# LR Network

Discussion in 'Homework Help' started by sitting_duck, Apr 20, 2010.

1. ### sitting_duck Thread Starter New Member

Apr 18, 2010
14
1
I hope people don't mind me asking all these questions.

The problem is the first image below.

I have transformed the original circuit with SW1 closed to a Norton equivalent. Hopefully correctly. See below.

I have the formula from my notes for this sort of circuit;

i(t)-I=(i(0)-I)e^(-t/GL)

and I have L=10, G=1/20, I=2 for the general formula. How do I get i(0)?

I think it must be very simple, i just can't see how.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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There is no need to derive the Norton equivalent - why do you think that is necessary?

You can determine the current at t=0 purely by inspection. Remember that the current in an inductor cannot change instantaneously - what was the current in the inductor just before the switch SW1 was closed?

Last edited: Apr 21, 2010
3. ### sitting_duck Thread Starter New Member

Apr 18, 2010
14
1
Well the formula i have is for the equivalent, i.e. it uses the current and the conductance of the resistor in parallel.

Is the current equal to 0 before then switch is closed?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
One of the problems of just applying a formula is that you may not always have a formula to fit the specific situation.

I am surprised that you don't have a worked example showing how you would deal with a problem of this type - solving the circuit without recourse to the Norton transformation.

Once the switch closes at t=0 you have a simple series circuit comprising a DC voltage source of 40V, a total series resistance of 20Ω and a 10H inductor. Have you not been shown how to solve such a 'classic' circuit problem?

The general form of the solution will be

$i(t)=\frac{E}{R} \left(1-\epsilon^{\frac{-R}{L}t} \right)$

5. ### sitting_duck Thread Starter New Member

Apr 18, 2010
14
1
Well to tell the truth what we always did is transform all dynamic circuits with inductors to current sources and questions with capacitors to voltage sources. Im sure it's possible to solve without doing this, then one would have to go back to the differential equation for the capacitor/inductor.

To answer your question then, I don't recall doing an example where i couldn't transform the circuit, then again i amn't very good at this subject, I am after all repeating the exam after failing it the first time.

Thanks for the help, using that formula I got i@t=0 to be 0.

i(0)=(40/20)x(1-e^((-R/L)xt))

i(0)=(40/20)x(1-1)

i(0)=0

is this correct? I have a feeling i made a wrong assumption of something. Isn't the formula only valid when t>0?

I know that your formula is also the answer to the question. But is there anyway you could simply explain to me how to get i@ t=0? My apologies if you have already done so and I haven't understood.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Using the formula you correctly showed that i(0)=0. That's one way of finding i(0).

Before the switch is closed there is no current in the inductor - unless you are explicitly told that is not the case, which you aren't. There is no initial condition.

Again, if you review what you would have been taught about the behaviour of an inductor with respect to current change you might remember that the current cannot instantaneously change. Recalling [from Faraday's Law] the ideal inductor emf e=L(di/dt), it should be clear this is the case in a physical circuit. If di/dt is instantaneous then emf 'e' must be infinite - which is improbable in a physical system. So if the current in L was zero before the switch is closed then at the instant the switch closes to complete the circuit, the current will still be zero.

In principle the full DC source voltage of 40V appears across the inductor at t=0. The rate of change of current in the inductor however is not zero. You can show this mathematically by differentiating the general equation for i(t) given.

$\frac{di}{dt}=\frac{E}{L}\epsilon^{-\frac{R}{L}t}$

So at t=0

$\frac{di}{dt}=4 A/s$

And at t=∞

$\frac{di}{dt}=0 A/s$

Last edited: Apr 21, 2010
7. ### sitting_duck Thread Starter New Member

Apr 18, 2010
14
1
Thanks a million t n k.

The expression i have got or the current then is:

i(t)=2(1-e^-(2t))

that's using the equivalent circuit above and the formula also.

the question then continues, see the jpg below.

ii) I was then asked for the rate of current increase, is this correct? I know it looks for initial increase, but is this the correct expression to use?

di(t)/dt = (I-i(t))/GL

iii)

current @ .3 seconds

i(.3)=2(1-e^-2(.3)) = .9 amps

iv)

Do i short circuit the inductor and find the current going through that wire?

using current division with the equivalent circuit:

i= (1x2)/((1/20)+1)= 1.9amps

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8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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(ii)

$i(t)-I=(i(0)-I)\epsilon^{- \frac{t}{GL}}$

Differentiate this to give ....

$\frac{di}{dt}=-\frac{I}{GL}(i(0)-I)\epsilon^{-\frac{t}{GL}}$

and since i(0)=0

$\frac{di}{dt}=\frac{I}{GL}\epsilon^{-\frac{t}{GL}}$

Substitute in this to find di/dt at t=0

(iii) 0.9A is correct

(iV) The inductor behaves as a short at steady state conditions (i.e. as t->∞) - so all the source current flows in inductor L. It doesn't divide as you suggest.