1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple RL Circuit

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data

    The switch has been closed for a long time.
    169hnc1.jpg

    a) Current flowing through inductor. Voltage across it. IS any current flowing through the resistor.

    b) The switch opens at t = 0s. Find an equation for the inductor voltage as a function of time for t > 0. How long does it take for the inductor to reach 10% of its maximum value?

    2. Relevant equations

    Kirchhoff's Voltage Law (KVL)

    3. The attempt at a solution

    a) 1A is flowing throw through the inductor as the circuit is stable and the inductor is replaced with a short circuit this causes the resistor to have no current through it and the inductor to have no voltage across it.

    b) This is where I am not sure about:

    t > 0

    ru4d8g.jpg

    I used the KVL equation around the circuit to get:

    [tex]-(100)10^{-3}\frac{di_{L}}{dt} + 1i_{L} = 0[/tex]

    I am not sure if this is the right way to do this.
     
  2. jcsd
  3. Jan 30, 2010 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    -L *di/dl is the emf arising if the current change. So your equation is

    [tex]
    -0.1\frac{di_{L}}{dt} = 1i_{L}
    [/tex]


    The starting condition is [tex]
    I_{L}(0)=1 A
    [/tex]

    The equation is easy to solve.

    ehild
     
  4. Jan 31, 2010 #3
    Ok I attempted this and here is what I got:

    KVL:

    [tex]iR + v = 0[/tex]

    [tex]iR + L\frac{di}{dt} = 0[/tex]

    [tex]\frac{di}{dt} = -\frac{iR}{L}[/tex]

    [tex]-\frac{di}{i} = \frac{R}{L}dt[/tex]

    [tex]-lni + C = \frac{R}{L}t[/tex]

    [tex]I_{L}(0)=1 A[/tex]

    [tex]-ln1 + C = \frac{R}{L}0[/tex]

    [tex]C = 0[/tex]

    [tex]-lni + 0 = \frac{R}{L}t[/tex]

    [tex]0 = \frac{R}{L}t + lni[/tex]

    [tex]1 = e^{\frac{R}{L}t} + i[/tex]

    [tex]i = 1 - e^{\frac{R}{L}t}[/tex]

    [tex]v = L\frac{di}{dt}[/tex]

    [tex]v = L\frac{d(1 - e^{\frac{R}{L}t})}{dt}[/tex]

    [tex]v = -Re^{\frac{R}{L}t}[/tex]
     
  5. Jan 31, 2010 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Is is correct up to here, but wrong from here.

    You have a "* " instead of "+"

    [tex]i = e^{-\frac{R}{L}t}[/tex]

    [tex]v = \frac{di}{dt}[/tex]

    [tex]v = L\frac{d(e^{-\frac{R}{L}t})}{dt}[/tex]

    [tex]v = -Re^{-\frac{R}{L}t}[/tex]

    ehild
     
  6. Jan 31, 2010 #5
    Ok I got it thanks for the help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple RL Circuit
  1. RL circuit (Replies: 9)

  2. RL Circuit (Replies: 3)

  3. Simple RL circuit (Replies: 2)

  4. RL circuit (Replies: 3)

Loading...