when capacitance is small, or frequency low, small amount of charge needed...

[PG1995]

Hi

Please have a look on this scanned page from a book (or, you can check the attachment):
https://img683.imageshack.us/img683/2193/imgdec.jpg

Please focus on where it says under the topic "Effect of alternating current (a.c.) on a capacitor":

If the capacitance is small, or the frequency low, only a small amount of charge needs to flow onto the capacitor to equal the supply p.d. In other words, the current will be very low. High capacitance and high frequency gives high current.

I understand that if the capacitance is small, then only a small amount of charge (from the supply) needs to flow into the capacitor before the capacitor's voltage equals the p.d. of the supply. But what I don't understand is frequency part. I know it can be proved easily with some math expressions. I'm unable to create a conceptual picture of this. Why would a small amount of charge be needed to equal the p.d. of the supply when the the frequency of the a.c. current is low?

Could you please help to conceptually understand it? Thanks.

Regards
PG

 

[Sceadwian]

All things being equal as AC frequency increases, so does charge flow, same current + more cycles == greater charge flow.

 

[Ratchit]

PG1995,

It would be nice if you told us what p.d. means.

Ratch

 

[carbonzit]

Potential difference? Just a guess. (Never saw it abbreviated it that way; what country is this text from?)

 

[PG1995]

It's from the UK. That is the book for A Level.

 

[Ratchit]

PG1995,

If p.g. means potiential difference, then you are right, it doesn't make sense. Can you confirm what p.g. means from the book?

Ratch

 

[alec_t]

pd = potential difference. The graph is dual-purpose (see the Y-axis), showing how both the charge (Q) and pd (V) decay.

 

[PG1995]

p.d. is potential difference.

 

[PG1995]

Thank you for your replies.

Let's talk about the capacitor a little more. I do have made some progress!

When a battery is connected to a capacitor an instantaneous current flows into the the capacitor. Now how much current (which is charge per unit time) flows into the capacitors depends on the capacitance of the capacitor. If capacitance is large, then it would take more current (hence charge) to reach the state where voltage of the capacitor counterbalances the voltage supplied by the battery.

A battery is also limited by a maximum current it can supply. Suppose we have two batteries which can supply the same voltage but battery_A can supply half the maximum current of battery_B. If a capacitor is connected to batter_B (it can supply the double the maximum current of battery_A) then the capacitor will take half the time to reach the stage where it can counterbalance the battery's voltage as compared to when it is connected to the battery_A. Do I make sense?

Now we will talk about AC and frequency - the main point of original discussion.

Just like a battery an AC source also has maximum current limitation, I believe. Let's say a capacitor is connected to an AC source, and needs 10 coulombs of charge to match the peak voltage of the source, and the maximum current the supply can deliver is 1A (1A current is flow of one coulomb of charge per second). If we keep on increasing the frequency of the source and reach a point when time taken for the voltage to change from 0 value to the peak value becomes 5 seconds. In 5 seconds only 5 coulombs of charge would flow but the capacitor needs 10 coulombs to match the supply's voltage. This means that the capacitor will start discharging even before it has been fully charged. Do I make sense? Please keep your reply simple.

Thank you for your help.

Regards
PG

 

[Diver300]

Mains supplies do not limit the current. If you overload a mains supply, the breaker or fuse will blow.

However, the mains does not change instantaneously. The rate of change is well defined. Mains voltage is a sinewave, and the rate of change of a sinewave is also a sinewave (or cosine wave if you want to be exact). The peak value of the rate of change is proportional to the voltage and the frequency. The current in a capacitor is proportional to the rate of change of voltage and the capacitance.

So the current in a capacitor that is connected to a mains supply is proportional to the voltage, the frequency and the capacitance.

An hydraulic analogy would be the tidal flow into or out of an estuary. The flow rate is proportional to the range of the tide (analogous to voltage), the frequency of the tide, and the area of the estuary (analogous to the capacitance). If the tide were to go in and out twice as often, twice as much water would have to flow in and out, which is like the current being twice as much.

 

[Ratchit]

PG1995,

I like to think of capacitors charged with energy instead of charge carriers. That is because when one plate gains a abundance of charges, the opposite plate loses an equal quantity of charge carriers, thereby resulting in a net change of zero charge carriers (electrons). It takes energy to cause this imbalance of electrons, so I believe capacitors should be referred to as energized when their plates are imbalanced. If a capacitor is energized to 100 volts, it contains the same net amount of charge as it did at zero volts.

If a capacitor is connected to batter_B (it can supply the double the maximum current of battery_A) then the capacitor will take half the time to reach the stage where it can counterbalance the battery's voltage as compared to when it is connected to the battery_A. Do I make sense?

Keep in mind that energizing a battery with a constant voltage is not a linear operation. As the battery's voltage approaches the voltage of the power supply, it energizes at a lesser rate that it did at the start. If a battery is not able to keep up its voltage when loaded down with the capacitor, then that will increase the time to energize the capacitor to a final voltage, too.

This means that the capacitor will start discharging even before it has been fully charged. Do I make sense? Please keep your reply simple.

Yes, if you reverse the voltage before the capacitor has time to energize up to a designated voltage, it will first deenergize and then start to energize at the opposite polarity.

Keep in mind that a capacitor is a electrostatic energy storage mechanism. It stores energy, not charge. It is charged with energy (energized), not electrons.

Ratch

 

[alec_t]

It stores energy, not charge.

It stores both. The charge Q comes from the applied source and = C x V (where C is the capacitance and V is the voltage between the capacitor terminals. The energy = ½ x C x V².

 

[Ratchit]

alec_t,

It stores both. The charge Q comes from the applied source and = C x V (where C is the capacitance and V is the voltage between the capacitor terminals. The energy = ½ x C x V².

Not really. An imbalance of charge is not a storage event. The applied source receives back as much charge at its receiving terminal as it put out on its supply terminal. If I de-energize a capacitor with a shorting wire, where does it "stored" charge go? Answer: There was no stored charge. The capacitor had the same net charge before as it did afterwards. If I take some amount of money from your left pocket and put it in your right pocket, are you richer or poorer?

Ratch

 

[Sceadwian]

If I de-energize a capacitor with a shorting wire, where does it "stored" charge go? Answer: There was no stored charge.

How does that make logical sense? The charge is stored in the electric field between the two plates and in the case of a dead short is dissipated as heat in the bulk resistance of the leads and the capacitor itself...

 

[Ratchit]

Sceadwian,

How does that make logical sense? The charge is stored in the electric field between the two plates and in the case of a dead short is dissipated as heat in the bulk resistance of the leads and the capacitor itself...

No. Energy is stored in the electric field, not charge. The charge resides on the plates of the capacitor. Heat is the energy loss when the charges on the plates get shifted around, and electric field collapses.

Ratch

 

[Sceadwian]

The electric field induces charge flow on the opposite plate under the right cirucmstances... Charge is transferred, not directly but that's just an abstraction.

 

[Ratchit]

Sceadwian,

The electric field induces charge flow on the opposite plate under the right cirucmstances... Charge is transferred, not directly but that's just an abstraction.

The electric field represents energy. Voltage is the energy density of the charge in joules/coulomb. Charges move from a higher energy density (voltage) to a lower energy density because they repel each other, and want to live in the lowest energy density possible.

Ratch

 

[Sceadwian]

If the plates of a capacitor are both neutral in relation to each other in order for energy to be transferred between the two plates charge has to flow into or out of one of the two plates. The electric field requires the flow of charge in order to come into being from a neutral state. To say it's not a charge storage event (albiet indirectly) seems wrong. It might be important to note it for purely mechanical reasons that energy is being transferred between the two plates in the form of the electric field not actual charge carriers (much the same as in a transformer) but if charge exists on both sides and it's transferred via energy to say that no charge is transferred isn't correct. No pure charge is transferred (IE the same electrons stay on either side of the divide) but the energy flowing from one side to the other in the form of the electric field does cause charge transference to be the effective result.

 

[carbonzit]

Moderated: Deleted Off Topic

 

[Ratchit]

Sceadwian,

If the plates of a capacitor are both neutral in relation to each other in order for energy to be transferred between the two plates charge has to flow into or out of one of the two plates.

No, not one of the plates, but both of the plates. Whatever is stored on one plate is depleted on the other plate. That makes a charge imbalance between the plates, but a net change of zero for the whole capacitor.

The electric field requires the flow of charge in order to come into being from a neutral state.

An electric field requires the presence of a charge. Charge movement or flow makes a magnetic field, not a electric field.

To say it's not a charge storage event (albiet indirectly) seems wrong.

If the amount of charge contained in a capacitor does not change, then it cannot be said to have a net storage event.

It might be important to note it for purely mechanical reasons that energy is being transferred between the two plates in the form of the electric field not actual charge carriers (much the same as in a transformer) but if charge exists on both sides and it's transferred via energy to say that no charge is transferred isn't correct.

No mechanics involved, the energy is in the field and not transferred between the plates, if a charge exists on one side, it is depleted on the other side. A transformer works differently via magnetic fields, and has no correlation with a capacitor. Charge change is internal, but its net charge change is zero. It receives no net charge from outside the capacitor.

No pure charge is transferred (IE the same electrons stay on either side of the divide) but the energy flowing from one side to the other in the form of the electric field does cause charge transference to be the effective result.

Energizing a capactor means adding electrons to one side of the capacitor and removing them from the other side in equal amounts. It takes energy to do that, and that energy results in a electric field. Energy does not flow from one side to the other side. It exists within the electric field that extends from one plate to the other.

Ratch