# Finding the charge on capacitor

Discussion in 'Homework Help' started by Niles, Dec 26, 2008.

1. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
Hi all.

Please take a look at this circuit. Here V is constant (V = 1.5 V).

I have to find the charge on the capacitor when t -> infinity. Using Kirchhoffs loop rule on the "outer" loop, I find:

V = Q(t)/C.

Does this mean that Q(t) = VC for all times, i.e. it is constant? This does not seem right. Where is my error?

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2. ### DrNick Active Member

Dec 13, 2006
110
2
at infinity the Voltage across the capacitor should be V. SO, use Q = C*V to find the charge.

Now when the circuit is first "turned on" there is a transient that will oscillate for some time, but since there is a resistor in the resonant circuit, it should be damped (and therefor well settle to a constant V across the capacitor).

3. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
But then I am right? I.e. Q = CV for all times t?

4. ### mik3 Senior Member

Feb 4, 2008
4,846
69
Yes, but it is not constant because V varies.

5. ### Niles Thread Starter Active Member

Nov 23, 2008
56
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V is 1.5 V, it is direct current.

6. ### mik3 Senior Member

Feb 4, 2008
4,846
69
Yes, but if the source was not present and at an instant you placed it, the voltage across the capacitor would vary in some way and at infinite time it will stabilize at 1.5V.

7. ### Niles Thread Starter Active Member

Nov 23, 2008
56
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I agree 100% with your reasoning, but Kirchhoffs law is telling me that it is constant for all times. That's why I am confused.

8. ### mik3 Senior Member

Feb 4, 2008
4,846
69
Kirchhoff's law is true if the battery can supply infinite amount of current on zero time!!! as to be able to charge the capacitor instantly.

9. ### Niles Thread Starter Active Member

Nov 23, 2008
56
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Ok, the question is to find the charge for times t -> infinity, which is Q = CV, so my original question is answered.

But I still find this topic interesting: How would you find Q(t) for the capacitor then?

10. ### mik3 Senior Member

Feb 4, 2008
4,846
69
If you assume an ideal battery and ideal wires then the charge of the capacitor equals C*V and is constant. However, in reality this is not true.

11. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
So there is no way to find the transient behavior of the capacitor before it settles on the value Q=CV?

12. ### Ratch New Member

Mar 20, 2007
1,068
4
Niles,

Because you don't specify any resistance in series with the capacitor, a sudden turnon of 1.5 volts will cause a infinite amount of charge to be withdrawn from the battery for an infinitesimal amount of time. The capacitor will instantly have a charge imbalance of Q = CV, and stabilize at that voltage due to the theoretically perfect battery being able to lock in 1.5 volts regardless of the current needed. As in all capacitors, the net charge will be zero because the same amount of charge added to one plate will be subtracted from the other plate. If you want to show smooth current curves for the capacitor instead of a step function, then you need to insert some resistance in series with the capacitor. Of course, in the "real" world, resistance will be present and limit the ability of the capacitor to follow its driving voltage.

Ratch

Last edited: Dec 26, 2008
13. ### mik3 Senior Member

Feb 4, 2008
4,846
69
The transient response is just like an ideal step response because the capacitor charges instantly (ideal components).

Mar 27, 2008
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15. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0

I have one final question just to see if I have understood it correctly: When a capacitor is charging or discharging, we always have the following expression:

$
I_C(t) = \frac{d Q(t)}{dt},
$

where I_C is the current through the capacitor and Q(t) is the charge of the capacitor?

16. ### silvrstring Active Member

Mar 27, 2008
159
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Right. That's the actual definition of current.

So for a simple series RC example, i(t) = (-E/R)e^(-t/(RC)).

17. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
Great! It is so nice of you guys to take the time to help me.

Thanks to all!