# System Analysis RL circuit

Discussion in 'Homework Help' started by kenneth dean, Nov 19, 2014.

1. ### kenneth dean Thread Starter New Member

Nov 19, 2014
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0

I'm fairly new to circuits and i'm having a bit of trouble with these questions.
So far i've got: Vo(t) = Vr(t), Vr(t) = -Ri(t), Vo(t) = Vr(t) = L di/dt(t)
Therefore L di/dt(t) + Ri(t) = 0 , I know then I have to use laplace transformation to get the solution, but I don't know if i'm even close with what I have atm, so any help would be great thanks!

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
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I have a question regarding Vr(t)=-R*i(t).
Why is there negative sign in front of R? Where did it come from?

3. ### kenneth dean Thread Starter New Member

Nov 19, 2014
11
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I was talking it that "i" is flowing clockwise and Vr is negative to positive from bottom to top of resistor, and therefore the current is flowing from negative to positive across the resistor. But then again I might be wrong, i'm not sure

4. ### WBahn Moderator

Mar 31, 2012
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You can define i(t) to be anything you want -- the problem is that you didn't define it at all! Not only did you not define the direction, you didn't even define what component(s) it was flowing through.

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
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Have you studied current dividers yet?

Because R and L form a current divider. The I1 enters the current divider and gets divided into two portions. Portion One passes though R. Portion Two passes though L. Once the two portions rejoin, they form I1 again.

And since I now mentioned current divider. It seems to me that your i(t) is wrong. You are using same i(t) for resistor and for inductor. Which is wrong. I1 gets split into two different currents when it enters current divider. So i(t) through resistor CAN NOT be possibly be the same i(t) that passes through inductor.

6. ### kenneth dean Thread Starter New Member

Nov 19, 2014
11
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Wow i'm an idiot haha, its nearly 1am here so ill give it a go in the morning and let you know how i get on, thanks very much.

7. ### WBahn Moderator

Mar 31, 2012
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You should be able to recognize this as a first order circuit which will result in an exponential response going from an initial current (and voltage) to a final current (and voltage). So before you even set things up, you should be able to determine the three parameters that define each response: initial value, final value, and time constant. From that you can write down the solutions and set them off to the side to compare to the answers you get from working the problem explicitly. It's all part of asking if the answer makes sense.

8. ### kenneth dean Thread Starter New Member

Nov 19, 2014
11
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Ok so i've been pretty busy lately with other modules, but I had a look over this again and this is how far i've got:
Ir = Vo/R, Vo = L di/dt, Therefore Ir = (L/R)di/dt
I1 = Ir+Il, Il = - (L/R)di/dt +I1
I think that is the solution to part (i) a first order differential equation. Adding in the values given = Il = - 5 di/dt + 4.
However i'm not sure how to solve the equation?

9. ### WBahn Moderator

Mar 31, 2012
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What will the current in the inductor be :
Q1) just prior to the switch opening?
Q2) just after the switch opens?
Q3) a long time after the switch opens.

What is the general form of a first-order exponential response in terms of the initial current, I0, the final current, If, and the time constant, T0?

10. ### kenneth dean Thread Starter New Member

Nov 19, 2014
11
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General form is I (t) = Ae^(pt)
Il(o) = 0 which is given = Ae^0 = A
p= pole, which if my answer above is correct -5s+4 = 0, s = 4/5 = p
so, Il = Ae^(4/5)t ?

11. ### WBahn Moderator

Mar 31, 2012
17,715
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Close, but this isn't the general form because this for only works for a response that starts at A and decays to zero. What if it starts at A and decays to B? Keep in mind that A might be zero, B might be zero, or neither might be zero (the case where both are zero -- or more generally where A=B -- is singularly uninteresting).

So you are claiming that A=0 in this case?

Does this make sense? If A=0, then ....?

12. ### kenneth dean Thread Starter New Member

Nov 19, 2014
11
0
Yeah that would mean that Il = 0, but i'm not sure what the correct form is

13. ### Papabravo Expert

Feb 24, 2006
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Whatever you decide you can always check that decision by differentiation and substitution back into the original equation. Do this a few times and you'll really appreciate the mechanics of it all.

14. ### WBahn Moderator

Mar 31, 2012
17,715
4,787
Okay, so let's see if we can think our way from what you are starting with to what we need.

$
f(t) = Ae^{\frac{-t}{\tau}}
$

This let's us go from an initial value of A to a final value of 0.

How might you modify this if you wanted to just shift this response by B so that it goes from an initial level of (A+B) to final value of just B?

15. ### kenneth dean Thread Starter New Member

Nov 19, 2014
11
0
I've seen the i(t) = i(∞)+(i(0)-i(∞)).e^(-t/t0)

16. ### kenneth dean Thread Starter New Member

Nov 19, 2014
11
0
Ae^-t/tau + B?

17. ### WBahn Moderator

Mar 31, 2012
17,715
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That's the form I was leading you toward. Do you see how this form works? If you understand how it works, then you should never have to look it up ever again because you will be able to easily reconstruct it.

Getting back to how you expect the circuit to behave, you've correctly identified the initial current, but it is unclear how you did so. Could you describe that?

You also need to come up with the expected final current. Please describe how you came up with that, too. Neither of these should involve any equations or formulas, but merely reasoning about how this circuit has to behave.

Where I'm headed is to first come up with the solution without solving the differential equation directly (but rather relying on the general solution you've given above). Once that is done, you can verify that it is, in fact, a solution to the differential equation you came up with (assuming that the D.E. is correct, of course).

Then might be a good time to tackle solving the D.E. directly since you know where it has to lead and where the various pieces of the final solution need to show up in the D.E. to make it work out.

18. ### WBahn Moderator

Mar 31, 2012
17,715
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Yep.

Now, what if we wanted the initial value to be A (not A+B) but the final value to remain B?

19. ### kenneth dean Thread Starter New Member

Nov 19, 2014
11
0
It says in the question that the current is 0 prior to the switch opening. Is that what you mean

20. ### kenneth dean Thread Starter New Member

Nov 19, 2014
11
0
B + (A - B)e^-t/tau