# Capacitor Energy Storage Math Problems

Discussion in 'General Electronics Chat' started by Austin Clark, Aug 30, 2012.

1. ### Austin Clark Thread Starter Active Member

Dec 28, 2011
409
44
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?

2. ### Austin Clark Thread Starter Active Member

Dec 28, 2011
409
44
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: Aug 30, 2012
3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
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 ...

$i(t)=C\frac{de}{dt}$

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 ...

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

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