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?

 

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).

 

But then I am right? I.e. Q = CV for all times t?

 

Yes, but it is not constant because V varies.

 

V is 1.5 V, it is direct current.

 

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.

 

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.

 

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.

 

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?

 

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.

 

So there is no way to find the transient behavior of the capacitor before it settles on the value Q=CV?

 

Niles,

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?

So there is no way to find the transient behavior of the capacitor before it settles on the value Q=CV?

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any resistance in series with the capacitor, a sudden turnon of 1.5 volts will cause a infinite amount of charge to be withdrawn from the battery for an infinitesimal amount of time.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

 

So there is no way to find the transient behavior of the capacitor before it settles on the value Q=CV?

The transient response is just like an ideal step response because the capacitor charges instantly (ideal components).

 

Niles,

If you are interested in the math of it, here is a little bit on RC circuits.

http://www.wellesley.edu/Physics/phyllisflemingphysics/106_o_circuits.html

Check part VI

 

Thanks to all for replying.

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?

 

Right. That's the actual definition of current.

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

 

Great! It is so nice of you guys to take the time to help me.

Thanks to all!