1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: L-C-R Circuit - finding current

  1. Apr 7, 2008 #1
    [SOLVED] L-C-R Circuit - finding current

    1. The problem statement, all variables and given/known data
    In the figure below, V = 100.0 V, R1 = 40.0 Ohms, R2= R3 = 36.0 Ohms , and L = 17.0 H. No current flows until switch S is closed at t=0. Find the magnitude of the current i1 immediately after the switch is closed.

    http://calculus.unl.edu/edu/classes/JF05/LRC.gif [Broken]

    2. Relevant equations
    Kirchoff's loop rule

    Voltage across Inductor: [tex]L\frac{di}{dt}[/tex]

    3. The attempt at a solution

    Since [tex]R_{2}[/tex] and [tex]R_{3}[/tex] are in series I can add them:

    36 Ohms + 36 Ohms = 72 Ohms

    And applying the loop rule:

    [tex]V - iR_{1} - L\frac{di}{dt} - iR_{2+3} = 0[/tex]
    [tex]100 V - i(40 Ohms) - ???? - i(72 Ohms) = 0[/tex]


    I have two questions regarding this problem:

    Is applying the Loop rule the right way to go?

    And if so, what exactly is [tex]L\frac{di}{dt}[/tex]? Isn't it just [tex]\frac{\epsilon}{L}[/tex]?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Apr 7, 2008 #2
    It's much easier than you thought. "Immediately after the switch is closed" is the easiest kind of question to answer. It takes time for the current through an inductor to build up. (Or, if this were a capacitor problem instead of an inductor problem, I would be saying: it takes time for the voltage across a capacitor to build up).

    So right after the switch is closed, with i3 beginning at zero, the inductor is like an "open switch".

    But remember that this is true only for a point in time -- then i3 will increase asymptotically toward some final value.
  4. Apr 7, 2008 #3
    thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook