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By the phasor method for AC circuits, we know that the voltage in a capacitive circuit lags the current by $\frac{\pi}{2}$.

My book says that the current in a purely capacitive circuit is maximum when the instantaneous voltage is 0. How can this be? If there is no voltage, how can there even be a current in the circuit?

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It is because the current in the capacitor does not depend on the instantaneous​ value of voltage, rather it depends on the time rate of change of voltage or time derivative of voltage which is not zero even if instantaneous value of voltage is zero.

You can easily see this from the following equation:

$$q = CV$$

$$\frac{dq}{dt} = C\frac{dV}{dt}$$

$$i= C\frac{dV}{dt}$$

Even if the instantaneous voltage is zero, $\frac{dV}{dt}$ may not be zero and hence current can flow.

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  • $\begingroup$ The formula isn't clear. $\endgroup$ Mar 11 '17 at 7:57
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    $\begingroup$ @KunalPawar $$q = CV$$ $$\frac{dq}{dt} = C\frac{dV}{dt}$$ $$i = C\frac{dV}{dt}$$ I added it to the answer now. When you are studying physics at this level, you are expected to guess how you arrive at the formula for $i$ and not ask someone to explain them. $\endgroup$
    – Yashas
    Mar 11 '17 at 7:58
  • $\begingroup$ @YashasSamaga I only said that so that user75221 would correct his error. Unfortunately I'm not familiar with MathJax yet. $\endgroup$ Mar 11 '17 at 8:09
  • $\begingroup$ I think I misinterpreted your first comment. You were probably complaining about the answerer using $v$ instead of $V$. I'm sorry if that was the case. $\endgroup$
    – Yashas
    Mar 11 '17 at 8:16
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When the charge on the capacitor is zero, it offers no resistance. The circuit behaves as if it is short circuited. So, as no resistance is offered, a very small voltage will produce a very large current.

The relations $\mathcal{E}(t) = \mathcal{E}_{\mathrm{m}} \cos \omega t$ and $i = i_\mathrm{m} \sin \omega t$ are valid only after transients die away. So, initially, consider that the charge across the capacitor is zero. If we apply the voltage, the electrons trying to come to the capacitor from the voltage source will face no opposition. So, the current will be high. As the charges accumulate on the capacitor, they start to repel the electrons trying to come from the source. Thus, current decreases. As the voltage increases, the charges are not able to flow as the voltage across the capacitor plates increase.

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  • $\begingroup$ Thank you for the qualitative explanation. What are the transients that you've mentioned? $\endgroup$ Mar 11 '17 at 8:35
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    $\begingroup$ You can consider transients as the current varying erratically for a short period of time as soon as the voltage is applied. Consider it as you open a water tap. The water does not flow initially in a smooth manner. There will be some swirls and other motion. $\endgroup$ Mar 11 '17 at 8:39
  • $\begingroup$ Okay cool. So my book says something like the general solution is the sum of the transient and steady state solution. Is there any particular way to derive these add them up? $\endgroup$ Mar 12 '17 at 10:20
  • $\begingroup$ @Kunal Well, I've absolutely no idea how transient currents flow or behave. $\endgroup$ Mar 12 '17 at 12:55

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