Simple initial conditions problem

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Discussion Overview

The discussion revolves around a homework problem involving initial conditions in an electrical circuit with a capacitor and an inductor. Participants explore the implications of the initial voltage on the capacitor and the current through the inductor when a switch in the circuit is opened. The focus includes theoretical reasoning and mathematical relationships relevant to the behavior of inductors and capacitors in circuits.

Discussion Character

  • Homework-related
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant identifies the first initial condition as the voltage on the capacitor being 10V at t=0, but expresses uncertainty about the second initial condition related to the inductor's current.
  • Another participant questions what the inductor current would be after the switch has been closed for an extended period and emphasizes that inductors resist changes in current.
  • There is a suggestion that the inductor current might simply be the current through a 0.5 Ohm resistor, but uncertainty remains about whether there would be a sudden change in current when the switch opens.
  • One participant proposes that the capacitor will maintain 10V when the switch opens, leading to no immediate change in current, but acknowledges that the decay of the capacitor's voltage will eventually affect the inductor's current.
  • Another participant agrees with the idea that the inductor's current will transition from being supplied by the source to being supplied by the capacitor, leading to a changing voltage across the inductor.
  • There is a suggestion to write a differential equation involving current and voltage to analyze the circuit further.

Areas of Agreement / Disagreement

Participants express varying views on the behavior of the inductor and the implications of the capacitor's voltage. While there is some agreement on the relationship between the capacitor and inductor, the discussion remains unresolved regarding the exact nature of the current through the inductor when the switch opens.

Contextual Notes

Participants have not fully resolved the assumptions regarding the initial conditions and the behavior of the circuit components over time. The discussion includes references to mathematical relationships that have not been fully derived or agreed upon.

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Homework Statement


http://imgur.com/Mwin7dB

http://imgur.com/Mwin7dB

Homework Equations





The Attempt at a Solution



This is a fairly simple problem. My issue is that I can't identify the second initial condition. The first one is simple. At t=0, the voltage on the capacitor is 10V.

The second condition I feel should be related to the inductor. Perhaps to the current at 0 seconds. Not sure how to find out what the current through that inductor would be the moment the switch opens.

Anyone lend a hand?
 
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What is the inductor current after the switch has been closed for hours?

As for what happens when the switch opens... remember that inductors resist changes to the current flowing in them.
 
CWatters said:
What is the inductor current after the switch has been closed for hours?

As for what happens when the switch opens... remember that inductors resist changes to the current flowing in them.

Would it simply be the current running through the 0.5 Ohm resistor?

I'm not sure what exactly would happen with the inductor resisting changes to the current flowing through it. Would there even be a sudden change?

The capacitor will charge to 10V. So that when the switch opens, there will still be 10volts on one side of the inductor, with the same resistance. There should be no change. It's just that when the capacitor starts decaying out that the changing voltage will cause a current across the inductor to happen as per 1/L intergal(vdt) = I(t).Am I correct with that idea?
 
Learnphysics said:
Would it simply be the current running through the 0.5 Ohm resistor?
Yes.

I'm not sure what exactly would happen with the inductor resisting changes to the current flowing through it. Would there even be a sudden change?
There is never a "sudden" (as in instantaneous) change in current in an inductor, not without fireworks anyway.
 
Learnphysics said:
Would it simply be the current running through the 0.5 Ohm resistor?

I'm not sure what exactly would happen with the inductor resisting changes to the current flowing through it. Would there even be a sudden change?

The capacitor will charge to 10V. So that when the switch opens, there will still be 10volts on one side of the inductor, with the same resistance. There should be no change. It's just that when the capacitor starts decaying out that the changing voltage will cause a current across the inductor to happen as per 1/L intergal(vdt) = I(t).


Am I correct with that idea?

Yes. The current i thru the inductor, which had been coming from the supply before t = 0, now has to come from the capacitor. So the capacitor voltage drops, V across the inductor begins to change and therefore so does its current since for the inductorr di/dt = V/L.

Now, to answer the question you should write the differential equation in i and v, wit v replaced by a function of i. Or you can get 2 equations in v and i & solve them simultaneously. Same difference.

After you get v(t), let t = 2.5πe-6 s.
 

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