at the instant a cap begins to discharge, the charge volatge will be at the max and the current will be at the max, as the cap discharges both will follow a discharge curve until eventually (7 time constants for 99% discharged) cap charge voltage will reach zero and current wiil reach zero, the cap having dissipated the power into the load.
     
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