# What prevents a capacitor from changing voltage abruptly?

• ainster31
In summary, when connecting a capacitor to a DC battery of 10 volts, there is a constant voltage of 10V across the capacitor. When removing the wire connecting the battery, the voltage at the capacitor remains unchanged at 10V due to the lack of a closed loop to discharge it. In diagram A, the derivative of voltage is not defined at one point, but this does not cause any special behavior. The equation for capacitor current, i=C*dV/dt, shows that current is proportional to the rate of voltage change. If voltage is kept constant, no current will flow. The current curve is different from the voltage curve. At the black circle, there is no change in voltage and therefore no current flows. The behavior of
ainster31

What happens if I connect a capacitor to a DC battery of 10 volts? At this point, there is a constant voltage across the capacitor of 10V, right? Now what happens if I remove the wire connecting the battery to the capacitor? Isn't that an abrupt change of voltage because it drops from 10V to 0V?

Also, from diagram A, I notice that while voltage is continuous, the derivative of the voltage is not defined at one point. I know that the equation of the current is:

$$i=C\cdot \frac { dv }{ dt }$$

So what happens to the current in that instant where the derivative of voltage w.r.t. time is not defined (i.e. at the top of the corner)? Would current be undefined? Is that physically possible?

ainster31 said:

What happens if I connect a capacitor to a DC battery of 10 volts? At this point, there is a constant voltage across the capacitor of 10V, right?
Yes, but every real world voltage source will have internal resistance. So the capacitor voltage will not reach 10V immediately.

Now what happens if I remove the wire connecting the battery to the capacitor? Isn't that an abrupt change of voltage because it drops from 10V to 0V?
No, the voltage at capacitor will not change, we don't have any closed loop to discharge the capacitor and this is why the capacitor voltage will remain unchanged and still will be equal to 10V
Also, from diagram A, I notice that while voltage is continuous, the derivative of the voltage is not defined at one point. I know that the equation of the current is:

$$i=C\cdot \frac { dv }{ dt }$$

So what happens to the current in that instant where the derivative of voltage w.r.t. time is not defined (i.e. at the top of the corner)? Would current be undefined? Is that physically possible?
The infinite amount of current will need to flow if we have abrupt change in voltage.

Jony130 said:
The infinite amount of current will need to flow if we have abrupt change in voltage.

So diagram A is not allowed? Textbook error?

ainster31 said:
So diagram A is not allowed? Textbook error?
No, diagram B is not allowed.

Jony130 said:
No, diagram B is not allowed.

But I was referring to diagram A in that second part of the question:

ainster31 said:
Also, from diagram A, I notice that while voltage is continuous, the derivative of the voltage is not defined at one point. I know that the equation of the current is:

$$i=C\cdot \frac { dv }{ dt }$$

So what happens to the current in that instant where the derivative of voltage w.r.t. time is not defined (i.e. at the top of the corner)? Would current be undefined? Is that physically possible?

Capacitor current is proportional the rate of voltage change across it (proportional to how quickly the voltage across capacitor is changing).
The faster the voltage change (frequency of a AC signal is high) the large the current flow through capacitor (short at high frequencies).

I = C*dV/dt

This means that to sustain current through a capacitor the applied voltage must change. The more rapidly voltage changes the larger the current. On the other hand if voltage is kept constant no current will flow no matter how large the voltage.
Likewise if the current through a capacitor is found to be zero, this means that the voltage across it must be constant, not necessarily zero.

Diagram B is allowed as long as you say it's "not allowed". The textbook clearly states "not allowed".

C*dv/dt=it describes the current in any case of a AC voltage across a capacitor. It even describes the current in DC. However, it does not describe intial transient for DC...only steady state DC. Whereas in AC...Cdv/dt will describe any transient.

There is where one of a physics theory guys could help a bit with some theory because Cdv/dt=it doesn't help the begginer grasp the situation very well.

Jony130 said:
This means that to sustain current through a capacitor the applied voltage must change. The more rapidly voltage changes the larger the current. On the other hand if voltage is kept constant no current will flow no matter how large the voltage.
Likewise if the current through a capacitor is found to be zero, this means that the voltage across it must be constant, not necessarily zero.

I understand what happens when voltage is either constant or is changing or at least I think I understand this. However, what happens if voltage is neither constant nor changing?

Take the voltage at the black circle:

The voltage is not constant nor changing because the derivative of the voltage with respect to time at that instant is undefined.

ainster31 said:
I understand what happens when voltage is either constant or is changing or at least I think I understand this. However, what happens if voltage is neither constant nor changing?

Take the voltage at the black circle:

The voltage is not constant nor changing because the derivative of the voltage with respect to time at that instant is undefined.

If there is no change in VOLTAGE...then there is ZERO current just like Cdv/dt=it suggests at that exact instance. Remember, the current curve is way different than the voltage curve.

For an exercise, sketch the current at every point in the graph you show. The first part and last part of the graph (flat line) also will display the same behavior as the paragraph above.

ainster31 said:
I understand what happens when voltage is either constant or is changing or at least I think I understand this. However, what happens if voltage is neither constant nor changing?

Take the voltage at the black circle:

The voltage is not constant nor changing because the derivative of the voltage with respect to time at that instant is undefined.
Nothing special happened at that point. At the beginning we charge capacitor from constant current source and this is why we have a positive slope (ramp up). After some time (at the black circle) we change the the current direction, and now we start discharging phase. And this is why the voltage across capacitor ramp down.

Typically, almost always in reality, the VOLTAGE creates the current in the capacitor...not the other way around.

Dont forget as well in regards to the "bubble"...at that top point when the positive slope changes to negative slope...the reality is that the slope = 0 at that exact top point. Zero current.

Also, you mention instantaneous voltages. What happens if you put 10 volts across cap. Well, it is a huge change in voltage going from 0 to 10 volts...the slope is almost vertical. This is why a capacitor acts like a short when DC voltage is first applied. When the capacitor is getting full and the voltage across it stablizes in DC, the capacitor now acts like an open circuit...full voltage and zero current.

And again, a cap will act exactly like Cdvdt in AC!

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psparky said:
Typically, almost always in reality, the VOLTAGE creates the current in the capacitor...not the other way around.

Not. An external source is providing a voltage and current. Current in the capacitor results in voltage developing across plates. When the cap is disconnected from the source, it holds charge, voltage is non-zero while current is zero. If a resistance is then connected across the cap, the current at time t = 0, is merely V/R, so in this narrow case the voltage determines what the current is based on resistance. In general, voltage does not create current. Nor vice-versa. What creates current as well as voltage is an energy conversion process.

Claude

cabraham said:
Not. An external source is providing a voltage and current. Current in the capacitor results in voltage developing across plates. When the cap is disconnected from the source, it holds charge, voltage is non-zero while current is zero. If a resistance is then connected across the cap, the current at time t = 0, is merely V/R, so in this narrow case the voltage determines what the current is based on resistance. In general, voltage does not create current. Nor vice-versa. What creates current as well as voltage is an energy conversion process.

Claude

I believe what you are saying is true, but for learning purposes it just makes it even more difficult. You are taking a physics approach which is true, but clouds things...as mentioned in other threads.

What I'm saying is when you have a V and an R...an I is induced. R being impedance in this case...0 - J*(1/wc)

A car battery for example is 12 volts. When you hook a resistance to it a current is induced based off of the resistance. Sure the battery has a max current, but it's current is based off of V=IR until it reaches its max.

When I hook a AC source to a capcitor, like an outlet in a house, again the current is based off of V=IR or IZ...until it again reaches it's max current, then breaker trips. Unless there is a load, I cannot measure the current...I can only measure the voltage!
The current of these two examples does not drive the voltage. It's definitely related, but most sources are sitting at a specific voltage with current always being a result of the resistance.

Splitting hairs here...but throwin out my two cents.

Most questions in school are based off a voltage source, that's why I tried to put in simplest learning terms.

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psparky said:
I believe what you are saying is true, but for learning purposes it just makes it even more difficult. You are taking a physics approach which is true, but clouds things...as mentioned in other threads.

What I'm saying is when you have a V and an R...an I is induced. R being impedance in this case...0 - J*(1/wc)

A car battery for example is 12 volts. When you hook a resistance to it a current is induced based off of the resistance. Sure the battery has a max current, but it's current is based off of V=IR until it reaches its max.

When I hook a AC source to a capcitor, like an outlet in a house, again the current is based off of V=IR or IZ...until it again reaches it's max current, then breaker trips. Unless there is a load, I cannot measure the current...I can only measure the voltage!
The current of these two examples does not drive the voltage. It's definitely related, but most sources are sitting at a specific voltage with current always being a result of the resistance.

Splitting hairs here...but throwin out my two cents.

Most questions in school are based off a voltage source, that's why I tried to put in simplest learning terms.

I understand that but I just wish to point out the following. With an ac wall outlet, yes of course the utility company generates power as a constant voltage source, NOT current source. I only wanted to point out that this is a man made condition. If the power company generated constant current, where the voltage varies with load, losses would be much greater. Likewise, batteries work better in constant voltage mode of operation. Nuclear batteries favor current source operation, but I doubt that nuke cells will be allowed anytime soon, concerns over hasrvesting fissionable material from them would be great.

Anyway, I agree with you that in pure science, I & V are related, either one can come first with the other determined by resistance. In real world power sources constant voltage is how they are constructed since that results in better transmission of power and less loss. So it is convenient and easy to acquire the habit of thinking of the source voltage as the independent variable, and current being dependent on source voltage and load impedance. Anyway, we seem yo concur. I didn't mean to make too much of it. BR.

Claude

It would be possible, in principle, to supply the power from a constant current source efficiently but it would be very difficult to arrange to switch loads in and out. They would need to be connected in series and the voltage across the supply terminals could be enormous at times. Moreover, turning everything off in the home would involve applying a short circuit at the input terminals.

ainster31 said:
So what happens to the current in that instant where the derivative of voltage w.r.t. time is not defined (i.e. at the top of the corner)? Would current be undefined? Is that physically possible?
Mathematically, there's nothing disturbing about the derivative of a piecewise linear function being undefined at certain points. What you have to be careful about is how you relate that to practical circuits.

In practice, your capacitor, connecting wires etc., will have a small amount of parasitic inductance, which means the current through your capacitor cannot be discontinuous. It follows that it's physically impossible to have the kind of voltage waveform graphed in (a) across your capacitor.

It doesn't matter, as long as the time derivative is not infinite. All that's necessary is for the function to be continuous (i.e. no instant changes of Voltage). Nothing strange will happen to the current at the apex in graph a. Any Inductance will tend to alter the details of the actual voltage / time curve but, as with all circuit realisations, the graph can be made to follow graph a within any specified limits. I assume you are trying to relate this to real life.

sophiecentaur said:
It doesn't matter, as long as the time derivative is not infinite. All that's necessary is for the function to be continuous (i.e. no instant changes of Voltage). Nothing strange will happen to the current at the apex in graph a. Any Inductance will tend to alter the details of the actual voltage / time curve but, as with all circuit realisations, the graph can be made to follow graph a within any specified limits. I assume you are trying to relate this to real life.
I'm assuming this was directed at my post?

If so, are you saying that it's possible to have the piecewise linear voltage waveform shown in (a) across a physical capacitor? It's continuous after all.

That would seem odd to me since you also mention:
sophiecentaur said:
Any Inductance will tend to alter the details of the actual voltage / time curve ...

Possible, within limits. But that's the same for any practical circuit, surely. It's the limits that count. Any series L will introduce ringing, which can be ameliorated with suitable resistance. But the change in slope is much easier to achieve than the change in amplitude in graph b.

sophiecentaur said:
Possible, within limits. But that's the same for any practical circuit, surely. It's the limits that count. Any series L will introduce ringing, which can be ameliorated with suitable resistance. But the change in slope is much easier to achieve than the change in amplitude in graph b.
I understand this as you saying you can approximate the voltage waveform in (a) to some extent in a physical circuit, which I agree with. What I'm arguing is that you cannot have a truly piecewise linear voltage waveform across the capacitor - that would imply a discontinuous current waveform.

At a small enough timescale, the voltage across the capacitance will "round out", but I think you're arguing the same.

When you get down to it, it's a matter of bandwidth. Both waveforms have components at infinity but the HF components fall off faster in a.
So a is easier to realize.
We are not disagreeing.

On that note, I'm going to bed. Good night.

sophiecentaur said:
It would be possible, in principle, to supply the power from a constant current source efficiently but it would be very difficult to arrange to switch loads in and out. They would need to be connected in series and the voltage across the supply terminals could be enormous at times. Moreover, turning everything off in the home would involve applying a short circuit at the input terminals.

Not difficult at all. Loads are wired in series with switches across the load. Switch open means load is powered. Switch closed turns off the load. If all loads are powered down, the supply is shorted which is NO PROBLEM. With current source, short is a 0 voltage 0 power condition. With currnt source, open circuit is destructive as the source outputs high voltage.

Current source can be done, but not as efficient as voltage source. Also voltage source is constant frequency, current source variable frequency. Constant freq makes multiple generators in parallel easy to sync. Constant freq means synchronous motors run fixed speed. Current source does not provide that.

Claude

Summary:
\
Waveform A could be produced if you could switch instantly from a positive current to a negative current. To the extent that you cannot do that, the waveform would be different. Inductance makes that difficult.

Waveform B could be produced if you could generate an infinite current spike of the proper energy.
See http://en.wikipedia.org/wiki/Dirac_delta_function for the mathematical representation. Several things make that difficult.

When you connect an ideal capacitor to the supply, it will appear as a short. To the extent that the supply has a current limit, there will be a slope to the resulting voltage rise. When you disconnect the capacitor, the voltage across the capacitor remains where it was when disconnected.

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Electric charge is measured in coulombs. 1 coulomb = 6.24×1018 electrons.
Electric current is the flow of charge. 1 amp = 1 coulomb per second.
Capacitance is defined and measured as the ratio of charge to voltage, C = q / v.
The voltage on a capacitor is proportional to the charge in the capacitor, v = q / C.

The charge in a capacitor is the integral of the current over time.
The voltage on a capacitor will therefore ramp up or down depending on the current flowing.

An integral cannot have an instant step change because the current cannot be infinite.

There are tricks using switches that can make it look like there are step changes. If several capacitors are charged in parallel, then suddenly switched into series, there will be a step change in voltage across the circuit, but no capacitor will instantly change it's charge or it's voltage.

## 1. What is a capacitor and how does it work?

A capacitor is an electrical component that stores and releases electrical energy. It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied to a capacitor, it causes a buildup of opposite charges on each plate, creating an electric field. This electric field stores the electrical energy in the capacitor.

## 2. Why does a capacitor change voltage gradually?

A capacitor changes voltage gradually because of its ability to store and release electrical energy. When a voltage is applied to a capacitor, the electric field is created, and it takes time for the charges to accumulate on the plates. Similarly, when the voltage is removed, it takes time for the charges to dissipate, causing the voltage to change gradually.

## 3. Can a capacitor change voltage abruptly?

No, a capacitor cannot change voltage abruptly. Due to the nature of its design, it will always change voltage gradually. However, the rate at which the voltage changes can be controlled by altering the capacitance of the capacitor or the voltage applied to it.

## 4. What factors affect the rate at which a capacitor changes voltage?

The rate at which a capacitor changes voltage is affected by several factors such as the capacitance of the capacitor, the voltage applied, the type of dielectric used, and the resistance in the circuit. Higher capacitance, higher voltage, and lower resistance will result in a faster change in voltage.

## 5. Are there any risks associated with abrupt voltage changes in a capacitor?

Yes, abrupt voltage changes in a capacitor can cause damage to the capacitor itself and other components in the circuit. This is because sudden changes in voltage can create a surge of electrical current, which can overload and damage the components. It is important to properly design and use capacitors to avoid these risks.

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