System Analysis RL circuit

Thread Starter

kenneth dean

Joined Nov 19, 2014
11

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!
 

Thread Starter

kenneth dean

Joined Nov 19, 2014
11
I have a question regarding Vr(t)=-R*i(t).
Why is there negative sign in front of R? Where did it come from?
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
 

WBahn

Joined Mar 31, 2012
32,745
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
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.

Always define your variables, engineering is not about guessing.
 

shteii01

Joined Feb 19, 2010
4,644
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
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.
 

WBahn

Joined Mar 31, 2012
32,745
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.
 

Thread Starter

kenneth dean

Joined Nov 19, 2014
11
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?
 

WBahn

Joined Mar 31, 2012
32,745
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?
 

Thread Starter

kenneth dean

Joined Nov 19, 2014
11
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 ?
 

WBahn

Joined Mar 31, 2012
32,745
General form is I (t) = Ae^(pt)
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).

Il(o) = 0 which is given = Ae^0 = A
So you are claiming that A=0 in this case?

p= pole, which if my answer above is correct -5s+4 = 0, s = 4/5 = p
so, Il = Ae^(4/5)t ?
Does this make sense? If A=0, then ....?
 

Papabravo

Joined Feb 24, 2006
22,061
Yeah that would mean that Il = 0, but i'm not sure what the correct form is
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.
 

WBahn

Joined Mar 31, 2012
32,745
Yeah that would mean that Il = 0, but i'm not sure what the correct form is
Okay, so let's see if we can think our way from what you are starting with to what we need.

Your starting point is:

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

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?
 

Thread Starter

kenneth dean

Joined Nov 19, 2014
11
Okay, so let's see if we can think our way from what you are starting with to what we need.

Your starting point is:

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

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?
Ae^-t/tau + B?
 

WBahn

Joined Mar 31, 2012
32,745
I've seen the i(t) = i(∞)+(i(0)-i(∞)).e^(-t/t0)
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.
 

Thread Starter

kenneth dean

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