Calculating Inductance in an RL Circuit

[n.hirsch1]

Homework Statement

An RL circuit has a 9.0 V battery and a 5.5 ohm resistor. When the switch is closed, the current in the circuit is observed to increase from 0 to 0.30 A in 0.17 s.
A) What is the inductance?
B) How long after the switch is closed does the current have the value 0.47 A?

Homework Equations

τ = L/R
EMF = L*I / t
I = EMF / R *(1-e^-tR/L)

The Attempt at a Solution

I don't know how to find a time constant, my text doesn't explain it, which makes me think that this question may not even use this equation. I can't get to part B without an answer to part A. By using the second equation I got L = 5.1, but that isn't correct.

 

[xcvxcvvc]

Homework Statement

An RL circuit has a 9.0 V battery and a 5.5 ohm resistor. When the switch is closed, the current in the circuit is observed to increase from 0 to 0.30 A in 0.17 s.
A) What is the inductance?
B) How long after the switch is closed does the current have the value 0.47 A?

Homework Equations

τ = L/R
EMF = L*I / t
I = EMF / R *(1-e^-tR/L)

The Attempt at a Solution

I don't know how to find a time constant, my text doesn't explain it, which makes me think that this question may not even use this equation. I can't get to part B without an answer to part A. By using the second equation I got L = 5.1, but that isn't correct.

are you sure you don't know how?

 

[n.hirsch1]

No, I don't. Its asking for inductance, I have resistance, I have no idea how to get the time constant.

 

[xcvxcvvc]

No, I don't. Its asking for inductance, I have resistance, I have no idea how to get the time constant.

$$I(t) = \frac{v}{R} e^{\frac{-tR}{L}}$$
You know the value of current = 300mA at t = .17. v = 9 and r = 5.5. This leaves 1 unknown. Solve for it.