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I have read on stack exchange that at t=0 capacitor behaves as short circuit.Does short circuit here means electrons will flow through it or into it.I mean at t=0 if I have connected a d.c battery than capacitor and than a lamp will it glow initially.If it will glow than how can electron pass until we get whole capacitor charged.

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It's a strange question. The electrons won't pass through the capacitor but rather displace the electrons on the other plate but the end result will be the same. This will happen until the capacitor is charged. This is also why an AC current charges and discharges the capacitor making it in fact an apparent conductor. – Paulo Soares Jan 10 at 11:31

You are missing one important critical assumption, which is the intial state of the capacitor. For example, consider this circuit:

If the voltage on the capacitor is 12 V at T=0 when the switch is closed, no current will flow.

If the capacitor voltage is 0 when the switch is closed, then yes, you can think of it instantaneously being a short circuit. The initial current will be the same as 12 V applied directly to the R1-D1 series combination.

The difference is that as current flows thru the cap, charge builds up in it, which makes the capacitor look like a increasing voltage source that opposes the applied 12 V. This opposing voltage is the time-integral of the current flowing thru the loop, which is the same as the total charge that has flowed thru the loop. After a finite amount of charge, the cap will be at 12 V and no current will flow any more.

Note that the current that flows depends on the voltage applied to the rest of the circuit after the voltage dropped by the cap. As more current flows, that opposing voltage increases, which decreases the voltage on R1-D1, which decreases the current, which decreases the speed at which the opposing voltage builds up. The result is a exponential with the current asymptotically approaching 0.

Let's put some real numbers on this. Let's say C1 is 1 mF, R1 500 Ω, and D1 drops 2.1 V. If C1 starts out at 0 at t=0 when the switch is closed, then there will be (12 V)-(2.1V) = 9.9 V across R1. From Ohm's law, that means (9.9 V)/(500 Ω) = 19.8 mA will flow.

Since we are modeling D1 as a fixed voltage source in this example, the time constant is simply C1*R1 = (1 mF)(500 Ω) = 500 ms. The current will decrease by a factor of e every 500 ms (half a second). The equation for the current, I, as a function of T seconds since the switch was closed is then:

I = 19.8mA * e-T/500ms

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When starting with a capacitor voltage of 0, then the final voltage capacitor voltage will be 12 V minus the forward voltage of D1, e.g. 9.9V in your example. – Martin Zabel Jan 10 at 21:07

Current flows through the capacitor but not the electrons.
Some electrons flow into one terminal and accumulate as negative charge on one electrode and some others flow out of the other terminal and cause the other electrode to loose negative charge.

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For something to behave like a short circuit it must present an impedance that is zero. Of course, a zero ohm resistor technically does that and, a capacitor can present a dynamic impedance that is also theoretically zero. The current that flows into a capacitor is governed by this equation: -

I = \$C\dfrac{dv}{dt}\$

If the change in voltage (dv) happens in a very short period of time (dt) then the current can be very large. Clearly if the change in voltage is instantaneous then current is infinite and this is, in effect, "short circuit" behaviour.

It's the same when trying to short out a capacitor that is already charged to some voltage - you are trying to instantly change the voltage and this means infinite current flow.

Of course all of this is theoretical because voltage can never change instantaneously and real capacitors have parasitic components that limit current (ESR = effective series resistance).

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Assuming the capacitor is not initially charged, at t=0 a current will start to flow through it, but there is zero voltage across it (because it hasn't built up any charge). Ohm's law tells us that the resistance is then \$\frac{V}{I}=\frac{0}{I}=0\$ which is a short circuit.

As time progresses, the current charges C and hence the voltage across C increases. In line with the increase in voltage, the current decreases because the resistance seen by the source is increasing.

This continues until the capacitor is fully charged and its voltage is the same as the supply voltage. The current is then reduced to zero and the resistance is \$\frac{V}{0}=\infty\$, ie an open circuit.

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