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In the textbook I'm using for physics it says that the charge left on the plates of a capacitor after time $t$, that is discharging through a fixed resistor, is $Q=Q_0e^{-t/\tau}$ where $\tau=RC$ is the time constant. This is also the equation for the potential difference across the plates after time $t$.

For charging, the equations given are $Q=Q_0-Q_0e^{-t/\tau}$ and a similar one for voltage. I have no problems with these.

What I'm having trouble with is the next statement:

"The graph of current versus time for a charging/discharging capacitor through a fixed resistor is always exponential and decreasing, like the discharge curves for charge/voltage versus time."

However, no corresponding equation is given for this. Is the equation too complicated? I thought it wouldn't be because as it's exponential it should have a similar form to the ones for charge/voltage.

Edit: My main concern is why it would be always (for both charge and dischrage) be a decreasing exponential graph. I understand that you can work this out from physical considerations ("current through circuit is obviously always decreasing for both charge/discharge"), but how do we do prove this using an equation?

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Since, as you have agreed, $Q=Q_0-Q_0e^{-t/\tau}$ and since $V = Q/C$, you can see that the capacitor voltage is an inverse exponential. Since the charge/discharge resistor is connected the capacitor andeither the charging voltage or ground, the voltage across it, and the current through it, must also be an inverse exponential, but of the opposite sign.

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You can view the capacitor as a load while charging and a source while discharging. As the ideal capacitor charges, its load resistance increases to infinity, thus the load current goes to zero. As it discharges, its source potential goes to zero, so again, the current goes to zero. Hope that helps.

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