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Inductance and RL circuits

  • #1
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 cant 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----^^^^^--I


3. 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 A


What am I doing wrong?
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
5
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.
 
  • #3
i = V/R(1 - e^-t/L/R + V/5 = (30 - V)/5
i thought that was what i was doing
 
  • #4
rl.bhat
Homework Helper
4,433
5
12.5 volt across LR branch is not the voltage at t = 0.
 
  • #5
hmmm..... I calculated 16.52?
 
  • #6
rl.bhat
Homework Helper
4,433
5
Show your calculation.
 
  • #7
v=30-30(e^(-.244(1.4/5)
 
  • #8
rl.bhat
Homework Helper
4,433
5
No. It is not correct. Work out according to my post#2
 
  • #9
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?
 
  • #10
rl.bhat
Homework Helper
4,433
5
Use law of conservation of current.
 
  • #11
rl.bhat
Homework Helper
4,433
5
I(0.224) = 1.2*(1 - e^-2) = 1.038 A
How did you write this step?
 

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