Please Help people?

Question

The current through a 3.0mH inductor is being built up from zero by a 1.5V battery. the equivalent series resistance of the circuit is 6.0 ohms.

a) Calculate the inductive time constant?
____________s

b) Calculate the current 2.3ms after the swith is closed.

___________mA


when 250 mA flows through a 3.0 mH inductor switches are thrown so that the current begins to decay with an equivalent series resistance of 6.0 ohms and no battery. Calculate the current 2.3 ms after the switch is closed

_________________mA

Answer 1

OK, this is the way how I think things are

1) a) The inductive constant is L/R = 3*10^-3/6 = 5* 10^-4 H/ohm

b) t = 2.3 ms = 3*10^-3 s, L = 3.0 mH= 3.10^-3 H, R = 6 ohm

I = Io (1- e^(-t L/R)

I = Io (1- e^[(-2.3*10^-3)*(3*10^-3)/6] = Io (1-e^1.5*10^-6) aprox 0.75 Io

Now Io = E/R = 1.5 V/6ohm = 0.25 A

I = 0.25* 0.75 = 0.06 A

2) In this case you don't have a battery but you have Io, so, just plug it in the equation and do the same

Hope this is right. Sorry, I am just learning, but this is what I think is right. Perhaps you will get some other reply and confirm mine.

Ilusion

Answer 2

It's e^(-tL/R). You need to figure out how to use it ☺

Doug