LRC DC circuit

Homework Statement

http://javid.us/phys22/week14q28.jpg
E=50.0 V, R=250 ohm, and C=0.50 microF.
The switch is closed for a long time and no voltage is measured across the capacitor. Just after the switch is opened, the voltage across the capacitor is measured to be 150 V. what is the inductance L?

Inductor and capacitor are ideal.

Homework Equations

q=Qmax*cos(wt+phi)
i=dq/dt=-Imax*w*sin(wt+phi)
U=1/2*L*Imax^2=1/2*(1/C)*Vmax^2=1/2*L*i^2+1/2*(1/C)*v^2

The Attempt at a Solution

I am having trouble understanding (or my teacher is having trouble understanding) how this circuit will behave.

The way I understand it, when the switch has been closed for a long time, the inductor offers no resistance or emf, therefore the capacitor is shorted, and cannot have any charge built up on it, and therefore cannot have a voltage difference across it.

If at this point we are told that the switch is thrown, then told that the voltage reaches a maximum of 150 V, then 1/2*L*Imax^2=1/2*(1/C)*Vmax^2 , and we get our answer.

However, this is not what the problem states (although I have a hunch it is what was meant) and therefore I am stuck trying to solve what may be unsolvable with the information given. My professor claims that it is doable, and that my logic is flawed, so allow me to share my logic with you and please feel free to point out any errors.

At steady state, the current through the circuit will flow exclusively through the inductor, will be steady, and will have a value of E/R. Since all the components in the circuit are ideal, at steady state, the only component draining power is the resistor, which will then have a voltage difference across it equal in magnitude and opposite in polarity to the battery. Thus I conclude that the voltage across the capacitor and inductor is zero.

Here is where I have the issue. How is it that "just after the switch is opened", there is a voltage across the capacitor of 150 V.

I have also been discussing this on my professors forum, http://jinxer.com/topic.asp?TOPIC_ID=424 .

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Talked to my teacher today, and he claims that instantaneously after the switch is opened the current and voltage are at a maximum. I cannot understand how this is so, please help!

The voltage over the Cap when the switch is closed is zero because there was no current flowing through it, so I think what the question is getting at is after the switch is opened some kind of discharge is causing current to flow through the Cap... hope that helps with your confusion. You can get the Cap-Reactance from the known Capacitance, and with the V = 150 you can calculate this current.

I think from there you are looking for an L-value that at that current has an equal-but-opposite Reactance your Cap.

berkeman
Mentor
Talked to my teacher today, and he claims that instantaneously after the switch is opened the current and voltage are at a maximum. I cannot understand how this is so, please help!

pyrophreek, I believe that your professor is wrong, and you are using the correct intuition in your answer. When the switch is opened, that diverts the inductor current (which is initially defined by the source voltage and current-limiting resistor) to the capacitor, and with ideal L and C components, that creates an undamped LC tank resonator circuit. In that kind of circuit, the energy storage oscillates back and forth between the voltage across the capacitor and the current through the inductor.

Thank you for the response. I really couldn't understand how there could possibly be any voltage and an Imax at the same time. If anyone knows where I could find some information proving this I would really appreciate it, have looked at wikipedia and other sources, but most talk only about steady state, or start with the capacitor charged rather than the inductor.

berkeman
Mentor
Thank you for the response. I really couldn't understand how there could possibly be any voltage and an Imax at the same time. If anyone knows where I could find some information proving this I would really appreciate it, have looked at wikipedia and other sources, but most talk only about steady state, or start with the capacitor charged rather than the inductor.

If your professor still has a problem with it, please have him/her e-mail me directly or call me on my cell phone. Probably just a simple misunderstanding or miscommunication. I'll send you my e-mail/cell contact via PM.

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berkeman
Mentor
Oh, and please provide your professor with a web link to this thread as part of your discussion. That will probably help clear up the miscommunication all on its own. Thanks.

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berkeman
Mentor

Javid said:
Obviously the rest of the class is totally clueless.
Solee is completely right!
In the absolute beginning when the battery is connected, the inductor is open and the capacitor is short like a wire since it is charging. Imagine like 2 water tanks connected next to each other the water level is always the same. Therefore not only the voltage is the same for both of them (since they are parallel), but also the energy must be equally distributed between them. The resistor has nothing to do with these. It simply drops the voltage coming to these elements. Therefore, after the switch is opened 1/2 L*(Imax)^2 must be equal to 1/2 CV^2 but Imax=E/R and from here we'll find the inductance.
I think there is mainly a language or comprehension problem on the part of your TA (that cannot be a professor, IMO). Again, your intuition is correct. I *think* that your TA is trying to say the same thing, but his inflamatory statements at the start of each of his posts, like "What are you talking about? This question is from chapter 30 before we even have learned about AC circuits!!!!" are totally not helping, and are confusing you in your understanding of this elementary circuit.

Sigh. All I can recommend at this point is that you keep doing the good job that you appear to be doing in your studies, and keep coming to the PF for clarification. This TA's thread would ensure no company would hire him in the future, IMO. Lordy.

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- When the switch is closed, the current through the inductor is I0 = E / R. You can assume that immediately after the switch is opened, through the inductor will remain I0 for that infintessimal moment.
- Immediately after the switch is opened, the voltage across the capacitor is zero.
- If you solve the 2ND differential equation and apply the two initial conditions I just listed above, then you have:

I(t) = I0 * cos( t / t0 ) and V(t) = V0 * sin( t / t0 ) where t0 = sqrt(L*C).

- You know the relation between I(t) and V(t).

V(t) = - L * dI(t)/dt

which gives:

V(t) = sqrt(L/C) * I0 * sin( t / t0 )

So, the amplitude of V0 is: V0 = sqrt(L/C) * I0 = sqrt(L/C) * E / R.

So, sqrt(L/C) / R = 3.... You do the rest.

Note: the confusing part of this problem is that the voltage across the capacitor will not reach its peak value of 150V until t = (pi/2) * t0.

Everything that you have posted so far was my understanding of it when I went to my professor (and yes he is a professor, at UCLA none the less, or at least he claims to be, since I am taking a class with him at SMC) Honestly I am not the best at explaining things, I tend to leave out vital facts because I know them and I assume (usually incorrectly) that everyone else knows them.

FYI: this was a question on an exam he gave us, and I had to leave it blank because I could not understand how there was a Current and Voltage at the instant the switch was opened. Only one person that I know of got the question right, and I haven't talked to anyone in class who can explain this problem.

I just gave you the answer... what more do you want? For me to do the number crunching? OK... here you go kid.

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berkeman
Mentor
I just gave you the answer... what more do you want? For me to do the number crunching? OK... here you go kid.

<< answer deleted by berkeman >>

We do not give out answers here at the PF. That is against the PF guidelines: