capacitors

Discussion in 'General Electronics Chat' started by kaz21, Jul 10, 2011.

  1. kaz21

    Thread Starter New Member

    Jun 11, 2011
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    How to derive a mathematical relationship for the voltage across a capacitor in the transient stage of charging the capacitor??The voltage source is a dc voltage source.
     
  2. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    1,607
    A device that holds a charge Q and has voltage V across has a capacitance of:

    C = \fra{Q(t)}{V(t)}

    Solving for Q(t):

    Q(t) = C V(t)

    Noting that the total charge is the integral of current, or

     \int I(t) dt = Q(t)

    We can substitute for Q(t):

     Q(t) =  \int I(t) dt = C V(t)

    If we differentiate by t we get the traditional relationship between voltage and current:

     I(t) = C \fra {dV(t)} {dt}

    Solving for dV(t) :

     dV(t) = \fra {1} {C} I(t) dt

    Integrate both sides:

     V(t) = \fra{1}{C}\int I(t) dt (limits of -∞ to t)

    (I somehow suspect this is not the answer you were looking for.)
     
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