Energy in capacitors in complicated circuit

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SUMMARY

The discussion focuses on calculating the energy proportions between two capacitors in a complex circuit when a switch is closed. The relevant equations include capacitance (C = Q / V) and energy stored in capacitors (W = 1/2 * C * V^2). The key conclusion is that the energy ratio W1/W2 can be expressed as C1*R1^2 / (C2*R2^2). Understanding the steady-state DC flow and voltage distribution across the capacitors is essential for solving the problem accurately.

PREREQUISITES
  • Understanding of capacitor energy calculations using W = 1/2 * C * V^2
  • Familiarity with Ohm's Law and Kirchhoff's circuit laws
  • Knowledge of current division principles in electrical circuits
  • Basic concepts of steady-state DC analysis in circuits
NEXT STEPS
  • Study the principles of capacitor energy storage and discharge dynamics
  • Learn about steady-state analysis in electrical circuits
  • Explore current division techniques in complex resistor-capacitor networks
  • Investigate transient response in circuits with capacitors and switches
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Electrical engineering students, circuit designers, and anyone involved in analyzing capacitor circuits and energy distribution in electrical systems.

Mårten
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Okey guys, here's another difficult capacitor circuit...

Homework Statement


Find the proportions (the quotient) between the energies in the two capacitors when the switch is closed in the circuit below.
2rp64om.gif


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? :confused:

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))
 
Last edited:
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Mårten said:
Okey guys, here's another difficult capacitor circuit...

Homework Statement


Find the proportions (the quotient) between the energies in the two capacitors when the switch is closed in the circuit below.
2rp64om.gif


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? :confused:

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.
 

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