Energy Losses in a Circuit with a Switching Element

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Discussion Overview

The discussion revolves around the energy losses in a circuit with a switching element, specifically focusing on the transition from one steady state to another after a switch is opened. Participants explore the implications of this transition on energy dissipation, particularly in the context of resistors, capacitors, and inductors.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory
  • Debate/contested

Main Points Raised

  • One participant describes their approach to finding the energy converted to heat in the resistor after the switch is opened, using initial conditions and energy stored in inductors and capacitors.
  • Another participant suggests that to find the energy dissipated, it may suffice to calculate the difference in stored energy between the two steady states.
  • There is a discussion about the initial conditions, including the current through the inductor and the voltage across the capacitors before and after the switch is opened.
  • Some participants express uncertainty about whether the difference in stored energy accounts for all energy losses, considering the possibility of oscillations and current flow back through the battery.
  • One participant notes that the solution found is correct but questions the relevance of damped oscillations in the overall energy analysis.
  • There is a desire for a theoretical justification regarding why oscillations may not matter in the context of energy loss, with a preference for an engineering perspective over a purely mathematical one.

Areas of Agreement / Disagreement

Participants express differing views on the significance of damped oscillations and whether they affect the energy analysis. While some agree that the final state is what matters, others remain uncertain about the implications of oscillations on energy loss.

Contextual Notes

Participants note the complexity of the problem due to the presence of multiple energy storage devices and the potential for oscillations, which complicates the analysis of energy dissipation.

Who May Find This Useful

This discussion may be useful for students and practitioners interested in circuit analysis, particularly those dealing with transient responses in circuits containing resistors, capacitors, and inductors.

Ivan Antunovic
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Homework Statement


By the time t = 0 network has been in steady state. At time t = 0, the switch S
switches off. Determine the total amount of energy converted to heat in the circle of resistance to the establishment of a new steady state.

Homework Equations


DSC_0950.jpg

DSC_0951.jpg
[/B]

The Attempt at a Solution


Picture,attached.
Okay before the switch is opened, both C and C2 are short circuited,so initial current i(0)=E/R.
So I had this approach,okay let's find voltages on the capacitors vc(t) and from ic(t) = C* dv/dt , I will get a current and then it's easy to find the energy since it's integral of i*R^2 *dt (from 0 to infinity).
I wrote down KCLs and KVLs,and used laplace transform,but things got really complicated,I think my approach isn't right(atleast not the shortest one).
Are those laplace transforms that I wrote,even correct?
 
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Your images do not identify where the switch is located. Also, it is difficult to read and comment on workings in image form. It is much preferable to type them out so that helpers can quote and comment on individual lines. The built-in LaTeX syntax interpretation of PF can help there.

For this type of problem where you're looking for the energy dissipated between two steady states, it generally suffices to find the stored energy in each of the steady states and take the difference. So the first thing to do is identify the conditions at each of the steady states.
 
gneill said:
Your images do not identify where the switch is located. Also, it is difficult to read and comment on workings in image form. It is much preferable to type them out so that helpers can quote and comment on individual lines. The built-in LaTeX syntax interpretation of PF can help there.

For this type of problem where you're looking for the energy dissipated between two steady states, it generally suffices to find the stored energy in each of the steady states and take the difference. So the first thing to do is identify the conditions at each of the steady states.

I forgot to upload a picture,it was very late so I guess that's the reason,sorry.
So before the switch is opened:
We have both R and C2 short circuited,initial current is i(-0)= E/R,no current goes through a capacitor so i(-0) is actually current going through inductor,so stored energy in inductor is WL(-0)=0.5 * L* (i(-0))^2 = 0.5 * L * (E/R)^2 ,okay now if I make KVL for a first loop ,I get +E-i(-0)*R-vc(-0)=0,therefore
vc(-0)=E-i(-0)*R= E-R*(E/R)=E-E=0,which was obvious since there is no induced voltage since we are dealing with DC current(I am not quite sure If I can say 'no voltage drop across inductor',since physicists going to kill me because there is never electric field in the ideal inductor so equation ,integral of E*dl = 0 cannot be applied).
After the switch is opened and after the transient process is over,there is no current flowing, because both C and C1 are charged and they are connected in parallel since no voltage drop on the resistances,and inductor is a short circuit here,and I get Vc(infinity)=Vc2(infinity)=E.
Therefore stored energy in the capacitors Wc(infinity)=0.5 * C * E^2 , Wc1(infinity)=0.5*C1*E^2.\Delta E=( (0.5*E^2)*(C+C1) ) - (0.5*L (E/R)^2 ) = 0.5 * E^2 (C+C1-(L/R^2))

Should this be okay?
 

Attachments

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Ivan Antunovic said:
I forgot to upload a picture,it was very late so I guess that's the reason,sorry.
So before the switch is opened:
We have both R and C2 short circuited,initial current is i(-0)= E/R,no current goes through a capacitor so i(-0) is actually current going through inductor,so stored energy in inductor is WL(-0)=0.5 * L* (i(-0))^2 = 0.5 * L * (E/R)^2 ,okay now if I make KVL for a first loop ,I get +E-i(-0)*R-vc(-0)=0,therefore
vc(-0)=E-i(-0)*R= E-R*(E/R)=E-E=0,which was obvious since there is no induced voltage since we are dealing with DC current(I am not quite sure If I can say 'no voltage drop across inductor',since physicists going to kill me because there is never electric field in the ideal inductor so equation ,integral of E*dl = 0 cannot be applied).
After the switch is opened and after the transient process is over,there is no current flowing, because both C and C1 are charged and they are connected in parallel since no voltage drop on the resistances,and inductor is a short circuit here,and I get Vc(infinity)=Vc2(infinity)=E.
Therefore stored energy in the capacitors Wc(infinity)=0.5 * C * E^2 , Wc1(infinity)=0.5*C1*E^2.\Delta E=( (0.5*E^2)*(C+C1) ) - (0.5*L (E/R)^2 ) = 0.5 * E^2 (C+C1-(L/R^2))

Should this be okay?So basically that difference of energy in reactive elements is always lost energy,dissipated in the resistor?
 
I'm thinking about this. Not knowing the position of the switch initially, I thought perhaps the inductors and capacitors would end up isolated so that no new energy could enter the loop. But now I can see that there is the possibility of damped oscillations while the battery E continues to supply energy to charge the capacitors to their final value. I'm not sure that the difference in stored energy between the states will account for all the energy. Current can actually flow back through the battery during the oscillations, and current flowing in either direction still burns energy in the resistors.

I can also see that a complete solution via Laplace transforms will be messy since there are three separate energy storage devices. So I'm not sure what to recommend at the moment. :frown:
 
gneill said:
I'm thinking about this. Not knowing the position of the switch initially, I thought perhaps the inductors and capacitors would end up isolated so that no new energy could enter the loop. But now I can see that there is the possibility of damped oscillations while the battery E continues to supply energy to charge the capacitors to their final value. I'm not sure that the difference in stored energy between the states will account for all the energy. Current can actually flow back through the battery during the oscillations, and current flowing in either direction still burns energy in the resistors.

I can also see that a complete solution via Laplace transforms will be messy since there are three separate energy storage devices. So I'm not sure what to recommend at the moment. :frown:
Okay,found the solution and it's correct,well there will be damped oscillations since we have R,L and C,but it doesn't matter right?We only care about the final state of the network.
 
Ivan Antunovic said:
Okay,found the solution and it's correct,well there will be damped oscillations since we have R,L and C,but it doesn't matter right?We only care about the final state of the network.
That was my initial thought (which I'm happy to have confirmed), but I wish I could think of a theorem that proves that the oscillations don't matter here.. I hate being right but not knowing why :mad:
 
gneill said:
That was my initial thought (which I'm happy to have confirmed), but I wish I could think of a theorem that proves that the oscillations don't matter here.. I hate being right but not knowing why :mad:
Well,it's quite intuitive that oscillations don't matter here.I don't know about the mathematical proof,but I like to think more as an engineer than as a mathematician.Maybe some good physicist could help us about the theorem.
 

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