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RL circuit fun! am I drawing this circuit correct at different times?

  1. Apr 1, 2006 #1
    Hello eveeryone! exam time monday! Just doing some last checks to make sure i'm understanding this correctly. Here is the directions and circuit:
    THe circuit shown below has beeen in the form shown for a very long time. THe switchopens at t =0. FInd iR at t equal to (a)0-, (b) 0+, (c) infinity, and (d) 1.5

    I'm showing u my work for (a) and (b).

    FOr par (a)
    WHat i'm confused on is, U see that wire that says at time t =0, its going to switch open. At time t < 0, does that mean there is just a wire there?

    Does that mean all the current is going to go through that wire and say screw all the other compoents like the 60, and 40 ohm resistors? THe answer in the back of the book is iR = 0;

    But is it 0 because all the current will go into that wire with no resistance?
    I remember the professor said, if the inductor is shorted(in this case it would be because the circuit has been sitting there for a long time) then anything in parellel with that shorted inductor is also shorted out?


    For part (b)
    iR = 10mA
    Is this iR 10mA because when that wire is opened that means the conductor isn't shorted but acts as a huge resistor. Not wanting any current to go through it, so it makes all the current go through the 60 ohm resistor?

    For partt (d) 1.5ms

    I need to find x = L/Req,
    to find Req, ur suppose to "look" into the inductor and take out all power supplies, in this case I think you would be left with 60 and 40 in parellel, so i got Req = 4mA, L = .1H

    I know i'm going to use the equation:
    iL(t) = io*e^t/x, where x = L/R, L being the indcutor, R being the equivlent resistance, but what io do i use? io is the initial current i though, usually i use the io i found at time i(0-) but in this case its 0!

    The book is getting an answer of 5.34mA.



    Thanks!

    [​IMG]
     
    Last edited: Apr 1, 2006
  2. jcsd
  3. Apr 1, 2006 #2

    nrqed

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    Yes. The switch is closed which means it acts as an ideal wire.
    well, yes...but please be careful with the language. I might not be the only one who is easily offended :frown:
    Yes, anything parallel with a wire with no resistance is shorted

    Yes
    The current will be of the general form C_1 + C_2 e^(-t/tau). You must impose that at t=0 this reproduces the result at t=0+, so 10 mA. At t=infinity, find the current by replacing the inductor by a wire. That gives you a second condition which will fix C_1 and C_2. Then you can find the current at any time.
     
  4. Apr 1, 2006 #3
    Thanks for the responce, sorry about the language somtimes when I type i don't realize what i'm actually typing :redface:

    I had no idea there was another form like that. Makes sense though!
    I used 10ma as C1, and C2, i used the value i got by evaulating the circuit at t(infinity) and i got the following value of iR = (10mA)(40)/(60+40) = 4mA. Which is what the book has. But when i put it into the forumla:
    10E-3 + 4E-3*e^(-240*1.5E-3) = .012791A or 12.79mA
    but the book has: 5.34mA.

    Any idea where i misunderstood?
     
  5. Apr 1, 2006 #4

    nrqed

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    No problem, I am probably too sensitive
    Watch out. If you set t=0 in the equation, you get C_1 + C_2 = 10 mA. If you set t= infinity, you get C_1 = 4 mA (the exponential is zero). So C_2 = 6 mA.

    Don`t jump to the conclusion that C_1 is the current at t=0 and C_2 is the current at t=infinity!

    Pat
     
  6. Apr 1, 2006 #5

    nrqed

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    Also (I had not checked that part of your calculation) but if you take out all power sources as you said, the inductor will seetwo resistors in series so your equivalent resistance should be 1000 ohms.

    Soyou get I(1.5 ms) = 4 mA + 6 mA e^(-100/.1 * 1.5E-3) = 5.34 mA

    Patrick
     
  7. Apr 1, 2006 #6
    ahh thanks again! it worked out nicely and great explanation! But i'm having problems visualing what happens when u "look" through a inductor and simpify from there. For example, in this case would it look like this?
    [​IMG]
     
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