- #1
Andy_ToK
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A simple circuit consists of a battery(emf=V), a switch and a capacitor(C).
At the steady state, the energy stored in the capacitor is 0.5CV^2.
However, the work done by the battery is QV=CV^2 (move a total charge of Q across a potential difference of V).
If the energy dissipated in the wire is negligible, according to the law of conservation of energy the work done by the battery=the energy stored in the capacitor. But 0.5CV^2!=CV^2.
What's wrong with my reasoning above? Thanks!
BTW, assuming the battery is without inner resistence(r=0).
PS. When an resistor is connected in series with the capacitor, things are much easier. the rest of the work done by the battery (0.5CV^2) dissipates in the resistor.
At the steady state, the energy stored in the capacitor is 0.5CV^2.
However, the work done by the battery is QV=CV^2 (move a total charge of Q across a potential difference of V).
If the energy dissipated in the wire is negligible, according to the law of conservation of energy the work done by the battery=the energy stored in the capacitor. But 0.5CV^2!=CV^2.
What's wrong with my reasoning above? Thanks!
BTW, assuming the battery is without inner resistence(r=0).
PS. When an resistor is connected in series with the capacitor, things are much easier. the rest of the work done by the battery (0.5CV^2) dissipates in the resistor.