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I am confused about this question. Why is it that you use 1/2CV^2 in one situation and q^2/2C in the other?
Thanks!
Thanks!
TPR FL2 #35 C/P section
- Thread starter akimhaneul
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Q = VC
PE (capacitor) = 1/2QV (but you can make substitutions using the equation above).
When the battery is connected, V is constant so we want to use PE = 1/2V^2C.
When the battery is disconnected, Q is constant so we want to use PE = 1/2Q^2/C.
PE (capacitor) = 1/2QV (but you can make substitutions using the equation above).
When the battery is connected, V is constant so we want to use PE = 1/2V^2C.
When the battery is disconnected, Q is constant so we want to use PE = 1/2Q^2/C.
