good morning :-) @JohnRennie

@user8718165 morning :-)

@JohnRennie does the maximum charge on a capacitor in a standard RLC circuit depend on its capacitance? I was thinking... $Q_{max} = CV_{c} = CIX_{C} = CI(1/C{\omega}) = I/{\omega}.$

so is $Q_{max}$ only a function of amplitude of current and driving frequency?

I've never thought about it, but yes the current in a series circuit is the same through all components and the voltage is just $V = IZ$ so you can find the maximum voltage. The maximum charge is then $Q_{max} = CV_{max}$.

@JohnRennie but this doesn't seem to depend on the actual capacitance of the capacitor, only current and driving frequency

that seems strange

with $V_{max} = IX_{C}$

Well $Z_c = 1/(i\omega C)$ so we get $V = I/(i\omega C)$

yes

Ah, and the Cs cancel. Hmm, yes, that does seem odd.

yes exactly

I'd have to actually do a calculation, which I don't feel inclined to do right now.

haha

OK no worries

Interesting thought though ...

I thought so too :p

@kylecampbell Think about the mechanical analog. Does the maximum displacement of a damped harmonic oscillator depend on the spring constant?

@AaronStevens interesting, no

@kylecampbell The spring example might be easier to reason through

@AaronStevens can you expand?

@kylecampbell you are interested in a driven oscillator at any driving frequency?

a driven or a damped oscillator?

any driving frequency, sure

Well for sure it's damped. That's what the resistor does

right

@kylecampbell I did the calculation ...

okay

@JohnRennie and?

It is indeed independent of the capacitance. And actually Aaron's approach nicely explains this though i think I could put it in a simpler way.

Suppose you're pushing a swing and the swing is undamped.

@JohnRennie am curious about your explanation for sure

Then as you keep pumping energy in by pushing the swing amplitude just keeps increasing without limit.

what was the extent of the calculation? would you care to share? i'm just curious

But real swings always have some damping, so as you keep pushing and the swing amplitude keeps increasing there comes a point where the losses due to damping are equal to the energy you're putting in by pushing and the amplitude settles to a stable value.

The point is the amplitude depends only on the damping not on the mass of the swing.

Basically we're seeing the same thing here. The battery keeps trying to charge the capacitor and the resistor keeps damping it. The maximum charge depends not on the capacitor but on the amount of damping i.e. on the resistor value.

interesting

@JohnRennie how did you show this with the calculation?

@kylecampbell I just took an RC circuit and did the calculation setting $V(t) = V_0\sin\omega t$

The charge came out to be:

$$ Q(t) = - \frac{V_0}{\omega R} \cos\omega t $$

It's curious that the charge goes to infinity as $R$ goes to zero. Intuitively you'd think it wouldn't. I think it's because with no resistance the current goes to infinity.

@JohnRennie Does $V_{0}$ depend on capacitance?

@JohnRennie none of the prefactors depend on $C$?

@kylecampbell no, $V_0$ is just the supply voltage so it's whatever you set your power supply to deliver.

Nothing in the expression depends on $C$.

interesting

@JohnRennie I set up my RC circuit but I'm struggling to get your result. any tips?

I believe you, I'm just trying to prove it to myself

The input voltage is $V(t) = V_) e^{i\omega t}$

The combined impedance of the resistor and capacitor is $Z = Z_R + Z_C = R + 1/i\omega C$, and the current is just $I = V/Z$ giving:

$$ I = \frac{V_0 i^{i\omega t}}{R + 1/i\omega C} $$

@kylecampbell OK so far?

yeah

The voltage across the capacitor is $V_c = IZ_c$ giving:

$$ V_c = \frac{V_0 e^{i\omega t}}{R + 1/i\omega C} 1/i\omega C $$

And that simplifies to:

$$ V_c = \frac{V_0 e^{i\omega t}}{1 + i\omega CR} $$

Ah ...

I've just realised I made a typo when I was working this out.

I lost the $1$ in the denominator and managed to get:

$$ V_c = \frac{V_0 e^{i\omega t}}{i\omega CR} $$

which is of course wrong.

right

So the $C$ doesn't cancel ...

$$ Q(t) = \frac{V_0 e^{i\omega t}}{1/C + i\omega R} $$

ah darn

But it does mean in the limit of large $C$ the charge approaches a constant.

hello sir @JohnRennie

@user8718165 hi

@kylecampbell Sorry to leave you hanging. I fell asleep.

@AaronStevens can you actually explain the driven oscillator case anyways? I'm curious

@AaronStevens and no worries

@kylecampbell Ah so you determined it is not independent of $C$? Good. I was a little confused because $k$ should matter for the damped driven oscillator too

ah okay

yes

I figured it was suspicious too because the result I found was $Q = I/{\omega}$ and I forgot that $I$ depends on impedance which in turn depends on capacitance

@kylecampbell I got confused with the simple case of just a mass on a spring. Or for circuits an LC circuit