What is the Voltage on C1/C2 after the switch closes?

[Lapse]

Homework Statement

Homework Equations

V_f+(V_i-V_f)e^-t/RC

Tau = RC

The Attempt at a Solution

[/B]

Normally, I would use the eq'n mentioned above to find Vr in an RC circuit. However, this second capacitor is throwing me off. Is there a new equation I should use? How do I think about this circuit?
Additionally, how would I go about finding Tau? Do I add the capacitance of each capacitor:
(R x (C1+C2))?

Thanks.

 

[BvU]

Hi,

Your relevant equation applies to charging from a constant voltage, so not in this case. You need something better.

The context of the question is unclear to me: there is no single 'the voltage after the switch closes': as your plot clearly shows, V_r is time dependent. From the remainder of the exercise it seems they mean 'after a long time'. But you are using some simulation programme to plot the time dependence (which as said is not described by your relevant equation).

 

[NascentOxygen]

Are you able to set the time scale to be logarithmic for that plot? If the rate of voltage change reveals as a straight line on semilog axes then you can say the voltage is changing exponentially with time.

But all the question seems to require is that you allow the system sufficient time to settle at its new equilibrium, and determine a value for this final steady state voltage.

 

[gneill]

Hint: Total charge is conserved and the potentials across the two capacitors must be equal when the circuit reaches steady state.

Additionally, how would I go about finding Tau? Do I add the capacitance of each capacitor:
(R x (C1+C2))?

The capacitors and resistance (of the switch) are all in series.

 

[Lapse]

Hi,

Your relevant equation applies to charging from a constant voltage, so not in this case. You need something better.

The context of the question is unclear to me: there is no single 'the voltage after the switch closes': as your plot clearly shows, V_r is time dependent. From the remainder of the exercise it seems they mean 'after a long time'. But you are using some simulation programme to plot the time dependence (which as said is not described by your relevant equation).

I understand. So what would the relevant equation be?

 

[Lapse]

Hint: Total charge is conserved and the potentials across the two capacitors must be equal when the circuit reaches steady state.

The capacitors and resistance (of the switch) are all in series.

I noticed that both capacitor charges become equal (they converge to 0.6V).

I see that they are in series; however, what does this mean for finding Tau?

 

[Lapse]

Are you able to set the time scale to be logarithmic for that plot? If the rate of voltage change reveals as a straight line on semilog axes then you can say the voltage is changing exponentially with time.

But all the question seems to require is that you allow the system sufficient time to settle at its new equilibrium, and determine a value for this final steady state voltage.

The graph being set at a time stop of 200ps was set by the professor. This is why I am assuming that instead of showing a final state at 0.6V, he probably wants to see an equation for the voltage over each capacitor proving what the voltage would be if the time stop went beyond 200ps.
What is a relevant equation I can use to solve for the voltages over the capacitors?

 

[gneill]

I noticed that both capacitor charges become equal (they converge to 0.6V).

No. Charge is not voltage (look at the units). What's the relationship between charge and voltage for a capacitor?

The capacitors will not have equal charges for the same potential difference as they do not have the same capacitance.

I see that they are in series; however, what does this mean for finding Tau?

Well for one thing the capacitors are in series... What's the net capacitance?

Note that you don't need $\tau$ to answer this question. You only need to find the eventual state of the system (voltages, charges).

Charge is conserved. After a long time the potential differences across the capacitors must be equal. So how will the (conserved) charge be distributed?

 

[Lapse]

No. Charge is not voltage (look at the units). What's the relationship between charge and voltage for a capacitor?

Q = C x V

Well for one thing the capacitors are in series... What's the net capacitance?

C1 + C2 = 50f + 200f = 250f

Note that you don't need ττ\tau to answer this question. You only need to find the eventual state of the system (voltages, charges).

The eventual voltage state is 0.6V.
The eventual charge state, hmm...
Q₁ = 50f x 0.6V = 30 coulombs
Q₂ = 200f x 0.6V = 120 coulombs

Charge is conserved. After a long time the potential differences across the capacitors must be equal. So how will the (conserved) charge be distributed?

Once the capacitors are fully charged they behave like wires, correct? It would then make sense that what follows is an equal potential difference (this would be similar to how a current is equal on either end of a closed loop, correct?).
Q = C x V = 250f x 0.6V = 150 coulombs

 

[Lapse]

Ahh, tell me if I'm thinking of it correctly now!

The energy after the switch closes:
E₁ = 9J
E₂ = 36J

But before the switch closes:
E₁ = 25J
E₂ = 50J

After potential difference: 36J - 9J = 25J
Before potention difference: 50J - 25J = 25J

25J = 25J ...gneill, is this what you meant by:

After a long time the potential differences across the capacitors must be equal.

 

[BvU]

Once the capacitors are fully charged they behave like wires, correct

No !
(Ideal) wires have $\Delta V=0\ \ \forall I \$
(Ideal) capacitors have $Q = CV \Rightarrow I = C{dV\over dt}$ which is something completely different !

 

[gneill]

C1 + C2 = 50f + 200f = 250f

That is how to add capacitors in parallel. How do capacitors in series sum?

The eventual voltage state is 0.6V.
The eventual charge state, hmm...
Q1 = 50f x 0.6V = 30 coulombs
Q2 = 200f x 0.6V = 120 coulombs

Small "f" is not Farads (capital "F"). Small f designates the femto prefix (10^-15).

Ahh, tell me if I'm thinking of it correctly now!

The energy after the switch closes:
E₁ = 9J
E₂ = 36J

But before the switch closes:
E₁ = 25J
E₂ = 50J

You'll need to redo the above taking into account that the capacitor values are given in femtoFarads. Also, check your math for the "before the switch closes" value for E₂.

After potential difference: 36J - 9J = 25J
Before potention difference: 50J - 25J = 25J

Energy is not potential. Potential difference is measured in Volts.