Why is there inrush current in a capacitor?

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walter

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Jun 21, 2006
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Why is there inrush current in a capacitor?

I recently had a job interview where I was asked about an RC circuit. I thought I did a good job at explaining what happens when the switch is thrown. I stated that there is inrush current as the cap charges to almost the supply in five time constants, mostly in the first time constant. Then I was asked “Why is there inrush current in a capacitor?” I said because the capacitor is at zero volts and wants to be at supply level, but that was not the correct answer, and the message was that I need to be constantly learning and brushing up on what I have learned in the past. It was a busy week and I didn’t think much about it. A few days later I had a follow up interview where I was asked of this round of interviews which was the hardest question and if I could give a better answer now. I said “Why is there inrush current in a capacitor?”, and when prompted, I responded “Ohms law.” The rest of the interview seemed like a flaming death spiral. I have asked Six different people this question and received Four different responses. I found this video that shows the math around the RC circuit:
https://www.youtube.com/watch?v=OwbQLrNBDpY
At the end of the video, the equation is simplified to Ohms law, but that is not the Why, that is the math supporting what happens at t=0, the instant the switch is thrown. At the beginning and the end of this video, it is stated “at t=0 the uncharged capacitor behaves like a wire so I total = VB/R”. I suspect that there is inrush current in a partially charged capacitor as well as a completely discharged capacitor, so it seems like I am barking up the wrong tree. I have also read that a capacitor is open to DC (low frequency) and short to High frequency (A\C) and that a transition from 0v to DC supply is a high frequency event, turning a switch on. So the reason that there is inrush current is due to the supply voltage transitioning from Zero (or something) to max in a very short period of time. And when this short period of time is plugged into the equation: I=Io e^-t/RC it yields a high current. Is this correct?
 
Because when there is a voltage potential one of the metal plates is stripped of its electrons while the other plate is charged with electrons from the source. As this happens current flows. There's a point of saturation when the current stops, unless the potential is really strong (HV) and then the current will keep flowing and the cap will Blow!

Thats what I would have said at least...Probably wouldn't get the job though.

In case anyone doesn't know, I'll spit out my rudimentary knowledge regarding AC current and frequency:

As a low frequency current is rising to its peak, the capacitor plates are charging and being stripped. If the plates are large enough, they will be able to accept all of the incoming electrons as the voltage is rising. As the voltage reaches its peak and starts to reverse its direction, the electrons will leave that plate and start to build up on the opposite plate.

As long as the plates of the capacitor have enough area it will continue to charge and discharge as the current flows in one direction then the other. It will have no effect on the current except for a small time delay. So the plates never saturate.
If the plates are small and saturate before the voltage rises to its peak, current will stop flowing and the capacitor becomes an open circuit until the current changes direction.

So because a higher frequency means the direction is changing quicker, you don't need as big of plates as you do with a lower frequency current.

But when I try to picture a multiple frequency AC current pass through a capacitor like in audio, it kinds of boggles my mind....
 
walter said:
..............I have also read that a capacitor is open to DC (low frequency) and short to High frequency (A\C) and that a transition from ...........

Yes, people keep using this regularly but it is wrong,

DC is not a low frequency. It is zero frequency. That is why it is called DC.

 
If the capacitor is initially uncharged then there is no potential across its plates. When you throw the switch, the entire supply voltage appears across the resistor and the instantaneous current that flows is V/R; good old ohms law. However, the flowing current charges up the capacitor so its potential rises. This means the voltage across the resistor drops so the current though it drops until eventually it reaches zero when the charge on the capacitor equals the supply voltage V.

Cheers

Ian
 
sahib said:
DC is not a low frequency. It is zero frequency. That is why it is called DC.


:eek: :eek: :eek:

wouldnt it better "not a cycling frequency as 60 hz"  !

1 hz technically doesnt have a cycling frequency, as u coulnt measure...
or where exactly u would  measure such large (low)  frequency!

edit: or 200000000000000000000000 mega zilion hz





 
walter said:
Why is there inrush current in a capacitor?

I recently had a job interview where I was asked about an RC circuit. I thought I did a good job at explaining what happens when the switch is thrown. I stated that there is inrush current as the cap charges to almost the supply in five time constants, mostly in the first time constant. Then I was asked “Why is there inrush current in a capacitor?” I said because the capacitor is at zero volts and wants to be at supply level, but that was not the correct answer, and the message was that I need to be constantly learning and brushing up on what I have learned in the past. It was a busy week and I didn’t think much about it. A few days later I had a follow up interview where I was asked of this round of interviews which was the hardest question and if I could give a better answer now. I said “Why is there inrush current in a capacitor?”, and when prompted, I responded “Ohms law.” The rest of the interview seemed like a flaming death spiral. I have asked Six different people this question and received Four different responses. I found this video that shows the math around the RC circuit:
https://www.youtube.com/watch?v=OwbQLrNBDpY
At the end of the video, the equation is simplified to Ohms law, but that is not the Why, that is the math supporting what happens at t=0, the instant the switch is thrown. At the beginning and the end of this video, it is stated “at t=0 the uncharged capacitor behaves like a wire so I total = VB/R”. I suspect that there is inrush current in a partially charged capacitor as well as a completely discharged capacitor, so it seems like I am barking up the wrong tree. I have also read that a capacitor is open to DC (low frequency) and short to High frequency (A\C) and that a transition from 0v to DC supply is a high frequency event, turning a switch on. So the reason that there is inrush current is due to the supply voltage transitioning from Zero (or something) to max in a very short period of time. And when this short period of time is plugged into the equation: I=Io e^-t/RC it yields a high current. Is this correct?
You may be getting tripped up by semantics... The capacitor impedance isn't the issue but the voltage drop across the R changes as the capacitor charges up from the current flowing into or out of it. This changing voltage (divided by R) defines the current (ohms law).

The equation for this charging curve is e^(-t/RC)....  I am not a math head (trust me) but I memorized this equation because it very specifically defines how cap terminal voltage changes in RC circuits (and pocket calculators have an "e" button.)  ;D

The current is completely defined by the resistor and voltage across it. Don't be confused by how caps behave when driven by diodes or very low impedance sources. An RC has an "R" in series limiting the current.

JR
 
kambo said:
sahib said:
DC is not a low frequency. It is zero frequency. That is why it is called DC.


:eek: :eek: :eek:

wouldnt it better "not a cycling frequency as 60 hz"  !

No.

There are conditions where you can describe DC with reference to frequency, but  not within this context. For example, you rectify 50Hz AC. Now you can say that the fundamental of this signal is DC because it does not alternate below or above the ZERO point.


1 hz technically doesnt have a cycling frequency, as u coulnt measure...

Where did you get that?

As soon as a signal departs from ZERO point and starts to swing in both positive and negative extremes then it is AC no matter how small the frequency is. You just have to study electrical fundamentals a bit.

Measuring 1Hz has never been a problem.


or where exactly u would  measure such large (low)  frequency!



There are signal generators with DC to gigahertz range and scopes and frequency counters have never had any problem reading them.


 
if i was patient enough to watch, 0.0000000000001 hz i would c AC  movement on my scope. so its AC.
i wasnt arguing that way :)
i meant if the signal is DC enough or not. then u can apply DC rules to AC.
AC and DC has nothing with the frequency anyway, if its alternating or not, isnt it!
if u look at 20khz at very short time period it will look DC!

anyone can measure 1hz, i dint mean that way, i meant its footprint is too large to measure.

 
kambo said:
if i was patient enough to watch, 0.0000000000001 hz i would c AC  movement on my scope. so its AC.
i wasnt arguing that way :)
i meant if the signal is DC enough or not. then u can apply DC rules to AC.
AC and DC has nothing with the frequency anyway, if its alternating or not, isnt it!
if u look at 20khz at very short time period it will look DC!

anyone can measure 1hz, i dint mean that way, i meant its footprint is too large to measure.

I sincerely do not mean to offend but this is not really an intelligent argument. As I said before please obtain a good book in electrical fundamentals and start reading. However a short and simple response to get you start ticking in that direction;

1). There is no such thing as "if the signal is DC enough". It is either DC or not. However, a signal can have AC and DC components at the same time as I gave in the rectifier example.

2). The only time you can apply DC rules to AC is when you are diving a voltage over a resistor. Even then when the frequency is high enough the resistor no longer behaves like it did in, say, audio frequency.

3). DC has nothing to do with frequency as the name implies it does not alternate. But AC has everything to do with frequency.  If the frequency is low enough you may simplify your calculations but theoretically as soon as it starts to alternate  above and below its reference then all the frequency dependent factors come into play.

4). Sometimes I look at my aunt and see my uncle too. But she is still my aunt.

5). You already changed your mind on measuring 1Hz?



 
sahib said:
kambo said:
if i was patient enough to watch, 0.0000000000001 hz i would c AC  movement on my scope. so its AC.
i wasnt arguing that way :)
i meant if the signal is DC enough or not. then u can apply DC rules to AC.
AC and DC has nothing with the frequency anyway, if its alternating or not, isnt it!
if u look at 20khz at very short time period it will look DC!

anyone can measure 1hz, i dint mean that way, i meant its footprint is too large to measure.

I sincerely do not mean to offend but this is not really an intelligent argument. As I said before please obtain a good book in electrical fundamentals and start reading. However a short and simple response to get you start ticking in that direction;

1). There is no such thing as "if the signal is DC enough". It is either DC or not. However, a signal can have AC and DC components at the same time as I gave in the rectifier example.

2). The only time you can apply DC rules to AC is when you are diving a voltage over a resistor. Even then when the frequency is high enough the resistor no longer behaves like it did in, say, audio frequency.

3). DC has nothing to do with frequency as the name implies it does not alternate. But AC has everything to do with frequency.  If the frequency is low enough you may simplify your calculations but theoretically as soon as it starts to alternate  above and below its reference then all the frequency dependent factors come into play.

4). Sometimes I look at my aunt and see my uncle too. But she is still my aunt.

5). You already changed your mind on measuring 1Hz?

none taken :)

1. may be not in ur generation!
2. so u do apply dc rules to ac for ur calculations.
3. everything in this universe has its frequency,  even DC.  frequency is time related.  as i said if u measure
20khz in a very short period of time, you will c DC
4. never met ur aunt nor ur uncle! 
5. no i havent. go measure physically 1hz signal. 
 
For any still following along this threat has followed the trajectory of so many internet discussions. Instead of thoughtfully answering the OP's question, we get cranium dumps from posters sharing everything they think they know vaguely on topic, then they argue among themselves over unrelated details.

I hope Walter can find some useful wheat among all this chaff.  ;D

JR 
 
Why is there in-rush air when you break a vacuum-filled CRT in normal atmosphere?

Why is there in-rush water when you take a closed empty (air-filled) bottle to the bottom of a swimming pool and open it?

For that matter, how-come in an engine, when the piston goes down and the valve is open, air rushes in? (True, we normally open the valve when piston is near top, and the air flow gradually rises as the piston goes down.)

Nature does not just abhor a vacuum, she abhors a difference.

I don't know What The Frik those interviewers were expecting.

It is almost too obvious to explain.

It is a nice point in the math. "Ohm's Law" stumbles because the simplified/ideal starting point is Zero Ohms.

But if you are hired to do work, not to theorize, there's always impedance in source and cap. You can do practical thinking by assuming the maximum current is "maybe 100 times" the full load current. My 240V 100A power line, I only need breakers rated for 10,000A surge.

> capacitor is at zero volts and wants to be at supply level

OK, eXcept I would object to a cap that "wants" to do something. It *will* go to the new voltage, just like the empty soda-bottle at the bottom of the pool *will* fill with water. It will fill as FAST as possible, limited by pressure difference and flow restriction.
 
"In-rush" current generally describes some kind of response to a step function (like switching on a power amp) where current is limited by things like line cord resistance, transformer winding resistance, etc... not a typical RC circuit. .

If I read Walter's post correctly he mentioned "in-rush" in the context of a discussion about RC circuits, and was asked to explain "in-rush".  They found an excuse to gong him,  learn from it and don't make that mistake next time. 

RC circuits don't rush... e^(-t/RC)

JR



 
JohnRoberts said:
For any still following along this threat has followed the trajectory of so many internet discussions. Instead of thoughtfully answering the OP's question, we get cranium dumps from posters sharing everything they think they know vaguely on topic, then they argue among themselves over unrelated details.


True. I apologise to Walter for muddling up his post by responding to a garbage that somehow looking at a 10kHz signal on a scope long enough to see DC from somebody who clearly lacks basic electrical fundamentals. Again, my apologies.

On the other hand, however, I too

.... hope Walter can find some useful wheat among all this chaff.

JR

as my second post was intended to draw his attention to his comment;

walter said:
.......... I have also read that a capacitor is open to DC (low frequency)..............

which if he gave as  part of his answer, would swamp  all the good that he did.

 
sahib said:
True. I apologise to Walter for muddling up his post by responding to a garbage that somehow looking at a 10kHz signal on a scope long enough to see DC from somebody who clearly lacks basic electrical fundamentals. Again, my apologies.

ohhhhhh what a butterrrrr cup  ::) ::) ::) ::)

where exactly did you read that  :eek:

since u are so intelligent, let me break it down for you;
i was saying :
if you look at at ur AC signal at μs. setting  on ur scope, u will c flat a line : same as ur DC flat line.

 
Thanks guys for proving my point....

Low S/N  is not a problem unique to this forum,,, it's world-wide on many different topics.

I can be as guilty as everyone for my veers.

JR

PS: Again I am making general observations from decades of spending too much time around forums like this.  Don't take it personally or too seriously.
 
To get this back on topic..............

There are two very helpful capacitor equations, but only one is required to understand the inrush problem.

                                                                            Q=It

Where Q = the charge on a capacitor in Coulombs, I = the current in Amperes and t = the current in seconds.

At the instant of turn on, when t = 0, any number for Q divided by 0 is infinity.  Of course it can't be infinite current as the power supply will have some resistance to restrict the current.

The second equation is Q = CV where C = the capacitance in Farads and V the Voltage.

It becomes obvious then that CV = It as they both = Q

therefore CV/I = t and as V/I =R (Ohms law), we end up with the attack and release time constant equation CR = t

I hope that helps

DaveP
 
Thanks to all who replied. This gives me some more to chew on, and a bit to grit my teeth. Most of my job is soldering wires to test points. The interviewer is used to hiring engineers. The question was meant to determine if I am passionately curious about learning. My manager wants to hire me from contract to full time, so we will have a re-do now that I know their game. They want employees who can collaborate and articulate concepts, I did not find those qualities in that last interviewer -not in a positive way. Of the six people I asked this question, one was supportive and I thoroughly enjoyed the conversation, another became defensive and left me befuddled. It’s good to understand why there is inrush current, but seems more important to know that it does occur and what can be done to counter it if necessary.
 
walter said:
Thanks to all who replied. This gives me some more to chew on, and a bit to grit my teeth. Most of my job is soldering wires to test points. The interviewer is used to hiring engineers. The question was meant to determine if I am passionately curious about learning. My manager wants to hire me from contract to full time, so we will have a re-do now that I know their game. They want employees who can collaborate and articulate concepts, I did not find those qualities in that last interviewer -not in a positive way. Of the six people I asked this question, one was supportive and I thoroughly enjoyed the conversation, another became defensive and left me befuddled. It’s good to understand why there is inrush current, but seems more important to know that it does occur and what can be done to counter it if necessary.
Ian gave you the right answer, but I wonder if the questioner was hoping you'd get to something like using an NTC Thermistor to mitigate the initial current.
 

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