How Does Closing a Switch Affect Capacitor Charge Over Time?

[thunderhadron]

Homework Statement

The Problem is as follows :-





https://fbcdn-sphotos-a.akamaihd.net/hphotos-ak-snc6/250882_2277870163529_994629362_n.jpg

In the figure shown initially the switch is open for a long time. Now the switch is closed at t = 0. Find the charge on the rightmost capacitor as a function of time. Given that it was initially uncharged.

Homework Equations

& The attempt at a solution[/B]
When the switch is open if we talk only about the first capacitor (Left one), it is connected to the battery for a very long time hence it is fully charged and the potential difference between its plates become the potential difference of the battery. When the switch is closed at t = 0, the resistance of the capacitor (right one) is zero hence the total charge will flow only in the outward circuit. Hence it is surely the charging of the right one capacitor and hence its equation will become as q= cV(1-e^(-t/RC) ) which is quite obvious. But there is big problem the left capacitor was charged for a very long time and we can assume that now it’ll behave like a battery. There’ll be discharging of left capacitor and charging of right capacitor. The net charge on the right capacitor will be due to both – the battery and the discharging current of the left capacitor. Now how to proceed for the charge from the left one capacitor. Or m I doing wrong in this method, Should I go for some another method I am pretty confusion in doing this.
Please friends help me in solving this issue.

The answer of the question is as follows:
q= cV/2 (1-1/2 e^(-t/RC) )

 

[truesearch]

When the first capacitor is fully charged it will not be to the battery emf.
The 2 resistors form a potential divider across the battery. Do you see what I mean?

 

[Ratch]

Thunderhadron,

In some of your equations you use a lower case "c", and in others, you use a upper case "C". What is the difference?

You need to discern what the voltage the two caps have immediately after the switch is closed. Until you know that, you won't be able to solve your problem. It is an initial condition.

Ratch

 

[thunderhadron]

Thunderhadron,

In some of your equations you use a lower case "c", and in others, you use a upper case "C". What is the difference?

You need to discern what the voltage the two caps have immediately after the switch is closed. Until you know that, you won't be able to solve your problem. It is an initial condition.

For the left capacitor it'll be discharging and for the right capacitor it'll be charging.
Hence the equations are as follows:

For the left capacitor:
V= v₀ e^-2t/RC

And for the right cap:

V = V₀ e ^-t/RC

Now how to add these voltages after that.

 

[Ratch]

Thunderhadron,

When the switch closes, and the left cap energizes the right cap, the path the charges take does not go through R. Therefore R should not be a consideration for the initial voltage after the switch closes.

Ratcfh

 

[tiny-tim]

thunderhadron, look at the official answer …

The answer of the question is as follows:
q= cV/2 (1-1/2 e^-t/RC )

from that, you can see that the charge on the right capacitor starts (t = 0) at CV/4, and ends (t = ∞) at CV/2

how could you find those two results if you weren't given them?

 

[thunderhadron]

thunderhadron, look at the official answer …


from that, you can see that the charge on the right capacitor starts (t = 0) at CV/4, and ends (t = ∞) at CV/2

how could you find those two results if you weren't given them?

So what should I think, the given anwer of the problem is incorrect?

 

[NascentOxygen]

So what should I think, the given anwer of the problem is incorrect?

The given answer is correct. Upon switch closure, the two caps share their charge and thereafter act as one large capacitor.

 

[thunderhadron]

The given answer is correct. Upon switch closure, the two caps share their charge and thereafter act as one large capacitor.

Well, I m not getting one thing and that is before closing the switch the left capacitor is fully charged and charge on it is CV after closing the switch this charge will distribute on both as CV/2 & CV/2


But after closing the switch and long time after the right capacitor will also be fully charged and charge on it should be CV but as the answer indicates q= cV/2 (1-1/2 e^-t/RC ), the maximum charge on the capacitor is CV/2.

How it is happening?

 

[NascentOxygen]

Well, I m not getting one thing and that is before closing the switch the left capacitor is fully charged and charge on it is CV

Can you explain how you arrived at CV?

 

[tiny-tim]

hi thunderhadron!

Well, I m not getting one thing and that is before closing the switch the left capacitor is fully charged and charge on it is CV …

no

as truesearch says …

When the first capacitor is fully charged it will not be to the battery emf.
The 2 resistors form a potential divider across the battery.

… the circuit is the same as

…

so what will the voltage across the "second" resistor be? ​

(this question seems to be from page 6 of http://www.scribd.com/doc/49913344/capacitor-sheet, #II Q.16)

 

[thunderhadron]

hi thunderhadron!


no

as truesearch says …


… the circuit is the same as

…

so what will the voltage across the "second" resistor be? ​

(this question seems to be from page 6 of http://www.scribd.com/doc/49913344/capacitor-sheet, #II Q.16)

Ok I got it so the voltage across left capacitor is V/2 so maximum charge on it before the switch closure is CV/2. and after closing the switch the right capacitor is also in parallel with the left capacitor hence its max voltage will be V/2 and max charge on it will be CV/2.

Thank you friend for making me understand this. But I still don't get the correct answer. as for the value of time constant somewhere I read just short circuit the battery and solve.
When I short circuit the battery the two resistors are in parallel hence R/2 and the cap. are in parallel hence 2C. Hence the time constant is RC.

But from where this 1/2 is appearing on the answer.

(1-1/2 e^-t/RC )

 

[tiny-tim]

hi thunderhadron!

Ok I got it so the voltage across left capacitor is V/2 so maximum charge on it before the switch closure is CV/2 …

yes

and after closing the switch the right capacitor is also in parallel with the left capacitor hence its max voltage will be V/2 and max charge on it will be CV/2.

nooo

that makes a total charge (on both capacitors) of CV …

where has the extra charge come from?? ​

 

[thunderhadron]

Hi tiny-tim!
Fiend I don't get this. Well Isn't the charge on right capacitor CV/2. Well the extra charge might come from the battery. After closing the switch the right cap is also connected with battery.
Isn't it so?

Please don't mind my foolish questions.

 

[tiny-tim]

… the extra charge might come from the battery. After closing the switch …
Please don't mind my foolish questions.

no, i don't mind at all …

you're now stating your understanding clearly, and that makes it easy to help you

(your original post, unfortunately, did not reveal how you got there )​

no, the charge can't come from "outside", it would have to pass through the resistors in an "infinitely" short time, which makes it an "infinitely" large current and power

impulsive currents can't go through resistors

when the switch is suddenly closed, the total charge on the capacitors is (temporarily) stuck there

 

[thunderhadron]

Yes I get this after a very long time as the question states the left cap is fully charged hence no charge will flow through it. hence after the switch closing all the charge will flow through the right cap. and the ckt will look like as you posted in post #11.
The same situation will happen for solving the potential through it with the right cap. But at this instant what is the role of left cap.?

 

[tiny-tim]

Yes I get this after a very long time as the question states the left cap is fully charged hence no charge will flow through it.

ahh! that's where you're going wrong …

no charge will flow into it

but charge can flow out of it! ​

 

[thunderhadron]

Yes and out of it there is right cap which is totally uncharged before the closing the switch. so the circuit will look like the same ckt as you showed me.
And the P.D. across the right capacitor will become V/2 due to battery. M I going wrong?

 

[tiny-tim]

yes the voltage across the two capacitors will be the same so long as the switch is closed (and will be the same as across the "second" resistor)

so what are the charges on each capacitor when the switch is first closed?

 

[thunderhadron]

yes the voltage across the two capacitors will be the same so long as the switch is closed (and will be the same as across the "second" resistor)

so what are the charges on each capacitor when the switch is first closed?

As the voltage across the two capacitors and the second resistor is same as V/2 then the charges on both the capacitors according the relation Q = CV are CV/2 & CV/2
?

 

[tiny-tim]

Ok I got it so the voltage across left capacitor is V/2 so maximum charge on it before the switch closure is CV/2

As the voltage across the two capacitors and the second resistor is same as V/2 then the charges on both the capacitors according the relation Q = CV are CV/2 & CV/2
?

no, if the charge on the left capacitor was CV/2 immediately before closing the switch (and 0 on the right capacitor), how can it be CV/2 on both capacitors (total of CV) immediately after?

the total charge has to be the same … impulsive current cannot flow through resistors

(the maximum possible charge, ie the relation Q = CV, is irrelevant)

 

[thunderhadron]

Do u mean to say that the charge on both will become CV/4 ?

 

[tiny-tim]

Do u mean to say that the charge on both will become CV/4 ?

yes … it's the same total charge, and the capacitances are the same, so it will be equally distributed

does that make the rest of the problem clear?​

 

[thunderhadron]

OK.. I am getting this. But please tell me one thing: At the time of closing the the switch the right cap is connected with left capacitor hence the charge distribute equally on both. Bit the right cap is also connected to the battery at this instance so what is the role of battery at the moment?

 

[tiny-tim]

… Bit the right cap is also connected to the battery at this instance so what is the role of battery at the moment?

at that moment (between "immediately before" and "immediately after" the switch closes), the battery and the resistors have no role …

the battery and the resistors need time to do anything, but the charge rearranges itself "instantaneously" …

comparing it with mechanics: the switch gives an impulsive force, and non-impulsive forces (such as gravity, or the battery) have no role​

 

[thunderhadron]

But some times when I solve some problems of charging- discharging, take the example of a simple circuit when a cap and a resistor are connected in parallel and a battery is connected with this circuit in parallel.
Many times the problem says at the instance of closing the switch : Current will have two ways on the junction of resistance and cap. but the current will flow through capa at that time, it acts like zero resistance or superconducting wire at t = 0 but the time goes on the cap stores charge by leaps and bounce and resistance of it keeps on increasing.
And after a very long time when it gets fully charged it no charge flow through it, it acts like
R→∞.
Is this wrong, so please tell me what is the actual concept but mind one thing I've solved ample number of problem keeping this concept in my mind, If you want I can show you some examples.

 

[NascentOxygen]

But some times when I solve some problems of charging- discharging, take the example of a simple circuit when a cap and a resistor are connected in parallel and a battery is connected with this circuit in parallel.
Many times the problem says at the instance of closing the switch : Current will have two ways on the junction of resistance and cap. but the current will flow through capa at that time, it acts like zero resistance or superconducting wire at t = 0 but the time goes on the cap stores charge by leaps and bounce and resistance of it keeps on increasing.

That's correct. The capacitor offers very low impedance under those circumstances. But if there is a resistance in series with the capacitor. then that resistor limits the current and hampers the capacitor in doing what it might otherwise do. Only where the effective series resistance is very low can the capacitor branch act like a "short circuit". You have to examine the circuit very closely to see which resistances are in the relevant charge/discharge current paths and under which circumstances.

 

[tiny-tim]

… at the instance of closing the switch : Current will have two ways on the junction of resistance and cap.

i don't understand that

… but the current will flow through capa at that time, it acts like zero resistance or superconducting wire at t = 0

you do know that no current ever flows through a capacitor?

talking about the "resistance" of a capacitor is just an analogy

a capacitor does not have a resistance (R), it has a reactance

ie it is not Ohmic, with I proportional to V, it is reactive, with I proportional to dV/dt

but the time goes on the cap stores charge by leaps and bounce

(you mean "leaps and bounds"?)

no, the charge flows smoothly …

it starts fast, and ends slow, but the change is smooth

immediately after you close the switch, the charge on the capacitor is still zero, but the rate of charge is non-zero​

 

[thunderhadron]

Current will have two ways on the junction of resistance and cap. but the current will flow through capa at that time

I am sending you an img. http://img3.orkut.com/images/milieu/1340682480/1340707691932/204964818/ln/Zt96paa.jpg?ver=1340707704

Well I mean to say that at the instance of closing the ckt the current has two ways at point c, one from capacitor link and another from resistor through wire cd. So the resistance sorry capacitive reactance of the topmost capacitor is very less at this time hence the charge will flow form capacitor.

 

[thunderhadron]

Friend please make me understand some according to this fig. Is this fig is correct. Will the ckt look like this?

 

[NascentOxygen]

What do you mean by "the topmost capacitor"?

 

[Yukoel]

I am sending you an img. http://img3.orkut.com/images/milieu/1340682480/1340707691932/204964818/ln/Zt96paa.jpg?ver=1340707704

Well I mean to say that at the instance of closing the ckt the current has two ways at point c, one from capacitor link and another from resistor through wire cd. So the resistance sorry capacitive reactance of the topmost capacitor is very less at this time hence the charge will flow form capacitor.

Hello thunderharon,
Look at the rectangular loop formed by the right and left capacitor.Apply Kirchhoff's Voltage rule.The capacitances being same they ought have the same charge initially.The left capacitor achieved steady state in the time before switch was closed ,true but now it has to achieve the steady state once again in accordance with Kirchhoff's rules.
regards
Yukoel

 

[tiny-tim]

Many times the problem says at the instance of closing the switch : Current will have two ways on the junction of resistance and cap. but the current will flow through capa at that time, it acts like zero resistance or superconducting wire at t = 0 but the time goes on the cap stores charge by leaps and bounce and resistance of it keeps on increasing.
And after a very long time when it gets fully charged it no charge flow through it, it acts like
R→∞.

ok, i understand now …

you're talking about problems with a resistor and a capacitor (or more than one
capacitor) in parallel, so that the current (or charge) can flow both ways

then yes, you're correct (apart from the "leaps and bounds"): the current (which before the switch closed was flowing entirely towards the resistor) will start flowing entirely towards the capacitor, but will then flow more and more towards the resistor until finally it flows entirely towards the resistor

(as if the capacitor started with zero resistance, and ended with infinite resistance)

 

[thunderhadron]

::Yukoel:: So like "tiny-tim" was telling me Its P.D. will become V/4 now so as the P.D. of the right capacitor. Is this correct?

 

[thunderhadron]

What do you mean by "the topmost capacitor"?

Sorry friend that was a slip of hand. I mean to say Rightmost..:shy:

 

[thunderhadron]

(apart from the "leaps and bounds"): the current (which before the switch closed was flowing entirely towards the resistor) will start flowing entirely towards the capacitor, but will then flow more and more towards the resistor until finally it flows entirely towards the resistor

(In aspect of "leaps and bounds"): I mean to say slowly sorry if I use wrong words.

So is this theory doesn't fit for this problem also?

 

[tiny-tim]

Many times the problem says at the instance of closing the switch : Current will have two ways on the junction of resistance and cap. but the current will flow through capa at that time, it acts like zero resistance or superconducting wire at t = 0 but the time goes on the cap stores charge [slowly] and resistance of it keeps on increasing.
And after a very long time when it gets fully charged it no charge flow through it, it acts like
R→∞.

So is this theory doesn't fit for this problem also?

yes it does, the two capacitors behave as one capacitor, which at first has zero reactance, with all the current going through it and none going through the resistor, and after a very long time it has infinite reactance, with all the charge going through the resistor

the only difference is that before all this happens, the charge rearranges itself from one capacitor to the other (the same charge)

 

[thunderhadron]

yes it does

the charge rearranges itself from one capacitor to the other (the same charge)

i.e. CV/4 , CV/4

And now charges have been redistributed in both of them just after the switch closed. But now the right cap is connected with battery also. Will it accumulate some more charge also? surplus?

 

[tiny-tim]

i.e. CV/4 , CV/4
…
But now the right cap is connected with battery also. Will it accumulate some more charge also? surplus?

the left and right capacitors are now one capacitor

just cross them out in the diagram, and replace by one capacitor!

(even if their capacitances were different, they would still form one capacitor, but the initial charges would be CV/6, CV/3, say, and then the charges would stay in that proportion)

 

[Yukoel]

::Yukoel:: So like "tiny-tim" was telling me Its P.D. will become V/4 now so as the P.D. of the right capacitor. Is this correct?

Yes .As tim says you have to make one capacitor out of the two and technically they have been corrected in parallel.A prerequisite for parallel connection is same P.D. right?After it is done you have capacitor with an initial charge (adjust the charge in it according to the potential difference across the ends).Find the charge on this new capacitor as a function of time and then again using potential difference across it as a function of time you can retrieve what charges subside on the individual capacitors.

regards
Yukoel

 

[thunderhadron]

Hi friends,
As you guys told me to take the capacitors in parallel and try to find the time dependent equation, I am attaching an image of what I did to solve the ckt.
http://img4.orkut.com/images/milieu/1340682480/1341024902370/204964818/ln/Z96sxil.jpg?ver=1341024924

Now this charge will distribute on both the capacitors equally.
Hence the equation of the charge on both the capacitors will be

q = CV/2 (1 - e^-t/RC)

But still this doesn't matches with the answer of the main problem.
Please tell me where I am doing wrong?

Thank you in advance for the reply.

 

[Yukoel]

Hi friends,
As you guys told me to take the capacitors in parallel and try to find the time dependent equation, I am attaching an image of what I did to solve the ckt.
http://img4.orkut.com/images/milieu/1340682480/1341024902370/204964818/ln/Z96sxil.jpg?ver=1341024924

Now this charge will distribute on both the capacitors equally.
Hence the equation of the charge on both the capacitors will be

q = CV/2 (1 - e^-t/RC)

But still this doesn't matches with the answer of the main problem.
Please tell me where I am doing wrong?

Thank you in advance for the reply.

Hello thunderhadron,
The image is very small and is not magnified hence difficult to see.Anyways note that at t=0 q=_________? Your equation doesn't yield that .Does it?
Now you have unified the capacitors .The new capacitor will have a different value of charge for the same p.d isn't it?Relate that to each capacitor and get the correct equation.Maybe this would help.
regards
Yukoel

 

[tiny-tim]

hi thunderhadron!

(your picture is too small to read )

Now this charge will distribute on both the capacitors equally.
Hence the equation of the charge on both the capacitors will be

q = CV/2 (1 - e^-t/RC)

yes, the total charge on the double-capacitor starts at CV/2, so the left part is correct

but the double-capacitor does not have capacitance C, it has a new capacitance which must go into the time constant …

what is that capacitance?​

 

[thunderhadron]

hi thunderhadron!

(your picture is too small to read )


yes, the total charge on the double-capacitor starts at CV/2, so the left part is correct

but the double-capacitor does not have capacitance C, it has a new capacitance which must go into the time constant …

what is that capacitance?​

2C ??

Well sorry for the size of img friends. Well I am attaching the same img. Please check it out.

 

[tiny-tim]

2C ??

yes, of course (two capacitors in parallel, so you add the capacitances)

Please check it out.

sorry, but i don't understand the lettering

j and i seem to be the same currents, and i don't understand when you're using i and when i₁

can you please rethink your diagram, and then type your equations directly onto the forum?

(and I'm now going out for the rest of the day)

 

[thunderhadron]

yes, of course (two capacitors in parallel, so you add the capacitances)


sorry, but i don't understand the lettering

j and i seem to be the same currents, and i don't understand when you're using i and when i₁

can you please rethink your diagram, and then type your equations directly onto the forum?

(and I'm now going out for the rest of the day)

i₁ = dq / dt

Applying Kirchoff's law for loop C D E F C,

q / 2C - (i - i₁)R = 0

\Rightarrow q / 2C - i.R + i₁. R = 0 --------------eqⁿ (i) let

Applying Kirchoff's law for loop A B F G A,

V = i. R + (i - i₁)R ----------------------------eqⁿ (2) let

Solving eqn (1) & eqn (2),

eqⁿ (1) X 2 + eqⁿ (2),

q / C + i₁. R = V

\Rightarrow i₁. R = V - q / C - - - -- - - - -- --eqⁿ (3)

but, i₁ = dq / dt

hence putting i₁ = dq / dt in eqⁿ (3) we'll get ,


dq / dt = (VC - q) / RC

\Rightarrow dq / (VC - q) = dt / RC

Integrating, above equation

∫ dq / (VC - q) = ∫dt / RC

I don't find definite integral sign here so explaining,

Left integral : Lower limit : 0
Upper limit : q


Right integral : Lower limit : 0
Upper limit : t


\Rightarrow ln [ (VC - q) / VC ] = - t / RC

\Rightarrow (VC - q) / VC = e^{- t / RC}

\Rightarrow ( VC - q ) = VC . e^{- t / RC}

\Rightarrow q = VC - VC . e^{- t / RC}

\Rightarrow q = VC ( 1 - e^{- t / RC} )


Now this charge will distribute in both the capacitors equally as, q = VC / 2 . ( 1 - e^{- t / RC} )

But answer is not the same.

I am also attaching another image. Please check it out.

Please make me clear.

Thank you.

 

[Yukoel]

i₁ = dq / dt


Integrating, above equation

∫ dq / (VC - q) = ∫dt / RC

I don't find definite integral sign here so explaining,

Left integral : Lower limit : 0
Upper limit : q


Right integral : Lower limit : 0
Upper limit : t

Hello thunderhadron,
I think you have misused the limits here.As tim explained at t=0 each capacitor gets a CV/4 amount of charge and thus the resultant 2C capacitor gets ________C of charge,isnt it?
So at t=0 do you really have q=0?Or the value of q is the same as that intended in the blank?
Try substituting the correct value of q at t=0.The rest of your method is correct .With the right value of q at t=0 the answer is also correct.

regards
Yukoel

 

[tiny-tim]

hi thunderhadron!

(just got up :zzz: i had a good day yesterday! )

dq / (VC - q) = dt / RC

Integrating, above equation

∫ dq / (VC - q) = ∫dt / RC

I don't find definite integral sign here so explaining,

Left integral : Lower limit : 0
Upper limit : q


Right integral : Lower limit : 0
Upper limit : t


\Rightarrow ln [ (VC - q) / VC ] = - t / RC

i wouldn't do the integral like that, with limits

for a ln integral, i'd say …

∫ dq / (q - VC) = -∫dt / RC

so |q - VC| = constant*e^-t/RC …​

and then all you need do is find what the constant is (at t = 0)

(apart from that, as Yukoel says, that's fine!)

 

[thunderhadron]

Oh My God!
That's a blunder mistake I've done here. Thank you guys I'll get back to you soon.:!)

 

[tiny-tim]

a thundermistake!