Time it takes for current in an inductive circuit to reach a steady value

Discussion in 'Homework Help' started by Thekosherpickle, Oct 21, 2009.

1. Thekosherpickle Thread Starter New Member

Oct 21, 2009
1
0
Hello to all, I am new to this forum, Im an electrical engineering Tech student in lexington, Ky i stumbled upon this website tonight while frustrated with a homework problem. This website appears to be a great studying resource.

I have a question that im having trouble with.

I apologize ahead of time if i havent provided enough info or a attachment of the circuit; this is my first time posting so any advice is appreciated.

I am givin a simple circuit with a power source of 180V with a switch, a resistor R=60ohm, and an inductor L=3H.

The question is how long is it until current reaches its steady value?

1. The first step i took was finding a time constant T=L/R T=3H/60ohm =50ms

2. I then found source current of the circuit; Is=V/R Is=180v/60ohm = 3A

3. Then I tried setting the source current equal to this equation we learned in class; 3A=(1-e^-x) and x=t/T i dont know what to put in for t.

The solution is 250ms.

My question is, am I using the correct formula? am I attacking this correctly? Am I overthinking this? thank you for any help.

2. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
You are doing well ...

You have correctly indicated for the series R-L circuit that

i(t)=Im(1-exp(-t/T))

In your case Im=3A, T=50ms

Generally the exponential term is considered (by convention) to have "vanished" at exp(-5). [Since exp(-5) = 0.006738 is small cf 1]

So for t/T=5 means that t=5T=250ms. In effect the response is taken as being at steady state when t>=250ms.