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Homework Help: Charging a capacitor with AC voltage

  1. May 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Can somebody explain how it is possible to charge a capacitor using ac voltage? The way I see it is the capacitor wouldn't have time to charge as the voltage is constantly switching. In the case of 60Hz - 120 times a second.

    I seen a schematic of a tesla coil where the capacitor is in series with a primary coil the source is a transformer. A spark gap is in parallel with the source and capacitor/transformer. There were no diodes.

    2. Relevant equations

    3. The attempt at a solution
    Last edited: May 27, 2012
  2. jcsd
  3. May 27, 2012 #2


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  4. May 27, 2012 #3
    Capacitor stores charge and not current. In genereral,you must convert AC into DC using rectifier diode and then charge the capacitor.

    But, if u connect a capacitor with AC source,then it will alternately get charged and discharged determined by frequency of the source. This is obviously because the current isnt in one direction. So it cant be charged i.e charge cant be retained/stored in capacitor in AC curcit
  5. May 28, 2012 #4
    just consider for example the AC voltage where the the voltage is given by V(t) = Vsin(t), for the duration of the positive part of the wave from 0 to pi/2 the capacitor would be 'charging' for that whole process and be at maximum charge at pi/2.

    then from pi/2 to pi, the voltage will be decreasing across the capacitor and so the charge will be decreasing. this will be the same process for the other bit of the wave except it will be charged the other way round eg the plate that was originally positive will be negative from pi to 2pi .

    So the capacitor is still charging just for very short intervals depending on the frequency
  6. May 28, 2012 #5
    Yep. I get the picture now. I just didn't appreciate the speed at which this is happening. Basically the spark gap must be firing twice per cycle. I wonder why the cap is there in the first place? Is it because of w=.5cv^2? A quick release of energy into the coil.
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