# Mechanical and electrical power of discharging a capacitor

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?

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

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.