Uncategorized

Charging of capacitor in RC circuit

Charging of capacitor in RC circuit

Ask QuestionAsked 2 days agoActive yesterdayViewed 480 times6

When a capacitor is charged in a first order RC circuit, it charges exponentially. I understand this behavior via equations. But can anyone explain the physical reason?capacitorchargingcircuitsshareedit  follow  flag asked 2 days agosumita sahu7766 bronze badges

  • 3Simply notice that at the beginning when the voltage across the capacitor is 0V. All the input voltage will be present across the resistor. Thus, in the beginning, the charging current is the largest. But as the voltage across the capacitor increases the voltage drop across the resistor is reduced (VR = Vin – Vcap), thus decreases the charging current. So, the large the voltage across the capacitor is the smaller the charging current is. And smaller the charging current will be, the more time is needed to charge the capacitor. – G36 2 days ago
  • 1For the physical reason, think about the discrete time approximation to this problem. For each time-step, the charge transferred is proportional to the resistor voltage. Iterating over the time steps we see that the charging is exponential (but probably not if the form ektekt, but of the form aKTaKT). Making the time steps smaller and smaller leads to the exponential. – AJN 2 days ago
  • 5The problem touches the question why we find very often an exponential function in nature. This e-function is very often the solution of diff. equations because the 1st derivation equals the origial function.: – LvW 2 days ago
  • 2I don’t understand questions like this. You state that you understand this behavior via equations. So, what does it mean to ask “what is the physical reason?”. The physical reason for what?. Physics consists essentially of finding mathematical models that adequately describe physical phenomena in the appropriate approximation. Are you asking why physical capacitors can be approximately described mathematically such that the (ideal) RC circuit has the characteristic exponential charge / discharge solutions? If not, please elaborate on what you’re actually looking for in an answer. – Alfred Centauri yesterday 

add a commentstart a bounty

6 Answers

ActiveOldestVotes7

schematic

simulate this circuit – Schematic created using CircuitLab

The current is determined by the voltage across the resistor, which is V1-Vc. As the capacitor charges, Vc increases while V1 stays the same, so the current decreases. The rate at which a capacitor charges is directly proportional to the current, so the rate at which it charges decreases proportional to its current state of charge–the classic differential equation for an exponential decay.shareedit  follow  flag answered 2 days agoHearth11.6k22 gold badges2727 silver badges6262 bronze badges

  • 1Yes, I understand that the Vc increases with time but why exponential? why not linear or some other nature? (I understand this mathematically but I want to know physical significance) – sumita sahu 2 days ago 
  • 1I’m not sure how to explain that any more clearly than the mathematics already does. It’s baked into how physics and mathematics work. – Hearth 2 days ago
  • 1Someone asked me this during an interview and I was blank, I searched a lot but can’t find anything. – sumita sahu 2 days ago
  • 1And explaining the derivation from the differential equation didn’t work? – Hearth 2 days ago
  • 1No, he said he wants a physical reason, no equations. – sumita sahu 2 days ago
  • 1Because the charging current is not constant. Thus the voltage across the capacitor will not increase linearly because V = Q/C = (I * t)/C. – G36 2 days ago
  • 4If someone asks for a physical reason with no equations, I question what kind of physics they’ve learned…. – Hearth 2 days ago
  • 1I rather think that a good engineer/physician is able to explain basic things even without formulas and equations. I rememeber that there is even a book about electronics – without any equations. – LvW 2 days ago
  • 1@LvW I bet you that book did have equations; they were just written differently. – wizzwizz4 yesterday
  • 1@LvW, An engineer that is an inventor even will not try to explain why the relation is not linear; it will simply make it linear with a clever trick:) This is the great idea of the negative feedback – it does not care what the disturbance is; it simply compensates for it with an equivalent anti-disturbance. See my answer… – Circuit fantasist yesterday 

add a comment5

When a series RC circuit is applied across a fixed DC voltage, the capacitor begins charging. It begins charging from 0 volts and, at that instant, the current that charges the capacitor is defined by the DC voltage and the value of the series resistor. That’s simple ohm’s law (if you are allowed to use that).

As the capacitor charges, the voltage across it rises from 0 volts and this means that the voltage across the resistor must reduce. Again, using ohm’s law, if the resistor voltage reduces then, the charging current must also reduce. This is because R and C are in series.

So now, because the charging current has reduced, the rate at which the capacitor voltage charges also reduces. I don’t know if you are allowed to use the charge formula in making an explanation but I guess, if you accept that current is the mechanism that forces a capacitor to charge up in voltage then, a reduction in charging current has to mean a slower rate in the rise of capacitor voltage.

Hence, the voltage rate of climb from 0 volts is starting to reduce as the capacitor charges. And, as the voltage climbs more there is even less voltage across the series resistor. In turn that means the charging current becomes even less and the rate of charge voltage across the capacitor slows down more.

More time passes; the rate at which voltage increases becomes less and the current into the capacitor is also less. Ultimately, as the capacitor voltage approaches the fixed DC voltage supply, the current through the resistor is getting very tiny indeed and so the rate of change of voltage of the capacitor is also very tiny.

Eventually (and being practical) the rate at which voltage rises across the capacitor is seen to virtually stop and, the current into the capacitor is virtually zero. An “engineering” equilibrium is reached where the capacitor voltage is virtually the same value as the fixed DC voltage.shareedit  follow  flagedited 2 days agoanswered 2 days agoAndy aka318k1818 gold badges248248 silver badges553553 bronze badgesadd a comment4

Imagine a steel pressure vessel you are trying to charge with compressed air of constant pressure. This vessel will be your capacitor, the capacity — amount of air mass it can store, being the capacitance. The compressor is the power source, outputting a constant air pressure — the voltage.

There is a restriction valve on the pipeline between your compressor and the pressure vessel, which restricts the movement of air, thus becoming a resistor. The flow rate — amount of air mass traveled through the pipeline per second is the current. Because of this restriction valve, the flow cannot be infinite.

As you charge the pressure vessel through the compressor and the restriction valve, the pressure in the vessel will gradually increase. Since the compressor only outputs a constant pressure, the pressure increase on the destination site causes the flow rate to decrease, reducing the speed at which the vessel is charged as it is being charged, until after an infinite amount of time (as in steady state), the compressor output pressure has equalized with the pressure of the vessel, and charging can no longer proceed.

The process of air mass increase slowing down is confirmed to be mathematically equivalent to the exponent representation.shareedit  follow  flag answered 2 days agoJohn Doe16622 bronze badges New contributor

add a comment2

You can think of the capacitor to be a voltage source.In the beginning when the capacitor is completely uncharged there isnt any voltage between the plates of the capacitor because no charge has come to sit on the plates and create a voltage difference. While the capacitor is being charged more and more charge sits on the plates and the result is a voltage differential. Now this opposes the voltage source which charged the capacitor and therefore less current must flow. This process will happen until the voltage of the capacitor becomes equal with the source which charged the capacitor.shareedit  follow  flag answered 2 days agoThe Force Awakens9977 bronze badges

  • 1No, the capacitor is not a voltage source. Saying such things will only cause more confusion later. – Elliot Alderson 2 days ago
  • 1@Elliot Alderson, For the purposes of this excellent intuitive explanation, a capacitor is a voltage source… like a battery…. but rechargeable battery. What is confusing is your formal and sterile thinking. It would be good to gain some practical experience to start feeling what is happening in circuits … – Circuit fantasist yesterday

add a comment2

You obviously see the circuit theory as a kind of symbol game which is disconnected from the physics. Actually you are right. Circuit theory doesn’t care what voltage and current mean, they are only quantities which depend on time and the circuit. Voltage and current are physical in the sense they present the state of something which exists and which isn’t only an imagined relation.

Electrodynamics based on Maxwell’s field theory and some properties of materials is the physics behind the circuit theory. From there come such things as Ohm’s law, Kirchoff’s laws and equation I=C(dU/dt) for capacitors.

If it happens that you like to see a mechanical system which you understand intuitively and which is analoquous with the RC charging circuit think for example heating a mass. The voltage source is there some heating power, the resistor is the not perfectly heat conducting medium between the source and the mass to be heated and the capacitance is the heat capacity of the heated mass. It’s temperature is the charged voltage.shareedit  follow  flagedited 2 days agoanswered 2 days agouser28700116.8k22 gold badges88 silver badges2929 bronze badges

  • 1I don’t like how you state that quantities of a circuit are not physical. – The Force Awakens 2 days ago
  • 1@TheForceAwakens I don’t see that anyone said they are not physical. The statement was that circuit theory doesn’t care what they mean. In other words, circuit theory operates at a higher level of abstraction when compared to physics. Which is just fine. – Elliot Alderson 2 days ago
  • 1@ElliotAlderson Comments can be opinions. The Force Awakens writes about his feelings, he claims nothing about the rightness of my writings. – user287001 2 days ago
  • 1@user287001 The Force Awakens made a statement “you state that quantities of a circuit are not physical”. I did not see such a statement anywhere in your answer, so as far as I can see The Force Awakens made an objectively false statement, not just an opinion. – Elliot Alderson 2 days ago
  • 1Ok. Voltages and currents are physical in the sense they present the state of something existing which we do not consider to be only a relation. Voltages and currents can even be measured, so you are right. – user287001 2 days ago

add a comment-2

Why not linear or some other nature?

My answer will be a little unexpected for you because I will answer not “why not linear” but I will show how it can be made linear. In fact, this is the goal in most cases of practice; exponential relation, with few exceptions, is undesirable.

The trick is extremely simple if only you can guess. The voltage VC across the capacitor does not linearly change because it is subtracted from the input voltage (it is a loss) and the current decreases – I = (VIN – VC)/R. So we have to compensate this voltage drop.

For this purpose, we connect a variable voltage source in series to the capacitor and with the same polarity as the input voltage source (travelling the loop)… and adjust its voltage equal to the voltage drop across the capacitor. As a result, the voltage drop will be removed and the current will be as in the beginning – I = (VIN – VC +VC)/R = VIN/R.

If we feel bored doing this tedious job, we assign it to an op-amp. This is the idea of the op-amp inverting integrator.

Op-amp RC integrator

EDIT: This is a great idea (removing disturbance by anti-disturbance) that we can see everywhere… even in SE EE. There are contributors here for whom the powerful explanations of others are disturbance for his ego; so they try to compensate it with an anti-disturbance (-1)…shareedit  follow  flagedited yesterdayanswered yesterdayCircuit fantasist5,62611 gold badge99 silver badges2626 bronze badges

  • 2I’m not the downvoter but you really need to stop taking it as a personal attack when people downvote you. I would guess the downvoter did so because this doesn’t actually answer the question that was asked, instead going off on a tangent to explain an interesting concept instead. And yes, linearization via feedback is an interesting topic, but it’s not an answer to the question that was asked here. – Hearth yesterday
  • 1So you declare that you have no intention whatsoever of answering the actual question, and you are then surprised when someone says “This answer is not useful”? You could have written a detailed and “powerful” explanation of quantum theory but it would still be useless to the OP. And if I were you I would not say anything about anyone else’s ego. – Elliot Alderson yesterday
  • 1@Hearth, Thanks for the interpretation. There is no problem; we all know each other very well and understand what we are talking about. A little humor is never superfluous. I just saw the downvoter’s reaction and made the connection with the great principle. BTW it can be implemented without negative feedback, e.g., by a “negative capacitor”. Really, this is not the exact answer but it is closely related to it; this is the answer to the next question that logically follows, “How do we make the capacitor charge linearly?” In your place, I would first admire the idea and then criticize… – Circuit fantasist yesterday
  • 1@Elliot Alderson, My answer is closely related to this question; so it is useful. If you take the trouble to follow the link above, you will see a 5-step scenario; the third step is dedicated to this question. Showing how something nonlinear can become linear is an indirect (and more original) way to explain what causes this nonlinearity over time… – Circuit fantasist yesterday 
  • I gave a down vote, because the answer does not explain why the capacitor is charged exponentially, as asked by the OP, but not linearly, quadratically, or cubically, …. – tlfong01 17 hours ago    
  • 1@tlfong01, I think it has become clear that this cannot be done through intuition but through mathematics. Unfortunately, intuition only tells us that the current is progressively decreasing but it can’t tell us exactly how … But nevertheless, I keep thinking about this phenomenon of “communicating vessels”. But now the question arose how to make a -1 ohm resistor. Interesting topics have no end… – Circuit fantasist 17 hours ago
  • @Circuit fantasist, Ha, I very much disagree with you saying that “Unfortunately this cannot be done through intuition, … when [charging] current progressively decreasing [to a very small amount], .. But I very much agree with you saying that “but through mathematics, ..”. Here you hinted that mathematics, or more precisely Calculus gives the answer to the OP’s question: “Why exponential? Now let me show you the picture that explains the origin of the weird thing “e“: i.imgur.com/Yggwixk.jpg. Ah, locking down supper time. Chat later. Cheers. – tlfong01 14 hours ago    
  • 1@tlfong01, Nice resource… I have a rule: qualitative things should be explained by qualitative means and quantitative things should be calculated by quantitative tools. I have noticed that processes in nature where the difference between two pressure-like quantities determines through a flow-like quantity the second pressure-like quantity, lead to an exponent. – Circuit fantasist 14 hours ago 

add a comment

Categories: Uncategorized

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.