Find resistance, inductance, and the time constant for an RL circuit.

[rail2k]

Homework Statement

A 5.0V battery is attached to an RL circuit. The current is given by the formula: I = V₀/R(1-e^-t/τ), where I₀ is the steady-state current, and τ is the time constant for the circuit.

Using the data table provided, determine τ, R, and L for this circuit.

 I(mA) | t(μs)
---------------
   0   |   0
 3.94  |   1
 6.32  |   2
 7.77  |   3
 8.65  |   4
 9.18  |   5
 9.50  |   6
  ...  |  ...
 10.0  |   ∞

Homework Equations

I = V/R
τ = L/R
ε_L = -ε₀e^-Rt/L
I = ε₀/R(1-e^-Rt/L)

The Attempt at a Solution

So I started by concluding that ε₀ = V₀ = 5 V and I₀ = 10 mA.

Using that I determined that R = V₀/I₀ = 5/0.01 = 500 Ω which gives me resistance.

I know that once I find τ or L I can easily find the other one, but I can't figure out how to find either of them. At first I thought I could simply work I = V₀/R(1-e^-t/τ) algebraically to give me τ, but I end up getting undefined from using the natural log with a negative number.

I feel like I have to use the data to find one of them (probably τ?) by using the relationship between two of the data points but I'm sort of stuck at this point.

 

[rude man]

When t = 1 time constant, what is the current?
Then, what is the time constant?
Then, expess the time constant in terms of L and R.

 

[rail2k]

Ah, so when t = τ, I = 5/500(1-e^-1) ≈ 6.32 mA.

So τ = 2 μs from the data and therefore L = 500*0.000002 = 1 mH.

That seems correct, right?

 

[rude man]

Ah, so when t = τ, I = 5/500(1-e^-1) ≈ 6.32 mA.

So τ = 2 μs from the data and therefore L = 500*0.000002 = 1 mH.

That seems correct, right?

It does to me!
Good going.

 

[rail2k]

Thanks a lot! Can't believe the answer was so close, lol.