Charging a partially charged capacitor

In summary, the problem statement is that the authors wrote equations for charging a completely discharged capacitor and discharging a capacitor, but the equations don't make sense. The solution is to replace the capacitor with an equivalent model that consists of an empty capacitor in series with an ideal voltage supply. If you want to find out what the charge on the "real" capacitor is at any time, then you convert back to the "real" model (total voltage across the model multiplied by the capacitance).
  • #1
cupid.callin
1,132
1
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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  • #2
That isn't very clear and there is no picture . . .
Please give the exact question so we can be sure what it is.
 
  • #3
I attached it ! :)
 
  • #4
It's just the principle of superposition.
 
  • #5
SammyS said:
It's just the principle of superposition.

how?
 
  • #6
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
So you mean that charging and discharging of a capacitor are two independent processes?
 
  • #8
cupid.callin said:
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
But this does not make any sense.

Why would a charged capacitor loose charge if charge given to it < CE
 
  • #10
cupid.callin said:
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
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.
 
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  • #12
cupid.callin said:
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
SammyS said:
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!
 

What is a partially charged capacitor?

A partially charged capacitor is a device that stores electrical energy in the form of an electric charge. It consists of two conductive plates separated by an insulating material, and when a voltage is applied, one plate becomes positively charged and the other becomes negatively charged.

Why would you need to charge a partially charged capacitor?

A partially charged capacitor may need to be charged in order to reach its maximum capacity and store the maximum amount of electrical energy. This is particularly important in applications where the capacitor is used to power a device or circuit.

How do you charge a partially charged capacitor?

To charge a partially charged capacitor, you will need to connect it to a power source, such as a battery or power supply, with the correct voltage. The capacitor should be connected with the positive side of the power source to the positive plate of the capacitor, and the negative side of the power source to the negative plate of the capacitor.

What happens if you overcharge a partially charged capacitor?

If a partially charged capacitor is overcharged, it can potentially lead to the capacitor becoming damaged or even exploding. This is because the capacitor is not designed to handle excessive amounts of charge, and the excess charge can cause the insulating material to break down.

Can you discharge a partially charged capacitor?

Yes, it is possible to discharge a partially charged capacitor. This can be done by connecting the positive and negative plates of the capacitor with a resistor. This will allow the stored charge to dissipate gradually and safely.

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