Simple initial conditions problem

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SUMMARY

The discussion centers on solving a simple initial conditions problem involving a capacitor and an inductor in an electrical circuit. The first initial condition is established as a voltage of 10V across the capacitor at t=0. The second condition pertains to the current through the inductor when the switch opens, which is influenced by the inductor's property of resisting changes in current. The participants conclude that the current through the inductor will initially be the same as that through a 0.5 Ohm resistor and that the differential equations governing the circuit must be formulated to analyze the voltage and current over time.

PREREQUISITES
  • Understanding of basic electrical circuit components: capacitors and inductors
  • Knowledge of initial conditions in differential equations
  • Familiarity with Kirchhoff's laws for circuit analysis
  • Ability to solve differential equations involving voltage and current
NEXT STEPS
  • Study the behavior of inductors in circuits, focusing on the concept of inductance and current continuity
  • Learn how to formulate and solve differential equations for RLC circuits
  • Explore the impact of initial conditions on circuit behavior over time
  • Investigate the relationship between voltage and current in capacitors and inductors using the equations V = L(di/dt) and Q = CV
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing transient responses in RLC circuits will benefit from this discussion.

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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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