Capacitor Energy Storage Math Problems

Thread Starter

Austin Clark

Joined Dec 28, 2011
If I try and calculate the energy (J) stored in a capacitor algebraically, I get J = CV^2
However, I know that in reality J = (CV^2)/2
I know this because that's the answer I get when I solve it graphically.
What's causing this contradiction? Correct mathematics will always come out consistently, so what am I doing wrong?

Thread Starter

Austin Clark

Joined Dec 28, 2011
I think I answered my own question.
I was confusing instantaneous voltage with average voltage, or something along those lines. I realized that, even though J = QV (where Q is charge), the Voltage across the capacitor isn't constant, so each charge receives a different amount of energy as the capacitor charges.

Even still, I'd like to see the math worked out, It'd make more sense then. Anyone care to give it a shot?
Last edited:


Joined Mar 6, 2009
If a source is charging a capacitor and at any instant the current is i(t) and the voltage is e(t) then the instantaneous power delivered by the source would be

\(p(t)=e(t) \times i(t)\)

We also know the relationship between capacitance C, capacitor current i(t) & instantaneous capacitor terminal voltage e(t) is ...


The energy delivered by the charging source to the capacitor Wc is the integral of the instantaneous power over time or ...

\(W_c=\int_{-\infty}^t p(t) dt=\int_{-\infty}^t {e(t) \times i(t) dt}=\int_{-\infty}^t {e(t) \times C\frac{de}{dt} dt}\)

It's normally reasonable to permit a change in the limits of integration with the change of variable [from t to e] within the integration. If the capacitor voltage at time t=-∞ is zero and at time t it is E volts, then one may write ...

\(W_c=\int_0^E C e(t) de=\frac{1}{2}CE^2 \ Joules\)