# Inductance, RL Circuits

## 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))

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gneill
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## 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?

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

gneill
Mentor
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?

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

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
Mentor
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).

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?

Also, the answer given is 8.7 millihenries

gneill
Mentor
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.

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

gneill
Mentor
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##.

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
Mentor
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!