- #1

Zahid Iftikhar

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I want to know if there is any physical explanation of this effect. Please guide me on this.

Regards

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- Thread starter Zahid Iftikhar
- Start date

- #1

Zahid Iftikhar

- 116

- 24

I want to know if there is any physical explanation of this effect. Please guide me on this.

Regards

- #2

DaveE

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The current that flows to charge the capacitor decreases as the capacitor voltage increases because the voltage across the resistor is the difference between the input voltage and the output voltage. So the more the capacitor is charged, the slower the rate of charge.

- #3

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You should carefully derive and solve the equation of motion for the circuit . At time ##t=0## you close a switch connecting the capacitor charged at voltage ##V_0## with a resistor.

Hint: The equation of motion reads

$$V+R C\dot{V}=0.$$

- #4

Zahid Iftikhar

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Summary:Mathematically it can be proved that time constant for charging and discharging of a capacitor is t=RC and it is time in which 63% of the capacitor fills up. During next time constant 63% of the left-over capacitor is filled. I want to know its physical explanation.

Statement of problem is given in the summary. Mathematics proves time constant is calculated from the equation V= V0 exp t/RC

I want to know if there is any physical explanation of this effect. Please guide me on this.

Regards

I am absolutely thankful for your time to address my query. What I meant from my question was how a capacitor knows it should charge to 63% of the equilibrium charge, not 70% or 80% or any other value. Please see the clipping I have attached if my question is still not clear.

The current that flows to charge the capacitor decreases as the capacitor voltage increases because the voltage across the resistor is the difference between the input voltage and the output voltage. So the more the capacitor is charged, the slower the rate of charge.

High regards.

Zahid

- #5

Nugatory

Mentor

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It doesn’t “know” anything. The greater the resistance R the more slowly charge moves into the capacitor, the greater the capacitance the more charge needs to move, and when you work through the math you’ll see that these effects combine so that RC is the time it takes for the capacitor to reach 63%. If you increase R or C it will take the capacitor longer to charge to 63%, but the value of RC will also be greater to compensate.What I meant from my question was how a capacitor knows it should charge to 63% of the equilibrium charge,

And of course the capacitor doesn’t stop charging after time RC - it keeps on accepting more charge until eventually it reaches equilibrium.

- #6

Mister T

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What I meant from my question was how a capacitor knows it should charge to 63% of the equilibrium charge, not 70% or 80% or any other value.

But it doesn't stop charging at that point! It continues to charge, asymptotically approaching 100%.

- #7

TurtleMeister

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I understand your question, but I don't know the answer. I'm only responding because I find it interesting. It could be that this is just one of those "why" questions that physics is not very good at answering. Keep in mind that the 63% is an approximation. I don't know, but I suspect, that it is an irrational number. The farthest I've seen it measured or carried out is 63.21%.What I meant from my question was how a capacitor knows it should charge to 63% of the equilibrium charge, not 70% or 80% or any other value.

- #8

DaveE

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It's 1-1/e. Again, from the solution to the differential equation. It's more of a definition than a mystery.I understand your question, but I don't know the answer. I'm only responding because I find it interesting. It could be that this is just one of those "why" questions that physics is not very good at answering. Keep in mind that the 63% is an approximation. I don't know, but I suspect, that it is an irrational number. The farthest I've seen it measured or carried out is 63.21%.

- #9

Tom.G

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Let's try a physical experiment, one where you can actually

You may have used a siphon sometime in your life. That is useful when you want to remove a liquid from a container that you can not pick up and pour the water out. You fill a hose with water and block at least one end so it stays full. Then put one end of the hose in the liquid you want to remove.

The other end of the hose is put in the receiving container,

Now remove the blockage(s) from the hose ends. The water will flow thru the hose, up over the edge of the big container, and into the the receiving container.

In the experiment, the big container is the battery, the hose is the wire and resistor, and the receiving container is the capacitor.

At first, the receiving container is filling fast. As its water level rises, it fills more slowly until its water level matches that of the big container. When the levels are the same, the flow stops. Try it. Lift the small container so the levels are the same. If you raise it higher than the big container the flow will reverse.

If that is too messy or too much trouble, you can do the same thing with a double sink if you have one in your kitchen. The same approach works even if you do not have a hose; you can use a wet dish towel instead.

You can raise and lower the outlet end of the hose, or dish towel, to see how the flow changes as it nears the water level in the full sink.

A capacitor charging acts the same way; as its voltage (level in receiving container) approaches the battery voltage (liquid level in the source) there is less voltage (height difference) across the resistor (hose) resultiing in less current (water flow).

That fancy capacitor voltage equation

V

was figured out to fit what happens.

Well, kind of long winded. Hope it helps.

Cheers,

Tom

- #10

anorlunda

Staff Emeritus

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.63 is (1-1/e)

Do you know what e is?

Do you know how we calculate (1-1/e)?

- #11

RPinPA

Science Advisor

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The things like the meaning of the RC time constant and the 63% are consequences that are derived from that instantaneous equation. They aren't fundamental physical principles. The capacitor doesn't know it. But we know the long term behavior of systems with that instantaneous behavior.

- #12

TurtleMeister

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Thanks. That helps a lot. I'm not a math person and only use it as a means to an end for whatever I'm dealing with. I had heard of Euler's number before, and even unknowingly used the concept in calculating compound interest, but never studied it. So my guess about it being an irrational number was correct..63 is (1-1/e)

- #13

Zahid Iftikhar

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- #14

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It is utmost important to state a problem clearly. So what's definitely the setup you want to study? If it's charging an uncharged capacitor by hooking it (via a resistance) to a DC voltage source of voltage ##U_0##, the differential equation to solve can be derived, in the quasistationary approximation.

Integrating along the circuit, which is a series of a capacitor ##C## with a battery and an resistor ##R##. Let the momentaneous voltage difference on the capacitor be ##U##. Then the momentaneous current through the circuit is

$$i=\dot{Q}=C \dot{U}$$

and thus integrating the electric field along the circuit you get

$$U+R i-U_0=0 \; \Rightarrow \; U + R C \dot{U}=U_0.$$

This is a linear differential equation. The general solution is the sum of the general solution of the homogeneous equation and one special solution of the homogeneous equation. As the latter we can obviously take

$$U_{\text{inh}}(t)=U_0=\text{const}.$$

The homogeneous equation has the general colution

$$U_{\text{hom}}(t)=A \exp[-t/(RC)] \; \Rightarrow \; U(t)=U_{\text{hom}}(t)+U_{\text{inh}}(t)=U_0 + A \exp[-t/(RC)].$$

Here ##A## is an integration constant, which we have to determine from the initial condition, ##U(0)=0##, leading to ##A=-U_0##. Then we finally get

$$U(t)=U_0 \left [1-\exp\left (-\frac{t}{RC} \right) \right].$$

Thus you see that the capacitor is asymptotically charged to the full voltage of the battery. It doesn't stop at 63%. I have no clue, where you got the idea from that a capacitor would stop charging at this value.

That this must be wrong you can also get from the physics of the problem without any calculation. Connecting the uncharged capacitor with the voltage via the resistor at the beginning enables the charges of the voltage source to flow (with some resistance due to friction of the electron's motion through the resistor) to the plates of the capacitor. Due to charge conservation, one plate must carry some charge ##+Q## and the other the opposite charge ##-Q## at any time. Now when the capacitor is charged, there's an electric field built up, which works against the further charging of the capacitor. However, the battery keeps charging until the electric field between the capacitor is strong enough. Now when stops further charging? Obviously at the moment, when the charge on one plate is such that the voltage difference ##U## of this electric field between the capacitor plates reaches the value ##U_0##.

The above calculation quantitatively confirms this. Note that formally you need an infinite time to fully charge the capacitor to ##U=U_0## since obviously ##U(t)## is monotonously growing for all time and only for ##t \rightarrow \infty## you reach ##U \rightarrow U_0##.

Now it's also interesting, how fast the capacitor charges, and this is obviously given by the typical time ##\tau=RC##. I guess that's where your 63% come from. For ##t=\tau## you have

$$U(\tau)=U_0[1-\exp(-1)] \simeq 0.632 U_0.$$

But that's just to give a measure for typical timescale for how quickly the capacitor charges.

- #15

Zahid Iftikhar

- 116

- 24

That is a brilliant analogy Sir, very helpful to understand capacitor working.

Let's try a physical experiment, one where you can actuallyseewhat is happening.

You may have used a siphon sometime in your life. That is useful when you want to remove a liquid from a container that you can not pick up and pour the water out. You fill a hose with water and block at least one end so it stays full. Then put one end of the hose in the liquid you want to remove.

The other end of the hose is put in the receiving container,with this end below the liquid level of the big container.

Now remove the blockage(s) from the hose ends. The water will flow thru the hose, up over the edge of the big container, and into the the receiving container.

In the experiment, the big container is the battery, the hose is the wire and resistor, and the receiving container is the capacitor.

At first, the receiving container is filling fast. As its water level rises, it fills more slowly until its water level matches that of the big container. When the levels are the same, the flow stops. Try it. Lift the small container so the levels are the same. If you raise it higher than the big container the flow will reverse.

If that is too messy or too much trouble, you can do the same thing with a double sink if you have one in your kitchen. The same approach works even if you do not have a hose; you can use a wet dish towel instead.

You can raise and lower the outlet end of the hose, or dish towel, to see how the flow changes as it nears the water level in the full sink.

A capacitor charging acts the same way; as its voltage (level in receiving container) approaches the battery voltage (liquid level in the source) there is less voltage (height difference) across the resistor (hose) resultiing in less current (water flow).

That fancy capacitor voltage equation

V_{CAP}= V_{BATT}(1-e^^{(t/RC)})

was figured out to fit what happens.

Well, kind of long winded. Hope it helps.

Cheers,

Tom

I use another while teaching to my classes. Two identical water tanks connected thru a valve, one tank remains full and is filled automatically if its level changes (Battery) and the other is empty (Capacitor). The valve does the job of a resistor for water flow. Both the tanks are placed with their bases at the same height. Now the flow of water resembles current which is high at start and gradually drops as the empty tanks keeps getting filled. I explain time constant by variation of size of the empty tank and the valve.

During this experiment it often comes to my mind whether the receiving tank also follows some kind of 63% filling rule. I am not sure about it but it is interesting to experiment it.

- #16

Zahid Iftikhar

- 116

- 24

Thanks Sir. You said that it was more of a definition. To my understanding definitions are man-made meant to simplify the things and once decided may be helpful to understand laws and principles.(I may be wrong). I have studied in this thread that it is not a Physics law or Principle (time constant). May we change this definition and say time constant is the time in which a capacitor charges to 70% of its equilibrium charge. I think we can't. If we are bound to say it is 63% always, isn't kind of law or sort of thing. I apologize for my little knowledge.It's 1-1/e. Again, from the solution to the differential equation. It's more of a definition than a mystery.

- #17

Zahid Iftikhar

- 116

- 24

It is very kind of you Sir to come for help. I am sure you will encourage to ask rather silly questions. I am sure these will bridge some knowledge gaps. Regards.I think the thread has reached the maximum-entropy state of maximal confusion of the OP. So I think there's no other way than to present the solution. I know, "frontal teaching" is out of fashion, and rightfully at PF we try to get students to find the solution themselves with some help, but at one point this pedagogical ideal fails due to enhancing the student's confusion rather than to help. I think this point is reached here. So here's the solution:

- #18

Zahid Iftikhar

- 116

- 24

Thanks indeed for the detailed explanation. I am sure I have lacked in exactly communicating my question and I added a clipping to support me to clarify. I understand in all you wrote above. I understand a capacitor keeps charging for infinite time and in every time constant there is an increment of 0.63 times the equilibrium charge to what it already had.Thus you see that the capacitor is asymptotically charged to the full voltage of the battery. It doesn't stop at 63%. I have no clue, where you got the idea from that a capacitor would stop charging at this value.

That this must be wrong you can also get from the physics of the problem without any calculation. Connecting the uncharged capacitor with the voltage via the resistor at the beginning enables the charges of the voltage source to flow (with some resistance due to friction of the electron's motion through the resistor) to the plates of the capacitor. Due to charge conservation, one plate must carry some charge +Q+Q+Q and the other the opposite charge −Q−Q-Q at any time. Now when the capacitor is charged, there's an electric field built up, which works against the further charging of the capacitor. However, the battery keeps charging until the electric field between the capacitor is strong enough. Now when stops further charging? Obviously at the moment, when the charge on one plate is such that the voltage difference UUU of this electric field between the capacitor plates reaches the value U0U0U_0.

The above calculation quantitatively confirms this. Note that formally you need an infinite time to fully charge the capacitor to U=U0U=U0U=U_0 since obviously U(t)U(t)U(t) is monotonously growing for all time and only for t→∞t→∞t \rightarrow \infty you reach U→U0U→U0U \rightarrow U_0.

My actual question is about the physical interpretation of this figure of 63%. Why does a capacitor follow this figure? May we explain it without going into the differential equations? Regards

- #19

Nugatory

Mentor

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Because of the mathematical relationship between R, C, and the rate at which charge flows into the capacitor.My actual question is about the physical interpretation of this figure of 63%. Why does a capacitor follow this figure?

Your question is like asking how a ten liter container "knows" to be half full if we pour in 5/N liters every day for N days - there's a mathematical relationship between the amount we add each time and the number of times that tells us how much total we've added.

No, or at least not in any more detail than the hand-waving explanation I gave in post #5 of this thread.May we explain it without going into the differential equations?

Last edited:

- #20

DaveE

Science Advisor

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You can if you like. Personally, I prefer τ = RC, but there's no reason you couldn't define τ = 1.2⋅RC.May we change this definition and say time constant is the time in which a capacitor charges to 70% of its equilibrium charge.

I think you are missing the point with these questions. The equation for the solution is all you need to know.

U(t)=U0[1−exp(−t/RC)]

No, not precisely. If anyone does come up with a different method, I would argue that they are just doing differential equations in an unusual way.May we explain it without going into the differential equations?

- #21

Zahid Iftikhar

- 116

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Thanks indeed. This is the best part of the discussion. I am way satisfied. Thanks to everybody who helped. You all are great people.You can if you like. Personally, I prefer τ = RC, but there's no reason you couldn't define τ = 1.2⋅RC.

I think you are missing the point with these questions. The equation for the solution is all you need to know.

U(t)=U0[1−exp(−t/RC)]

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