# Energy storage in a capacitor

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#### v9260019

##### Member level 2 If let p=d/dt, 1/p=∫da from -∞ to t then the power absorbed by a capacitor is
p(t)=v(t)i(t)=v(t)Cpv(t)=p[1/2*Cv(t)^2]=pw(t)
Look at the last two terms and apply the 1/p operator from the left. Then ,assuming that w(t) is zero for t<=0 along with voltage and current, we have
w(t)=1/2*Cv(t)^2
"Then ,assuming that w(t) is zero for t<=0 along with voltage and current"
why need this assumption ????

#### gliss Probably to tell you there is no previous charge on the capacitor.

#### v_c Note that you are performing the integral from $-\infty$ to $t$ and that integration is being split into two integrals
$W(t) = \int_{-\infty}^{t} v_C (t) i_C(t) dt = \underbrace{\int_{-\infty}^{0} v_C (t) i_C(t) dt}_{Initial Energy} + \int_{0}^{t} v_C (t) i_C(t) dt$
The result of the first integral is the initial energy (which can be denoted as $W_C(0)$), which need not necessarily equal zero. But if the initial capacitor voltage is zero then it will be. The second integral is solved by using the fact that $i_C(t) = C \frac{dv_C(t)}{dt}$.
$W(t) = W_C(0) + \int_{0}^{t} v_C C \frac{d v_C}{dt} dt = W_C(0) + C \int_{0}^{t} {v^2_C} C dv_C = W_C(0) + \frac{1}{2} C {v^2_C(t)}$
You could also define change in the capacitor energy as $\Delta W_C(t) = W_C(t) - W_C(0)$ which would just give you $\frac{1}{2} C {v^2_C(t)}$ no matter what the initial energy is since it is a difference.

Best regards,
v_c

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