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Derivation of energy in capacitor

Electrical Engineering Asked on January 4, 2022

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 $$i_c(t) = Cfrac{dv_c}{dt}$$, this becomes $$p_c = v_c(t)cdot Cfrac{dv_c}{dt}$$

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 $$Cfrac{dv_c}{dt}cdot v_c(t)=frac{d}{dt}[frac{1}{2}Cv_c^2(t)]$$ on the right hand side of the equation?

One Answer

Just use the chain rule of calculus:

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

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

Answered by rpm2718 on January 4, 2022

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