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Consider a parallel plate capacitor connected to an AC. The current in the circuit leads the emf by $\pi /2$.

My question is, what does it mean that the current lead the emf by $\pi/2$? Does it mean that current start developing in the circuit before the emf or what?

Another query of mine is that why does the capacitor stop direct current but allow alternating current?

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2 Answers 2

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$Q=CV \Rightarrow \frac {dQ}{dt} = I = C \frac {dV}{dt}$

The voltage across the capacitor is equal to the voltage of the supply.
So whatever the voltage of the supply does the voltage across the capacitor exactly follows.

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At time $t_A$ the capacitor is uncharged and the voltage across the capacitor is zero.
However at this time the voltage of the supply is increasing most rapidly so the change of charge on the capacitor, the current, is a maximum.

As time goes by the rate of change of voltage decreases so the current decreases until at time $t_B$ the voltage has reached a maximum and the capacitor is fully charged which means that the current is momentarily zero.

At time $t_B$ the voltage is a maximum but at a time which was a quarter of a period before that, at time $t_A$, the current was a maximum.
Whatever the current does the voltage does a quarter of a period later - the current leads the voltage by $\frac \pi 2$.

The voltage now starts to drop and so the capacitor starts to discharge, the current direction reverses and at time $t_C$ the capacitor is totally discharged but the rate of change of voltage and hence the magnitude of the rate of charging, the current, is a maximum.
The current was zero at time $t_B$ whilst the voltage was zero a quarter of a period later at time $t_C$ - the current still leads the voltage by $\frac \pi 2$.

And so the cycle continues.


As long as there is an alternating voltage the charge on the capacitor will change and so a current will flow.
Now imagine that at time $t_A$ the voltage starts to rise but this time the voltage reaches a maximum at time $t_B$ and then does not change any more.
The capacitor is fully charged and as the voltage across is not changing then the charge on its is not changing so there is no current - this is the DC situation - constant voltage across the capacitor with no current flowing.

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Alternating circuit voltage and current periodically reach their respective maximum values after a certain time interval. Consider the current I=I(m)sin(wt) and Voltage V=V(m)sin(wt+pi/2)= v(m)cos(wt). What this phase difference signifies is that if you start the circuit at t=0, then the voltage at t=0 v=V(m) while i=0, meaning the circuit will gain maximum voltage value before the maximum current value is reached. Also in dc capacitor circuits, current does flow, but it eventually becomes zero. But in ac circuits a current flows in both directions, thereby not allowing the impedance of the capacitor to completely block the current. There is some resistance though, but the high frequency of such current direction makes it hardly noticeable. (sorry, but according to your question the current will lead, thus it will gain its maximum value first)

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  • $\begingroup$ Suppose the AC voltage has the instantaneous value, V=V(m)sin(wt). Then current will be I=I(m)cos (wt). In this case current will be ahead of emf. Does it mean that current achieve its maximum value even if the voltage in the circuit is zero? How is it possible? $\endgroup$ May 12, 2016 at 14:34

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