Energy in capacitors in complicated circuit

[Mårten]

Okey guys, here's another difficult capacitor circuit... :yuck:

Homework Statement

Find the proportions (the quotient) between the energies in the two capacitors when the switch is closed in the circuit below.

Homework Equations

(1) C = Q / V
(2) W = CV^2
(3) Current division (maybe?): I2 = I*R1/(R1+R2)
Ohm's and Kirchhoff's.
And maybe some others...?

The Attempt at a Solution

I tried to find the potential over each capacitor, because then I could use equation (1) and (2) above and find the solution. But I don't know how to get grip of the potentials. It's hard to do any potential walk here, cause there's so many branches all over the place...

The key said this:
W1 = 1/2 * C1 * U1^2 = 1/2 * C1 * (R1*I)^2
W2 = 1/2 * C2 * U2^2 = 1/2 * C2 * (R2*I)^2
=> W1/W2 = C1*R1^2 / (C2*R2^2)

4. Questions
a) According to the key, U1 and U2 (which I suppose are over the capacitors, the key doesn't tell) equals R1*I and R2*I respectively. How could that be? Why is not R3 included?

b) What's happening when the switch is closed? Is the battery driving current through the capacitors, or is the capacitors driving current, i.e., they are being discharged? How can you see that in that case? Isn't the battery still trying to hold the charges on the capacitors at place?

c) How to approach a problem like this in general... (if it's not evident from a) and b))

 

[LowlyPion]

Okey guys, here's another difficult capacitor circuit... :yuck:

Homework Statement

Find the proportions (the quotient) between the energies in the two capacitors when the switch is closed in the circuit below.

Homework Equations

(1) C = Q / V
(2) W = CV^2
(3) Current division (maybe?): I2 = I*R1/(R1+R2)
Ohm's and Kirchhoff's.
And maybe some others...?

The Attempt at a Solution

I tried to find the potential over each capacitor, because then I could use equation (1) and (2) above and find the solution. But I don't know how to get grip of the potentials. It's hard to do any potential walk here, cause there's so many branches all over the place...

The key said this:
W1 = 1/2 * C1 * U1^2 = 1/2 * C1 * (R1*I)^2
W2 = 1/2 * C2 * U2^2 = 1/2 * C2 * (R2*I)^2
=> W1/W2 = C1*R1^2 / (C2*R2^2)

4. Questions
a) According to the key, U1 and U2 (which I suppose are over the capacitors, the key doesn't tell) equals R1*I and R2*I respectively. How could that be? Why is not R3 included?

b) What's happening when the switch is closed? Is the battery driving current through the capacitors, or is the capacitors driving current, i.e., they are being discharged? How can you see that in that case? Isn't the battery still trying to hold the charges on the capacitors at place?

c) How to approach a problem like this in general... (if it's not evident from a) and b))

The most important thing to know about this problem is that the voltage is fixed. When the switch is closed there are transient currents and voltages at various points that will flow to allow the network to settle to the new equilibrium. Happily you may ignore them given the statement of the problem.

Hence what you are dealing with then is the steady state DC flow of the current at Equilibrium, which was what it was before the switch closed. The key voltage distribution then is V1 across R1 and C1 and V2 the voltage across R2 and C2. (With no current flowing in R3 once at the new equilibrium you may ignore its effect all together.)

Knowing the voltage across each capacitor then allows you to calculate the charge on each and hence to return the Quotient that the question requires.

 

[baba89]

I would approach this problem by first analyzing the circuit and identifying all the components and their connections. From there, I would use Kirchhoff's laws and Ohm's law to determine the current and potential differences in each branch of the circuit.

Next, I would use the equations (1) and (2) to calculate the energy stored in each capacitor. The key provided a way to do this, but I would also try to understand the reasoning behind it.

To answer your specific questions:

a) The key is using the fact that the current in a series circuit is the same throughout, so the current through R1 and R2 is also the current through C1 and C2. The potential difference across a resistor is given by Ohm's law (V=IR), so U1 = R1*I and U2 = R2*I. R3 is not included because it is in parallel with C2, so the current through R3 does not affect the potential difference across C2.

b) When the switch is closed, the battery is driving current through the circuit, including the capacitors. However, as the capacitors charge up, the current decreases until it reaches 0. At that point, the capacitors are fully charged and the battery is no longer driving current. The battery is still trying to hold the charges on the capacitors in place, but as long as the switch is closed, the charges will remain on the capacitors.

c) In general, approaching a problem like this involves understanding the fundamental principles of circuits and using Kirchhoff's laws and Ohm's law to analyze the circuit. It also helps to have a good understanding of how capacitors work and how they store energy. Practice with similar problems can also help improve problem-solving skills.