Charging a partially charged capacitor

  • #1
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1. The problem statement and My attempt
>>> Pic <<<

I got wrong answer!!!

the answer is that they just wrote the eqn's of charging a completely discharged capacitor and discharging a capacitor.

i.e. CE (1 - e-t/RC) + Q e-t/RC

Now does that make any sense?
______________________________

Edit:
Woops ... forgot to add picture
 

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Answers and Replies

  • #2
Delphi51
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That isn't very clear and there is no picture . . .
Please give the exact question so we can be sure what it is.
 
  • #3
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I attached it !!!! :)
 
  • #4
SammyS
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It's just the principle of superposition.
 
  • #6
gneill
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The superposition theorem allows you to model an initially charged capacitor as an "empty" capacitor in series with an ideal voltage supply equal to the initial voltage of the capacitor, Q/C.
 
  • #7
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So you mean that charging and discharging of a capacitor are two independent processes?
 
  • #8
gneill
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So you mean that charging and discharging of a capacitor are two independent processes?

Well, that's certainly true in this case since the first capacitor was given its charge before t = 0.

If you take a "snapshot" of a circuit at any given instant and "measure" the charge on each capacitor, you could model the circuit from that time on by replacing each capacitor with an equivalent model consisting of an empty capacitor in series with an ideal voltage supply.
 
  • #9
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But this does not make any sense.

Why would a charged capacitor loose charge if charge given to it < CE
 
  • #10
gneill
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But this does not make any sense.

Why would a charged capacitor loose charge if charge given to it < CE

The "real" capacitor doesn't lose charge just because you replaced it with an equivalent model. If you want to find out what the charge is on the "real" capacitor at any time, then you convert back to the "real" model (total voltage across the model multiplied by the capacitance).

But perhaps this is over complicating things in this case. If you can write an equation for the voltage across the capacitor w.r.t. time, then you can always find the charge on it via Q = V*C.

At time zero, the capacitor has a certain voltage on it. It's just been connected to a resistor in series with another voltage E. Current will flow until there are no voltage differences, so the capacitor voltage will be driven towards E. It'll be a typical RC circuit, so you can find the time constant and write the equation for the voltage (and hence the charge) by inspection.
 
  • #11
Delphi51
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In your integration step, I think you assumed Q was a constant but it varies with time. It looks like a differential equation for which the answer can be guessed and checked . .

EDIT: I see my mistake; confused between Q and q.
 
Last edited:
  • #12
SammyS
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1. The problem statement and My attempt
>>> Pic <<<

I got wrong answer!!! ( No! You got the right answer! )

the answer is that they just wrote the eqn's of charging a completely discharged capacitor and discharging a capacitor.

i.e. CE (1 - e-t/RC) + Q e-t/RC

Now does that make any sense?
______________________________

Edit:
Woops ... forgot to add picture

Your setup has the total charge on the capacitor as being: Q+q.

Your answer for q is:

[tex]q=\left(\mathcal{E}C-Q\right)\left(1-e^{-t/(RC)}\right)[/tex]

[tex]q=\mathcal{E}C\left(1-e^{-t/(RC)}\right)-Q+Qe^{-t/(RC)}[/tex]

So that:

[tex]Q+q=\mathcal{E}C\left(1-e^{-t/(RC)}\right)+Qe^{-t/(RC)}\,,[/tex]

Which IS the book answer.

Edit (additional stuff):

If you had defined q as being ths total charge on the capacitor, then your integration over q would have been:

[tex]\int_Q^q \frac{dq}{\mathcal{E}C-q}\,,[/tex] which also would have worked out fine.
 
Last edited:
  • #13
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Your setup has the total charge on the capacitor as being: Q+q.

Your answer for q is:

[tex]q=\left(\mathcal{E}C-Q\right)\left(1-e^{-t/(RC)}\right)[/tex]

[tex]q=\mathcal{E}C\left(1-e^{-t/(RC)}\right)-Q+Qe^{-t/(RC)}[/tex]

So that:

[tex]Q+q=\mathcal{E}C\left(1-e^{-t/(RC)}\right)+Qe^{-t/(RC)}\,,[/tex]

Which IS the book answer.

Edit (additional stuff):

If you had defined q as being ths total charge on the capacitor, then your integration over q would have been:

[tex]\int_Q^q \frac{dq}{\mathcal{E}C-q}\,,[/tex] which also would have worked out fine.

Hey yes!!!

You are right!!!!!!!!!!!
I totally forgot that my q was extra charge added to capacitor!!!!!!!!
WOW!!!! You saved me!!!!!!!
I was starting to think that i am a total dumb in this topic!!!!!!!!!!!!!!!!!!!!!!!

Thanks a lot bro!!!!!!!!!!!!!!!
 

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