# A question about capacitors and current draw **URL:** \ **Category:** Factual Questions\ **Created:** [June 26, 2003, 5:14am UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464 "2003-06-26T05:14:15Z")\ **Posts on this page:** 13\ **Page:** 1
**Author:** ![Trigonal\_Planar](https://avatars.discourse-cdn.com/v4/letter/t/0ea827/32.png) [@Trigonal\_Planar](https://boards.straightdope.com/u/Trigonal_Planar)\ **Post date:** [June 26, 2003, 5:14am UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/1 "2003-06-26T05:14:15Z")
Current drawn by a capacitor is the derivative of the voltage across it. Constant voltage = no current draw. The faster the voltage chanes, the higher the current draw (I think I’m correct so far). Now, suppose you are using a perfect battery is the power supply. Obviously, this battery can only supply a certain maximum current. So what if you changed the voltage across the capacitor so fast such as to produce a current draw that exceeds what the battery can supply? What would happen in such a situation? ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [June 26, 2003, 5:17am UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/2 "2003-06-26T05:17:47Z")
Better to think of the capacitor as an impedance. Int he case you describe, the impedance would eventually drop so low as to equal the internal resistance of the battery, and the current would stop rising with continued decrease in impedance level, just the same as with a large resistive load. Or am I misunderstanding your question? ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [June 26, 2003, 5:30am UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/3 "2003-06-26T05:30:33Z")
Batteries have internal resistance so even if you short-circuit the terminals, the current would be finite. If you connected an empty capacitor, it would be similar to short-circuiting the terminals for a short time (i.e. the time it takes to charge the capacitor at this max current). If this time is long enough (i.e. if the capacitor is _huge_), the battery may explode before the capacitor is fully charged. ---
**Author:** ![Trigonal\_Planar](https://avatars.discourse-cdn.com/v4/letter/t/0ea827/32.png) [@Trigonal\_Planar](https://boards.straightdope.com/u/Trigonal_Planar)\ **Post date:** [June 26, 2003, 5:32am UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/4 "2003-06-26T05:32:07Z")
You may or may not be misunderstanding my question, as I don’t really understand impedance. My knowledge of impedance doesn’t go beyond “a synonym for resistance” (which I know isn’t quite right). I’m just thinking mathematically here. You change the voltage at a certain rate, which causes a certain current to flow. What if this current value exceeds what the power source can supply? ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [June 26, 2003, 5:41am UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/5 "2003-06-26T05:41:08Z")
Impedance is the oppostion to AC current. It’s measured in ohms, just like resistance, and is frequency dependant. For capacitors, higher frequency yields lower impedance, while the opposite is the case for inductors. Again, the case you describe is no different than any other overload of a power supply. The current peaks at some maximum value, and the supply may overheat and burn oout/explode/whatever. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [June 26, 2003, 5:45am UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/6 "2003-06-26T05:45:31Z")
Actually, let me clarify slightly. Basically I described _reactance_. Impedance is reaslly the vector sum of reactance and resistance. ---
**Author:** ![Trigonal\_Planar](https://avatars.discourse-cdn.com/v4/letter/t/0ea827/32.png) [@Trigonal\_Planar](https://boards.straightdope.com/u/Trigonal_Planar)\ **Post date:** [June 26, 2003, 6:05am UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/7 "2003-06-26T06:05:34Z")
Okay, so lets say you have a battery, switch and a capacitor - no resistor. You press the switch so the capacitor is instantly connected to the battery. That’s a pretty fast dv/dt…do I risk damaging the battery by doing this? I’ve certainly done it with no (apparently) ill effects. (also, not sure why you introduced AC. I’m talking DC from a battery here.) ---
**Author:** ![Antonius\_Block](https://avatars.discourse-cdn.com/v4/letter/a/51bf81/32.png) [@Antonius\_Block](https://boards.straightdope.com/u/Antonius_Block)\ **Post date:** [June 26, 2003, 7:02am UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/8 "2003-06-26T07:02:08Z")
> [@](#): > > \*Originally posted by Trigonal Planar \* > **Okay, so lets say you have a battery, switch and a capacitor - no resistor. You press the switch so the capacitor is instantly connected to the battery. That’s a pretty fast dv/dt…do I risk damaging the battery by doing this? I’ve certainly done it with no (apparently) ill effects.** Except that the battery and capacitor both have internal resistances that are non-negligible. It is the battery’s internal resistance that _causes_ the current limit. Your “perfect battery” in the OP would not have a current limit! There is also some resistance in the wiring and in the switch, and inductance in all elements of the circuit, but these can _usually_ be ignored as their effects are swamped by internal resistance. The dv/dt when the switch is closed will be determined, to a first approximation, by the following parameters: 1. the battery voltage (or, if the capacitor is not totally discharged, the voltage difference between the two) 2. the capacitance value 3. the internal resistances (and, usually to a lesser extent the internal inductances) of the battery and capacitor. Your statement above: > [@](#): > > You change the voltage at a certain rate, which causes a certain current to flow. What if this current value exceeds what the power source can supply? Suggests that you think that you get to decide both the dV/dt **and** the current flow (which you want to be greater than a known maximum). In the real world, you _don’t_ get to define both; the values are not independent. > [@](#): > > Also, not sure why you introduced AC. I’m talking DC from a battery here. **Q.E.D.** brings up AC because the situation you are describing is not a “steady-state situation”. Whenever you have a change in conditions (i.e. switch closure leading to current flowing where there previously was none), you need to consider impedance and reactance. The terms DC and AC can be confusing because they serve “double duty”. Sometimes AC is used strictly to mean that the voltage alternates between positive and negative values with respect to zero, but it can also mean that it’s a varying voltage of constant sign. This is an example of the latter case. ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [June 26, 2003, 12:06pm UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/9 "2003-06-26T12:06:54Z")
Haven’t you asked the same question before, **Trigonal Planar** with your OP [Understanding capacitor charging](http://boards.straightdope.com/sdmb/showthread.php?s=&threadid=177843)? I’ll repeat my answer from last time: > [@](#): > > \*Originally posted by Desmostylus \* > \*\*Okay. > > You’re charging a capacitor with capacitance C. > > You’re charging from a source that has voltage V, and source resistance R. > > The voltage across the capacitor with respect to time is: > > _v_[sub]c[/sub] = V ( 1 - e[sup]-_t_/(RC)[/sup] ) > > The current is: > > _i_ = (V/R) e[sup]-_t_/(RC)[/sup] > > If there is no source resistance, it takes zero time to charge the capacitor, and the transient current is infinite. \*\* You’ve stated this time: > [@](#): > > _Originally posted by Trigonal Planar_ > \*\*Now, suppose you are using a perfect battery is the power supply. Obviously, this battery can only supply a certain maximum current. \*\* This is a contradiction. A “perfect” battery can supply infinite current. A real battery can’t. Either way, you just plug the battery’s internal resistance into the equation and out comes the answer, whether you use R=0 for a perfect battery or R≠0 for a real battery. And, as others have noted, real capacitors (and real wires) have non-zero resistances. They also have non-zero inductances. These factors also act to reduce the actual current that flows through the capacitor. ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [June 26, 2003, 12:23pm UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/10 "2003-06-26T12:23:10Z")
This reminds of the question of a ladder leaning against a wall that you drag the bottom end away from the wall at a constant velocity and calculate the downwards speed of the other end (the one in contact with the wall) as exceeding the speed of light at the very last instant before it hits the ground. The only explanation is that it is impossible to drag the other end at a constant velocity. In this case, there are no perfect batteries, no pure capacitances, no resistanceless wires and if you suppose there are why should the violation of other laws of physics be a surprise? ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [June 26, 2003, 12:32pm UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/11 "2003-06-26T12:32:05Z")
I thought the explanation was that the Earth’s gravity is incapable of accelerating an object beyond it’s own escape velocity, which is nowhere near the speed of light. ---
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [June 26, 2003, 2:24pm UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/12 "2003-06-26T14:24:31Z")
> [@](#): > > \*Originally posted by Trigonal Planar \* > **Okay, so lets say you have a battery, switch and a capacitor - no resistor. You press the switch so the capacitor is instantly connected to the battery. That’s a pretty fast dv/dt…** If the battery, switch, capacitor, and conductors are modeled as “ideal,” and there’s no resistor in the circuit, then the circuit would blow up and the universe would self-destruct. But let’s say you build this circuit anyway. You flip the switch and… nothing extraordinary happens. The fact that you’re still alive proves there _must_ be series resistance (and/or series inductance) in the circuit. > [@](#): > > \*Originally posted by Trigonal Planar \* > \*\* do I risk damaging the battery by doing this? \*\* Probably not, unless the capacitor was really big. I’d be more concerned about harming the capacitor. ---
**Author:** ![Trigonal\_Planar](https://avatars.discourse-cdn.com/v4/letter/t/0ea827/32.png) [@Trigonal\_Planar](https://boards.straightdope.com/u/Trigonal_Planar)\ **Post date:** [June 26, 2003, 6:08pm UTC](https://boards.straightdope.com/t/a-question-about-capacitors-and-current-draw/184464/13 "2003-06-26T18:08:33Z")
Okay, so when all is said and done, the problem lies in assuming ideal circuit components which doesn’t reflect “real life”. I get it. Thanks.