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I want to find the power of completely discharging a capacitor with capacitance $C$ during a time interval $\Delta t$.

  1. Using the mechanical definition of power as the rate of change of energy $W$ stored in the capacitor:

$P_\mathrm{mech}=\frac{\mathrm{d}W}{\mathrm{d}t}=\frac{\mathrm d}{\mathrm d t}\frac{1}{2}CV^2 $,

where $V$ is the voltage. Assuming that energy changes linearly with time and using $C=\frac{Q}{V}$ with charge $Q$:

$P_\mathrm{mech}= \frac{\frac{1}{2}CV^2}{\Delta t} = \frac{QV}{2\Delta t}$

  1. On the other hand I could use the electrical definition of power and assume that current $I$ flows uniformly across time:

$P_\mathrm{el}=VI = V\frac{\mathrm d Q}{\mathrm d t}=V\frac{\Delta Q}{\Delta t}=\frac{QV}{\Delta t}$

Why do these results differ by a factor of $2$, shouldn't they be the same?

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  • $\begingroup$ If you assume that energy changes linearly, then charge Q is not a linear function of time t and $dQ/dt \neq \Delta Q/\Delta t$. $\endgroup$ – Gec Dec 25 '18 at 13:23
  • $\begingroup$ @Gec Could you elaborate why this is implied? $\endgroup$ – Marvin Bana Dec 25 '18 at 22:35
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As the charge flows out of the capacitor, the voltage difference between the plates decreases.

If the current is constant, the voltage will decrease at a steady rate down to zero, so the average voltage is $V/2$ not $V$. That is why your result was twice what it should be.

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