Discussion in 'General Electronics Chat' started by gorgondrak, Jan 12, 2015.

  1. gorgondrak

    Thread Starter Member

    Nov 17, 2014

    I'm having difficulty grasping this new concept of capacitance value times voltage per second for total current. I would like to know how this relates to the method of calculating the current using the RC time constant and the universal time constant equation or calculating its peak current using its capacitive reactance in an ac circuit.
  2. WBahn


    Mar 31, 2012
    The defining relationship for a (linear) capacitor is

    C = Q/V

    Namely that the capacitance is, by definition, equal to the charge stored on the capacitor (+Q on one plate and -Q on the other) when a voltage V appears across it.

    Current is, by definition, the rate at which charge is flowing past a given point (or through a given area).


    I = ΔQ/Δt = dQ/dt (the latter being the derivative of charge with respect to time, the calculus way of expressing it).

    If we change the charge on a capacitor by a ΔQ, then the voltage will change by an amount ΔV and these will be related by

    ΔQ = C·ΔV

    or, in calculus terms:

    dQ/dt = C·(dV/dt)

    If we use the Δ notations, then we have

    I = (C·ΔV)/Δt = C·(ΔV/Δt)
  3. wayneh


    Sep 9, 2010
    In words, the rate of flow of charge - the current - is proportional to the rate of change in voltage across the capacitor, dV/dt, scaled by the capacitance of the capacitor C.

    This makes sense since Q and V are directly proportional for a linear capacitor, scaled by the capacitance.