Simple initial conditions problem

[Learnphysics]

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?

 

[CWatters]

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.

 

[Learnphysics]

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?

 

[NascentOxygen]

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.

 

[rude man]

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.