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A uncharged capacitor $C$ is connected to a battery with potential $V$. It becomes fully charged and has a charge $Q=CV$ stored on it.

Now the polarity of the battery is reversed. The capacitor will have the charge $Q$ still but with polarity reversed too.

My question is: What is the work done by the the battery?

According to me, during 1st charging it will do a work of $QV$. The energy of the capacitor is also not changing, then what is the work done to change its polarity?

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  • $\begingroup$ The energy of the capacitor ($0.5QV$) is only half the work done by the battery. You cannot use conservation, there is loss of energy everytime it is charged. Let's say it is path dependent, and total $W=W_1+W_2$ $\endgroup$ – Shubham May 22 '14 at 15:16
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A uncharged capacitor C is connected to a battery with potential V

Since the voltage across a capacitor must be continuous (for finite current), one cannot specify this and expect to get valid answers.

Remember, for the ideal capacitor

$$i_C = C \dfrac{dv_C}{dt}$$

But specifying that an "uncharged capacitor C is connected to a battery with potential V" is equivalent to specifying

$$v_C(t) = Vu(t)$$

But this is equivalent to specifying

$$i_C(t) = CV\delta(t)$$

which certainly ejects this problem outside of the realm of ideal circuit theory.

Now, there is an easy remedy for this - place a resistor in series with the capacitor and then the capacitor voltage becomes

$$v_C(t) = V(1 - e^{-\frac{t}{RC}})u(t)$$

and the current is finite

$$i_C(t) = \frac{V}{R}e^{-\frac{t}{RC}}u(t)$$

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Now, it is straightforward to show that the energy stored in the fully charged capacitor is less than the energy supplied by the battery with the difference being the energy dissipated by the series resistor.

Finally, $R$ cannot be reduced to zero - one can show that there is a fundamental radiation resistance associated with the circuit such that energy from the battery can be radiated away during charging.


My question is: What is the work done by the the battery?

When the polarity is reversed, the capacitor will initially discharge, doing work on the battery, until fully discharged and then the battery will again begin doing work on the capacitor.

Since there is loss during the charging / discharging process, one cannot equate the work done by the battery to the work done on the capacitor.


Perhaps the problem is to be solved assuming that the capacitor connected to a battery gets approximately a charge CV(in negligible time).

Including a series resistor allows one to compute the energy dissipated by the resistor which turns out to be independent of the resistance and equal to

$$W_R = \frac{CV^2}{2}$$

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  • $\begingroup$ Perhaps the problem is to be solved assuming that the capacitor connected to a battery gets approximately a charge CV(in negligible time). $\endgroup$ – Shubham May 22 '14 at 16:34

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