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While searching for something totally unrelated to this, I came a cross a website that derived it in this fashion: the instanteous power in a capacitor is given by $$p_c= v_c(t)\cdot i_c$$

since $$i_c(t) = C\frac{dv_c}{dt}$$, this becomes $$p_c = v_c(t)\cdot C\frac{dv_c}{dt}$$

No issues so far....but, he then proceeds to write: $$\frac{dw_c(t)}{dt}=\frac{d}{dt}[\frac{1}{2}Cv_c^2(t)]$$.

power is the derivative of energy, so I get the left hand side of the equation. However, how does $$C\frac{dv_c}{dt}\cdot v_c(t)=\frac{d}{dt}[\frac{1}{2}Cv_c^2(t)]$$ on the right hand side of the equation?

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  • \$\begingroup\$ Your question is Energy and you start off with power . Recheck.. Integrate (t) to get E and then you get E=1/2CV² \$\endgroup\$ Jul 24, 2020 at 21:49

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Just use the chain rule of calculus:

\$\frac{d}{dt}v_c^2(t) = 2v_c(t)\frac{dv_c(t)}{dt}\$

therefore \$v_c(t)\frac{dv_c(t)}{dt} = \frac{1}{2}\frac{d[v_c^2(t)]}{dt}\$

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  • \$\begingroup\$ What happens when the cap is charged up with DC and dV/dt goes to 0 with your result. You are showing the charger energy, not the cap. \$\endgroup\$ Jul 24, 2020 at 21:46
  • \$\begingroup\$ @Tony Not sure what you are getting at. \$v_c(t)\$ is the instantaneous voltage across the cap at time t, while charging. What do you mean by 'charged up with DC?' \$\endgroup\$
    – rpm2718
    Jul 24, 2020 at 22:16
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    \$\begingroup\$ @Tony This relationship was not the energy of the cap. This was a mathematical answer to his purely mathematical question of how to relate the LHS to the RHS of the OP's last equation. He already had the derivation of the cap energy, just needed help with a mathematical step. \$\endgroup\$
    – rpm2718
    Jul 24, 2020 at 22:32
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    \$\begingroup\$ @Tony My belief is that the OP understood that this was power, and considered it a trivial step to get the energy from there, and didn't bother mentioning it. \$\endgroup\$
    – rpm2718
    Jul 24, 2020 at 23:02
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    \$\begingroup\$ @Tony...@rpm2718 answered the question that was being asked. But, to respond to your statement.... since d(wc)/dt = d/dt[1/2Cvc^2(t)], if we integrate both sides, you get E=1/2CV^2 as expected -- there is nothing wrong with it. The question was about the step that required the use of the chain rule as @rpm2718 pointed out which I was not seeing. \$\endgroup\$
    – jrive
    Jul 26, 2020 at 16:58

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