# Help! Solving 2nd-Order RLC Circuits...

Discussion in 'Homework Help' started by Z.Jin, Apr 12, 2015.

1. ### Z.Jin Thread Starter New Member

Apr 12, 2015
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This is the practice problem 8.9 of Fundamentals of Electric Circuits 5th Edition.
Determine v and i for t>0 in the circuit:

Can someone help me with this question? I need a more detailed solution of this problem. Especially, what are the initial conditions. I though, by turning of the independent source, it should be i'' + 7i' + 10i = 0, and the initial conditions should be i(0)=0, and di(0)/dt=0. But my solution turns out to be incorrect. Can someone help me? Thank you in advance!

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
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i"?
7i'?

As far as I can tell, the initial conditions are:
i=0
v=0

3. ### Z.Jin Thread Starter New Member

Apr 12, 2015
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Yes! You're right! But what about di(0)/dt.
And the solution is i(t) = 3(1-exp(-5*t)). But, I can't get the right answer. Could you please walk me through your analysis process? Thank you so much!

4. ### Z.Jin Thread Starter New Member

Apr 12, 2015
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Thanks everyone for attention. I got it!

5. ### MrAl Distinguished Member

Jun 17, 2014
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515
Hi,

To get the initial conditions temporarily treat the capacitor as a short circuit and the inductor as an open circuit.

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6. ### Z.Jin Thread Starter New Member

Apr 12, 2015
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Hi, thanks! That's really a good point. It seems much easier to get the initial conditions by using the method you mentions than the one I figured out!

7. ### MrAl Distinguished Member

Jun 17, 2014
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Hi,

You are welcome, and it would still be interesting to hear how you got the initial conditions.

BTW, the final value can be found by assuming the opposite: the capacitor is open and the inductor is a short.

8. ### Z.Jin Thread Starter New Member

Apr 12, 2015
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I'm OK in finding final values. It's just that initial values of dv/dt and di/dt made me confused. I use KVL or KCL on the circuit assuming v(t) of capacitor and i(t) of inductor cannot change abruptly from v(0-) and i(0-). But my strategy is not as straight forward as yours. Again, thanks

9. ### WBahn Moderator

Mar 31, 2012
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In general, does shutting off a source seem like a good way to find the initial conditions? That implies that the initial conditions are independent of the sources in the circuit. Does that make sense?

In this case it is a poorly defined problem because it has an open-circuited current source prior to t=0, which invokes undefined behavior. About the best you can do is assume that the current source produces so much voltage (before t=0) that it arcs back from the output to the input (without damaging the source, of course).

In order to the initial conditions at t=0+, you need to find the conditions at t=0- and then carry across the t=0 boundary those conditions that can't change instantaneously, namely the voltage across the capacitor and the current in the conductor. That will give you i(t) and v(t) at t=0 for the quantities shown. You then can find the i(t) (in the capacitor) and v(t) (across the inductor) by analyzing the circuit at t=0+ subject to the constraints imposed by the conditions that carried across. With those in hand, you have everything you need to find the first derivatives of the capacitor voltage and the inductor current.

EDIT - typos

Last edited: Apr 12, 2015
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10. ### Z.Jin Thread Starter New Member

Apr 12, 2015
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Thanks for your help! That makes sense. I think I need more practice.

11. ### MrAl Distinguished Member

Jun 17, 2014
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515
Hi,

Some authors will show the current source shorted out until t=0 when it is open circuited at the same time as another switch is closed, or in this circuit it can be short circuited and then the opened at t=0.
Once in a while you sort of have to guess at what the author intended.

Also, there are other ways to analyze this without having to calculate the initial conditions, as long as you know there is zero energy in each reactive element first. If there is non zero energy before t=0 then you have to include that too.

12. ### Z.Jin Thread Starter New Member

Apr 12, 2015
12
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Could you please kindly tell me the names of those methods for analyzing it without having to calculate the initial conditions with zero initial energy? I would like to google them.

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I would think it isn't too difficult to estimate the time zero derivative values for the current and voltage parameters in this example.
As WBahn points out the problem statement lacks certainty, with an ideal current source having a series switch. Far better had the source been in parallel with a switch that opens at t=0 sec.
In any event, one can probably assume at t=0+ sec the 3A flows only in the capacitive branch. The 10 ohm resistor would then have a voltage drop of 30V at that instant. This 30V would appear across the inductor at t=0+ sec, suggesting a di/dt in the inductor of E/L [Amps per sec] or 30 [Volts] /2 [Henry] or 15 A/sec. A similar argument can be mounted for deriving the dv/dt value at t=0+ sec.

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14. ### Z.Jin Thread Starter New Member

Apr 12, 2015
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Inspiring! Thank you so much!

15. ### Z.Jin Thread Starter New Member

Apr 12, 2015
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I just found a video on Youtube discussing RLC calculation. It treats the capacitor and inductor at t=0+ as voltage source and current source. And zero initial conditions can be treated as special cases of this method. I think it's a more generalized solution and very easy to understand.

I'm just a bit confused that why the author of Fundamentals of Electric Circuits, which is the book that I'm reading right now, doesn't mention this convenient method. He did mention that students always having trouble with finding the initial conditions of derivative... And it's a very popular book..

16. ### MrAl Distinguished Member

Jun 17, 2014
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515
Hi,

Just to note, playback of that video on other websites has been disabled.

Initial conditions are a little confusing at first, but if you think about it that is much simpler than having to analyze the whole circuit. If all we ever had to do was find the initial conditions, circuit analysis would be so much simpler and faster.
So think about it that way

17. ### Z.Jin Thread Starter New Member

Apr 12, 2015
12
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Thanks for the note.

Yes, you're right. I'm just standing at the beginning point. A long journey is waiting for me