# What prevents a capacitor from changing voltage abruptly?

1. Mar 10, 2014

### 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?

2. Mar 10, 2014

### Jony130

Yes, but every real world voltage source will have internal resistance. So the capacitor voltage will not reach 10V immediately.

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

3. Mar 10, 2014

### ainster31

So diagram A is not allowed? Textbook error?

4. Mar 10, 2014

### Jony130

No, diagram B is not allowed.

5. Mar 10, 2014

### ainster31

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

6. Mar 10, 2014

### Jony130

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.

7. Mar 10, 2014

### psparky

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.

8. Mar 10, 2014

### ainster31

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.

9. Mar 10, 2014

### psparky

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.

10. Mar 10, 2014

### Jony130

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.

11. Mar 10, 2014

### psparky

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!!

Last edited: Mar 10, 2014
12. Mar 10, 2014

### cabraham

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

13. Mar 10, 2014

### psparky

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.

Last edited: Mar 10, 2014
14. Mar 10, 2014

### cabraham

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

15. Mar 10, 2014

### sophiecentaur

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.

16. Mar 10, 2014

### milesyoung

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.

17. Mar 10, 2014

### sophiecentaur

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.

18. Mar 10, 2014

### milesyoung

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:

19. Mar 10, 2014

### sophiecentaur

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.

20. Mar 10, 2014

### milesyoung

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.