I know the water analogy is not perfect but let's give it a try.

A pipe full of water is the analogy of the wire full of moveable electrons.

The capacitor is a larger cylinder with a pipe connection at each end and a flexible rubber membrane (the dielectric) across the center of the cylinder that keeps water (electrons) from continually flowing through the cylinder.
But if more pressure (voltage) is applied at one end as compared to the other, then water will continue to flow until the membrane elastic force balances the pressure and the flow stops (the capacitor is now charged).
So the water (electrons) can move into and out of one end of the cylinder with other electrons moving in and out the other end without any electrons ever crossing the membrane (dielectric).

So it should be apparent that if the pressure difference across the cylinder is alternately switched back and forth from one side to the other (AC), the current will flow in both directions, but if a steady pressure is applied (DC) only a limited current will flow until it stops.

And don't let the ground connection confuse you. The ground connection is just another wire current path, no different from any other wire except for its name.

Does that help?

 

I got it. But the question is about shunting to the Ground. Additional spikes will go towards the Ground. Even if we make positive plate as a reference point and make the changes only on the negative plate, in that case, in order to reach to the ground or a reference point the additional electrons are still required to cross the dielectric (Which is not possible). So if we see this analogy when the capacitor is required to discharge then it has to supply the electrons back to the circuit. Now it will supply the electrons towards the same circuit for which it is required to protect (As we know the rate of charging and discharging same). Ahhhhh I am so confused.

Hi again,

Well when you say "shunting to ground" that is just a circuit theory device intended to mimic what it LOOKS like is happening, even though it is not in reality actually happening. So for a cap that is connected to bottom ground with (-) on ground and (+) on top when current enters the top of the cap (+) it does NOT have to travel THOUGH the cap to get to the negative (-) terminal. All that has to happen is the negative (bottom) plate has to loose electrons. That makes it look like the cap is passing electrons, but it's not.

One of the theories that allows this viewpoint is the way the plates store charge, but another theory is that electrons are so elementary to nature that they have no individual identity. That means that if you could somehow 'see' an electron enter the top (+) plate you could never tell whether or not that SAME electron got stored on the plate or went through to the other side and came out the bottom. So if we could number the electrons and the one that went into the top we labeled #1 and the one that came out the bottom as #2, we could not tell if it was the same electron or a different one. So we can treat it either way we wish, we can say that it went through (basic network theory like 'the current in a series circuit is the same through each element') or we can treat it in the physical sense (charges stored on the plates).
In each case for every electron that enters the top plate one electron leaves the bottom plate, but since the electrons have no individual identity we can not state that the two electrons are different, so we can look at it as though only one went through the dielectric.
As you should know by now, most electrons dont pass though only some do which is due to leakage current which is considered small compared to the normal operating current in most cases.
There is however a faux current which is called the "displacement current" which is the current that also makes it seem like current is flowing through the cap and that is due to the field that is created which is the same field that would exit if there really was current flowing through the cap.

But the main idea is that for every one that enters the top, one exits the bottom and externally it doesnt matter if they went through or not, and a voltage can only be measured across two points not at a single point. This means that when electrons enter the top and leave the bottom the voltage difference will change.

Hope that helps a little more. If you still have questions it might help to do a diagram and label the different parts and explain what exactly you dont understand yet.

 

Try this: Think of an air pump. No tank, just the pump. As the piston pushes the air through the hose it pulsates. As the piston draws back to gather another puff of air, thus there is a lack of push through the air hose. Now, put a balloon on a tee. No, we're not pumping up the balloon, the balloon is just there to act as a buffer in the air line so that when the piston pumps air into the line the balloon expands. When the piston draws back for another puff the balloon contracts and is now supplying air to the air hose. The balloon expands slightly with the puff of air and contracts when the pump is drawing back.

The same is happening with electrons; and my analogy only relates to pulsating DC currents. With AC it's a slightly different story. But for the sake of understanding what's going on the capacitor is acting just like the balloon in that when there's a surplus of electrons it expands (figuratively speaking, not literally). When there's an absence of electrons the capacitor contracts. So a capacitor absorbs extra electrons when available and supplies electrons when there's a lack of electrons coming from the source. So when a power supply consisting of a transformer pulses and is rectified the capacitor smooths out the pulses for a more steady stream of electrons.

I don't think I can explain it any better than that. There may be better explanations, OR different ways of explaining it so you can understand. But I think I've done all I can to help. I hope I've helped enough.

 

At this point I’m not exactly sure what the OP does not understand.

usmansa1, in the case of “noise” consider the capacitor as a kind of wave shaper…the noise enters the circuit with a certain amount of energy, but is in the shape of a “spike” where it is higher than wide, the capacitor absorbs the “spike” and delivers it to the circuit as a wider and lower “bump” of the same energy.

So the circuit sees a smaller voltage increase, but for a longer time. And the opposite occurs if the spike was in the reverse voltage…the circuit will then see a small drop in voltage for a longer amount of time rather than a short quick sharp drop.

In the end the energy has to go somewhere, so the cap gives it back to the circuit, just differently then it received it.

 

TS is thinking too much about electron flow. TS needs to look at voltage also.

If TS understands that V = Q/C
then a large capacitor means that it will take a greater change in Q to cause V to change.

A large C softens the effect on V caused by changes in Q.

 

Hi every one,

I finally get some idea because of all of your explanations.And I am really thankful to all of you. There is one final doubt left. When the high voltage spikes come it will store more energy in the Electric field. Once the spike is over the capacitor is required to discharge. Now it will supply the current back to the load. If we see in that case the same spike is supplied back to the load. I understand there is one reply from Electric spide that shows that it will take supply high voltage for longer time. But in that case I dont understand this statement that " Capacitor protects from Electric Spikes "

 

Hi every one,

I finally get some idea because of all of your explanations.And I am really thankful to all of you. There is one final doubt left. When the high voltage spikes come it will store more energy in the Electric field. Once the spike is over the capacitor is required to discharge. Now it will supply the current back to the load. If we see in that case the same spike is supplied back to the load. I understand there is one reply from Electric spide that shows that it will take supply high voltage for longer time. But in that case I dont understand this statement that " Capacitor protects from Electric Spikes "

The nature of the spike is important. Most have a high AC energy component that is shunted away from the load by a capacitive noise filter system before the load. The DC/low frequency energy component of the spike is usually lower than the stored energy of the DC filter system so the change in actual voltage as a factor of time for that movement of charge in the DC filter system is small.

 

Here is my response to your continuing query on how capacitors mitigate current and voltage spikes. You are fixated on the flow of charge and the return of charge to the load.

You need to look at voltage also.

Voltage V = Charge Q divided by Capacitance C.

The larger the capacitance, the lower the voltage. (This needs some clarification when discussing very high frequency transients. We will not discuss this at this point.)

Think of a capacitor as a large storage tank. It mitigates any changes of voltage by damping the rise or fall of voltage.

You are also fixated in thinking that the charge spike is dissipated in the load. This is not true. The current spike is dissipated in the power supply feeding the load.