> You can charge a capacitor at any voltage as long as you cut the current off when it acquires the desired charge / terminal voltage.
Capacitors aren't charged with a voltage, they're charged with a current. The voltage is irrelevant until breakdown occurs. More current, faster charge, less current, slower charge.
Also, there's the practical constraint that a capacitor's physical construction limits the amount of current that can be applied without overheating the capacitor's materials.
The TL;DR: one does not apply a voltage to a capacitor, one applies a current. If you doubt this, try applying a constant-voltage source to a capacitor. See what happens.
> one does not apply a voltage to a capacitor, one applies a current
This is meaning less. You have to apply an EMF (https://en.wikipedia.org/wiki/Electromotive_force) to get the current flowing in the first place. There is no current without EMF (measured in volts).
When a capacitor is directly connected to a real world voltage source (with finite internal resistance) it shorts the power source and causes a massive amount of current to flow. I = VEmf/Rbatt (but only at t=0).
This current causes charge to accumulate on the plates of the capacitor causing the voltage across the plates to rise while reducing the current flowing through it at the same time.
The voltage across the capacitor rises (asymptotically) until it matches the potential of the power source. The voltage across the plates at any given time t is given by the equation: V = VEmf(1-e^(-t/(Rbatt * C))). See: https://en.wikipedia.org/wiki/RC_time_constant
Also the current at any given time is: I = VEmf/Rbatt * (e^(-t/(Rbatt * C)))
Thus an ideal capacitor can be charged by a voltage source of any value (VEmf) to any voltage (less than or equal to VEmf) across its plates. There is no theoretical limit imposed by physics. Practical capacitors will experience a dielectric breakdown above their rated voltages: https://en.wikipedia.org/wiki/Electrical_breakdown
So all you have to do is remember to disconnect the power source before the voltage across the capacitor crosses its safe operating area. So if a capacitor is rated at 50V max terminal voltage then you need to disconnect the power source when you get 50v across the cap. You will reach 50v earlier if you use a 200v power source instead of a 50v source. The only thing that changes is the the time required to charge the cap (and the current like you said). A higher voltage power source or one with lower internal resistance can drive larger currents.
In a circuit without resistance, current can flow with no potential difference, so the above claim is false. As to circuits other than ideal ones, I draw your attention to the behavior of superconductors, in which currents flow endlessly until interrupted.
> The voltage across the capacitor rises (asymptotically) until it matches the potential of the power source.
For a constant-current source, the rise is linear. For a constant-voltage source, the current at time zero is infinite, barring stray resistances.
> You will reach 50v earlier if you use a 200v power source instead of a 50v source.
False! The issue is current, not voltage. The voltage on a capacitor is the time integral of past applied currents.
Capacitors have an upper limit to inrush current, and to charge one, you stay below that limit. The ideal charging source for a capacitor is a constant-current source, and the voltage driving the source is irrelevant until the capacitor's voltage limit is reached.
Consider a capacitor that can tolerate 10 amperes of current and has a voltage limit of 50 volts. Let's say I want to charge it as quickly as possible. I will apply 10 amperes of constant current, and cut off the charge as the capacitor's voltage limit is approached. Will I be able to charge the capacitor faster from a source of 200 volts than from a source of 50 volts? Of course not.
> Also the current at any given time is: I = VEmf/Rbatt * (e^(-t/(Rbatt * C)))
I am mystified as to why you posted a battery equation, and/or reference to resistance, in this capacitor discussion, in particular when constant current sources are the current state of the art. Surely you realize one doesn't attach capacitors to voltage sources, or is this still not understood?
> A constant-current power supply is the preferred way to charge a capacitor wihtout destroying it.
Sure, I agree. I just wanted to use a simpler circuit to keep my explanation simple.
>As to circuits other than ideal ones, I draw your attention to the behavior of superconductors, in which currents flow endlessly until interrupted.
I think you have cause and effect mixed up. How do you get this current started in the super conductor? What is the magnitude of this current? Infinite amps? You will still need an initial jolt (voltage source) to get the current started.
>For a constant-current source, the rise is linear
You do realize there no such thing as a practical constant current source? A current source is just a voltage source which can adjust its terminal voltage to keep it's current output constant with changing load? Like I said you need to understand the cause and effect relationships here.
> Capacitors have an upper limit to inrush current
Sure practical capacitors, conductors, resistors, ... everything has an Imax. I did say raise the voltage as long as you are within the safe operating area. SOAs are defined in terms of Imax, Vmax and Pmax.
>I will apply 10 amperes of constant current
As I have already explained a constant current source is just a voltage source with current feedback. You are actually changing the terminal voltage to maintain your 10A. I agree charging at Imax is the fastest way of charging a capacitor but my circuit was simpler (constant voltage source with varying current vs your constant current source with varying terminal voltage).
> Will I be able to charge the capacitor faster from a source of 200 volts than from a source of 50 volts?
This is a resounding yes. A constant voltage source rated at 200V will charge a capacitor faster than a source rated at 50V by virtue of the capacitor charging equation (assuming internal resistances are comparable).
> Surely you realize one doesn't attach capacitors to voltage sources, or is this still not understood?
There is no reason why you cannot attach a voltage source to a capacitor. See: http://imgur.com/gallery/sOZ1F. Let me remind you a so called current source is also a voltage source. Current sources are a theoretical construct. Internal resistances have to be considered in a practical circuit otherwise you will have to deal with infinite currents at t=0.
> You will still need an initial jolt (voltage source) to get the current started.
Yes, but that wasn't what I replied to. Your claim was this:
> There is no current without EMF (measured in volts).
That's false. There can be massive currents without any potential difference. Again, superconductors show this behavior.
>> For a constant-current source, the rise is linear
> You do realize there no such thing as a practical constant current source?
Oh, really? So all those power supplies I designed for NASA were not really what they seemed? Constant-current sources are a normal part of electrical engineering practice and have been for decades.
Also, by using a switching inverter and manipulating the relationship between voltage and current in reactive elements, a constant-current source can be made very efficient. This is the ideal way to charge a supercapacitor, and it is standard practice. But I think I already said that.
> I did say raise the voltage as long as you are within the safe operating area.
Yes, but that's not how you charge a capacitor -- you use a controlled current, not a voltage. The voltage follows the current, not the other way around.
>> Will I be able to charge the capacitor faster from a source of 200 volts than from a source of 50 volts?
> This is a resounding yes.
Quite false. For a given current that a capacitor can tolerate, the charging rate is the same regardless of the source voltage. The reason? You don't charge capacitors with voltage, you charge them with current.
The example I gave earlier was a capacitor rated at 50 volts and two current sources -- one that can deliver 200 volts and one that can deliver 50 volts. The charging rate is the same. The reason? You don't charge capacitors with voltage, you charge them with current.
> There is no reason why you cannot attach a voltage source to a capacitor.
No reason at all. But don't be in the same room with a large capacitor and a voltage source that will supply substantial current to maintain a specified voltage. But, you know what? I already said that.
I've been designing power supply circuits for 40 years. My man-rated designs flew on the NASA Space Shuttle. I hold several patents. You're arguing with the wrong person.
> Current sources are a theoretical construct.
Current sources are an everyday, trivial design task that all competent designers must learn to be regarded as employable. See the linked schematic above.
> Internal resistances have to be considered in a practical circuit otherwise you will have to deal with infinite currents at t=0.
That is true for a voltage source. It is not true for a current source, and this is by design.
> They're charged with power, which is current x voltage.
No, they are charged by currents. The normal power source for capacitor charging is a constant current source, one in which the voltage follows the desired current. The least desirable source is a constant-voltage supply, which (in the ideal case) will deliver infinite current at time zero.
The voltage on a capacitor is the time integral of past applied currents. The ideal charging scheme is a constant current source, which cuts off as the capacitor's voltage limit is reached.
Capacitors aren't charged with a voltage, they're charged with a current. The voltage is irrelevant until breakdown occurs. More current, faster charge, less current, slower charge.
Also, there's the practical constraint that a capacitor's physical construction limits the amount of current that can be applied without overheating the capacitor's materials.
The TL;DR: one does not apply a voltage to a capacitor, one applies a current. If you doubt this, try applying a constant-voltage source to a capacitor. See what happens.