How Does Closing a Switch Affect Capacitor Charge Over Time?

[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!

 

[thunderhadron]

a thundermistake!

lol

 

[thunderhadron]

Hi friends,

There is a problem after integrating with proper limits its going in this manner,

∫dq/ (CV - q) = ∫ dt/ RC

Left limit ==> CV/4 to q &

Right limit ==> 0 to t

ln[(CV - q)/ CV] = - t/RC

Placing the limits,

ln[(CV - q)/ CV] - ln [ (CV - CV/4) / CV] = -t / RC

since, ln a - ln b = ln (a/b)

hence,

ln [(CV - q) / (3CV/4) ] = -t / RC

==> (CV - q) / (3CV/4) = e^{-t / RC}

==> (CV - q) = (3CV/4) . e^{-t / RC}

==> q = CV - (3CV/4) . e^{-t / RC}

==> q = CV ( 1 - 3/4 . e^{-t / RC} )

Now it'll distribute in both as,

q = CV/2 . ( 1 - 3/4 . e^{-t / RC} )


Still it is not the answer as given, q = CV / 2 . ( 1 - 1/2 . e^{-t / RC} )

 

[tiny-tim]

ln[(CV - q)/ CV] - ln [ (CV - CV/4) / CV] = -t / RC

no, you're getting confused …

your q is the charge on both capacitors combined, so at t = 0 that's q = CV/2

(and i really really REALLY think you should stop using this "limits" method, use standard integration as in my previous post )

 

[thunderhadron]

Oh My dear tim,
I got the correct answer. And I've understand the problem completely now. And that is only by din't of you & Yukoel.

Cordially Thank you very much guys.

I appreciate your help.:!)