# Capacitor across an ideal current source

How long will the capacitor below charge? Given that both the current source and capacitor are ideal.

If someone says the capacitor will be charging up to its capacity, what is the capacity of this capacitor?

simulate this circuit – Schematic created using CircuitLab

• In the ideal case, this circuit is very boring! It will charge to infinite voltage in infinite time and it will take Voltage (in Volts) * Capacitance (in Farads) seconds to reach any arbitrary voltage along the way up. – DrFriedParts Jun 6 '16 at 7:49

Ideal:
U = I * t / C
The voltage will be increasing over time as the current keeps charging.

In reality:
- The current source will stop charging because it's maximum output voltage is reached.
OR
- The capacitor has reached it's maximum voltage and will break.

• what will be the maximum voltage of the capacitor in this case upto which the capacitor will not break? – nikhil pandya Jun 6 '16 at 7:23
• An ideal capacitor does not break. – CL. Jun 6 '16 at 7:26
• @nikhilpandya Depends what it says on the label of the capacitor. – user253751 Jun 6 '16 at 7:55
• @immibis An ideal capacitor does not have a label. (It would just detract from the picture of the unicorn.) – CL. Jun 6 '16 at 8:27
• @JimDearden No, the entire circuit is immersed in a liquid unobtainium bath, preventing any arcing. (While solid unobtainium makes an excellent conductor, liquid unobtainium is an excellent insulator, for some reason) – user253751 Jun 6 '16 at 20:19

Ideal? For ever.

As of course there are no ideal current source in reality it will stop when the capacitor reach the maximum voltage that the current generator can provide.

F = C / V, so if your ideal current source is capable of generating an infinitely large voltage difference, it will also generate an infinitely large charge.

The capacitance of a capacitor tells you how much charge is required to get a voltage of 1V across the capacitor.

Putting a charge of 1uC into a capacitor of 1uF will result in a voltage of 1V across its terminals.

An ideal capacitor can take an infinite amount of charge resulting in an infinitely high voltage.

It's like an infinitely high bucket, where the diameter would determine the water level for a specific amount of water. The relationship between amount of water and water level could be likened to the capacitance of a capacitor.

The time for the capacitor to charge to 100% is infinity, however it can be considered for practical purposes charged when it reaches around 99 - 99.9% which will be 5 time constants

To calculate time constant (TC) use formula TC = R*C

Now usually you would have a resistor infront of the capacitor and this calculation would be easy. For 500 Ohms resistor calculation is 500*0.000001 = 0.0005 seconds for one time constant. Fully charged is 5 time constants so 0.0005seconds * 5 = 0.0025seconds fully charged.

In your case the only resistance is the wire. There will always be resistance. You could measure resistance of wire with multimeter or LCR meter. It will be very small value maybe 0.001 ohms and that will be used for formula.

If you want calculation for ideal you may be saying zero resistance, then the formula wont work, and the charge time might be instant (impossible) therefore its not good to think about ideal, you must instead give the wire some value of resistance to have result based in reality.

• No. The rules you describe apply when the supply is constant voltage, and with a resistor (indeed). Here, we have a constant current source. Nothing you say applies for the circuit OP submitted. And the fact there is no resistor should have gotten you thinking. And 99.9% of what, here, by the way ? – dim Jun 6 '16 at 7:41
• I didn't DV, but this answer, while largely correct about real circuits, doesn't apply to this question as asked (current source, ideal, et. al.). Please consider revising. Thank you! – DrFriedParts Jun 6 '16 at 7:43
• Thank you for your edits. Unfortunately, the core issue still remains. Namely, the "resistor" you keep talking about is actually inside the current source (as it's output resistance/impedance) and the value keeps changing (this is how the current source regulates its current output -- well, at least one way to model it). So your coarse formulas with fixed R, while correct approximations, don't really apply here (or at least aren't useful). – DrFriedParts Jun 6 '16 at 7:53
• Yea sorry. I edited typo before noticing these comments. My knowlege is still basic. I did not distinguish the current source from a voltage source hence my answer. – tylerdurden Jun 6 '16 at 8:23