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Suppose a capacitor having capacitance $C$ is being charged using a cell of emf $E$. By the time the capacitor is fully charged, the cell has supplied $QV$ energy while the potential energy of the capacitor is $QV/2$. So there is a net loss of $QV/2$ joules of energy. Where is the energy lost? Since it is an ideal circuit, there is no resistance and there should be no heat loss. Can anyone tell me which assumption do we make and where do we make it while setting up the ideal circuits and deriving their formulas which is the reason for the loss of energy?

Also, why isn't energy lost when we pull out a dielectric slab from between the plates of a charged capacitor which is connected across a cell? The slab is moved without changing its kinetic energy.

Akshit
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1 Answers1

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Even an ideal capacitor cannot be losslessly charged to a potential E from a potential E without using a voltage "converter" which accepts energy at Vin and delivers it to the capacitor at Vcap_current.

If you connect an ideal voltage source via a lossless switch to an ideal capacitor which is charged to a lower voltage, infinite current will flow when the switch is closed. If you introduce a very small resistance in the charging path the current will be large and large losses will occur despite the resistor's small resistance. As you increase the resistor the current will decrease but the increasing R causes increased power loss.

The closest real world approximation to lossless capacitor charging is to use a buck converter.

Russell McMahon
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