Derivation of energy in capacitor

Question

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?

rpm2718 · Answer

Just use the chain rule of calculus:
$frac{d}{dt}v_c^2(t) = 2v_c(t)frac{dv_c(t)}{dt}$
therefore $v_c(t)frac{dv_c(t)}{dt} = frac{1}{2}frac{d[v_c^2(t)]}{dt}$

Derivation of energy in capacitor

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