# Inductor current calculation in time.

Discussion in 'Homework Help' started by Winten, Nov 12, 2010.

1. ### Winten Thread Starter New Member

Nov 12, 2010
2
0
I am having problems calculating currents in time. Please help me to understand it, for example we have this circuit:

And during time in seconds.
0-1 (switch up)
1-5 (switch down)
5-6 (switch up)
6-6,5 ( switch down)

So in current circuit phase, then switch is up: I=5A, and at end time switch is down I=7,5A.

Time 0->1
I(change) = 2.5(1-1/(2.71^(-1)) = 1,58A

At zero time I = 2,5A, so I(1sec) = 2,5 + 1.58 = 4,08A (please tell if this is OK)

Now at 1-5 switch is down. At 5 sec its getting up again, so I(end) = 5A, and current I = 4,08A.

Ant tau = 1/3 (please tell if this is ok, either)

I(change) = (5 - 4,08)(1- 1/2.71^(5/ (1/3)) = ~ 0,92

So I at 5sec is I(5sec) = 4,08 + 0,92 = 5

Also current should have decreased in this time.

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
290
Rather than use "switch up" and switch down", try "switch open" and "switch closed". That is correct terminology, and not dependent on the orientation of the drawing.

For your figuring, Rtotal with the switch open is 3 ohms. It becomes 2 ohms with the switch closed.

Your currents are incorrect - it's 5 amps open and 7.5 closed.

3. ### mik3 Senior Member

Feb 4, 2008
4,846
67
Why at zero time I=2.5A?
Is it an initial condition?

If it is an initial condition it will change with time and thus the current at 1 sec is not 2.5+1.58.