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

In summary, we discussed how to find the inductance of an inductor in a series circuit with a resistor and a battery, given a specific time for the current to reach 90% of its final value. We used the equation I = (E/R) (1- exp(-t/(L/R)) to find the time constant, and then solved for the inductance using the formula τ = L/R. Through algebraic manipulation, we were able to determine the inductance to be 8.7 millihenries. It is important to practice and develop a "tool kit" of approaches to solve similar problems.
  • #1
a_patel32
10
0

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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  • #2
a_patel32 said:

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?
 
  • #3
Using the second equation? I don't think I can because I need the battery voltage and inductance right?
 
  • #4
a_patel32 said:
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?
 
  • #5
It would be I(final) = (E/1000)(1 - exp(-20/(L/1000)) ?
 
  • #6
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→[itex]\infty[/itex]). Then ask your self: on what time, I(t) will be 90% of the final current. Do some algebra and get the solution!
 
  • #7
a_patel32 said:
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).
 
  • #8
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?
 
  • #9
Also, the answer given is 8.7 millihenries
 
  • #10
a_patel32 said:
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.
 
  • #11
Sorry, I'm not sure how to continue beyond this?
 
  • #12
a_patel32 said:
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##.
 
  • #13
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?
 
  • #14
a_patel32 said:
Great thank you. i just got the answer.
You're welcome :smile:
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!
 

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

1. What is inductance?

Inductance is a property of an electrical circuit that describes its ability to store energy in the form of a magnetic field. It is measured in units of Henrys (H).

2. How does inductance affect RL circuits?

Inductance plays a crucial role in RL (resistor-inductor) circuits as it causes a time-varying current to flow, which can create a back EMF (electromotive force) that opposes the flow of current. This results in a delay in the current reaching its maximum value, causing a lag in the circuit's response.

3. How is inductance calculated?

Inductance can be calculated using the formula L = NΦ/I, where L is inductance (H), N is the number of turns in the coil, Φ is the magnetic flux (Wb), and I is the current (A).

4. What is the role of inductors in RL circuits?

Inductors are used in RL circuits to control the flow of current and to store energy in the form of a magnetic field. They also help to smooth out fluctuations in the current, resulting in a more stable circuit.

5. How do RL circuits affect AC and DC signals differently?

RL circuits have different effects on AC and DC signals due to the time-varying nature of AC signals. In DC circuits, the inductor will behave like a wire once it reaches a steady state, while in AC circuits, the inductor causes a phase shift between the current and voltage, resulting in a lag in the circuit's response.

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