How Does Inductor Behavior Affect Circuit Current at t = 0.224 s?

[christianerik]

pic : http://s982.photobucket.com/albums/ae310/christianerik/?action=view&current=untitled.jpg

If the switch is closed at t = 0, find the
current in the inductor at t = 0.224 s. Answer
in units of A.
&
Find the current in the switch at that same
time.

I know that i must use i = V/R(1 - e^-t/L/R) but can't seem to get the right answer.

Any help is appreciated...5 ohms
-------------^^^^^^^------I
I.......I
I...10 ohms...1.4 H...I
I------^^^^^^^^---&&&&--I
I.......I
I...S...30 V...5 ohms.I
I-------/..---I I----^^^^^--I3. The Attempt at a Solution

i = V/R(1 - e^-t/L/R + V/5 = (30 - V)/5

(V/10)( 1 - e^-(0.224*10/1.4) + V/5 = -V/5 + 6

0.1V(0.798)+0.4V = 6

(0.4+0.0798)V = 6

V = 6/0.4798 = 12.5 v

The inductor current at t = 0.224 s

i = V/R(1 - e^-t/L/R)

i = (12.5/10)[ 1 - e^-(0.224(10/1.4) ]

i = 0.9976 A

The switch current:

Is = i + V/5 = 0.9976 + 12.5/5 = 3.497 AWhat am I doing wrong?

 

[rl.bhat]

At t = 0, you have to find out the voltage across LR circuit.
When the switch is open, what is the total resistance of the circuit? Note that when there is no current or a steady current, L has no reactance.
At t = 0, when the switch is closed, find the current drawn from the source.
Then find the voltage across LR branch.
Then using relevant equation find the current t given time t.

 

[christianerik]

i = V/R(1 - e^-t/L/R + V/5 = (30 - V)/5
i thought that was what i was doing

 

[rl.bhat]

12.5 volt across LR branch is not the voltage at t = 0.

 

[christianerik]

hmmm... I calculated 16.52?

 

[rl.bhat]

Show your calculation.

 

[christianerik]

v=30-30(e^(-.244(1.4/5)

 

[rl.bhat]

No. It is not correct. Work out according to my post#2

 

[christianerik]

I got one answ... I(0.224) = 1.2*(1 - e^-2) = 1.038 A
::::
5 + 5*10/(5 + 10) = 8.33
30/8.33 = 3.6 A
5*10/(5 + 10)=3.33
3.33*3.6 = 12 V
12/10 = 1.2 A
so I have that now how would I find the current in the switch at that same
time?

 

[rl.bhat]

Use law of conservation of current.

 

[rl.bhat]

I(0.224) = 1.2*(1 - e^-2) = 1.038 A
How did you write this step?