How to Determine Inductance in RL Circuits for a Required Current Response Time?

[a_patel32]

Homework Statement

An inductor is connected in series with a 1000 ohm resistor and a battery. What mus be the inductance in millihenries of the inductor if the current in the circuit is required to reach 90% of its final value after 20 microseconds?

Homework Equations

I am not actually sure which equatins to use. All I know is E(naught) - L(dI/dt) = IR

The Attempt at a Solution

First I tried to just plug in the numbers but I am obviously missing some values. Another equation I was given was I = (E/R) (1- exp(-t/(L/R))

 

[gneill]

Homework Statement

An inductor is connected in series with a 1000 ohm resistor and a battery. What mus be the inductance in millihenries of the inductor if the current in the circuit is required to reach 90% of its final value after 20 microseconds?

Homework Equations

I am not actually sure which equatins to use. All I know is E(naught) - L(dI/dt) = IR

The Attempt at a Solution

First I tried to just plug in the numbers but I am obviously missing some values. Another equation I was given was I = (E/R) (1- exp(-t/(L/R))

Hi a_patel32. Welcome to Physics Forums.

That second equation is in fact the solution of the first equation (which is a differential equation in variable I), so that's a good place to start.

Have you worked out what the steady-state (final) value of the current will be?

 

[a_patel32]

Using the second equation? I don't think I can because I need the battery voltage and inductance right?

 

[gneill]

Using the second equation? I don't think I can because I need the battery voltage and inductance right?

Nope.

Suppose that you happened to have a value for E and a value for L. What would you write for the final current?

 

[a_patel32]

It would be I(final) = (E/1000)(1 - exp(-20/(L/1000)) ?

 

[cosmic dust]

The final current will be the value of the current at a very late time, so to find it you should take the limit I(t→$\infty$). Then ask your self: on what time, I(t) will be 90% of the final current. Do some algebra and get the solution!

 

[gneill]

It would be I(final) = (E/1000)(1 - exp(-20/(L/1000)) ?

Where did the "20" come from in the exp argument?

What is the behavior of exp(-x) when x is allowed to go to infinity (a very long time indeed).

 

[a_patel32]

Sorry the 20 was the time in the equation, but understand we are talking about the final current. In the exp(-x) it would eventually go down to zero. So it is just going to be I(final) = E/1000?

 

[a_patel32]

Also, the answer given is 8.7 millihenries

 

[gneill]

Sorry the 20 was the time in the equation, but understand we are talking about the final current. In the exp(-x) it would eventually go down to zero. So it is just going to be I(final) = E/1000?

Yes, that's right. So let's call that final current Imax. The current over time is then
$$I(t) = I_{max}\left(1 - exp(-t/\tau)\right)~~~~~~\tau = L/R$$
You're interested in a particular point in time when the value of I(t) will be 90% of the final value. So replace I(t) with (90/100)Imax and continue.

 

[a_patel32]

Sorry, I'm not sure how to continue beyond this?

 

[gneill]

Sorry, I'm not sure how to continue beyond this?

Just replace I(t) with (90/100)Imax (that's 90% of Imax). At this point it's just algebra. Cancel what you can, then solve what's left for the time constant $\tau$.

 

[a_patel32]

Great thank you. i just got the answer. This was very helpful. I don't know why I didn't see this before. How are you able to find a solution because when I am taking physics exam I'm not exactly sure how to go about solving problems like these?

 

[gneill]

Great thank you. i just got the answer.

You're welcome

This was very helpful. I don't know why I didn't see this before. How are you able to find a solution because when I am taking physics exam I'm not exactly sure how to go about solving problems like these?

It's a matter of practice. Eventually you build up a "tool kit" of approaches to these problems, and you'll quickly recognize how to attack them. Good luck!