Capacitance and energy

[OneManShow]

I'm confused about the formulas..
the first formula is
PE = 1/2*C*V^2
and the second formula is
PE = 1/2*(Q^2 / C)

I believe the 2nd formula is obtained by just inserting V = Q /C to the first formula...

Anyways..how come increasing capacitance decreases energy in the 2nd formula but increases energy in the 1st formula?

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[Rabolisk]

For a resistor, the power dissipated through it is
P = I^2*R
P = V^2/R

Increasing resistance increases power in the first formula but decreases power in the second formula.

My point is, it matters what you keep constant. Q, C, and V describe the "state" of a capacitor. Only two of them are independent; the third variable is not really a variable once you know the first two. In the same manner, PE is dependent on only two of the three variables, since the third is "redundant". You can't say how PE will change when you change only one of the variables, because the "state" of a capacitor can only be described by at least two of Q, C, and V. It's like having one equation and two unknowns.

 

[lolarogers]

W = 1/2 C V2 (1)
where
W = energy stored (Joules)
C = capacitance (Farad)
V = potential difference (Voltage)
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