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Simple R/L Circuit Current Problem

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data

    http://img202.imageshack.us/img202/8355/screenshot20130330at234.png [Broken]

    a) The switch closes at time t=0. What is the current flowing down through the inductor at time t=2ms?

    b) What time will it take for the inductor to reach maximum current?

    c) How long will it take for the inductor to reach current 2mA?



    2. Relevant equations

    i(t) = Ifinal + (Iinitial - Ifinal)[itex]e^{-t/\tau}[/itex]



    3. The attempt at a solution

    Current before the switch closes at t=0 is obviously 0A, so that means Iinitial = 0A.

    After the switch closes and the circuit has come to steady state, you can represent the inductor by a short-circuit. The short circuit being in parallel with the two 3k[itex]\Omega[/itex] (equivalent to a 1.5k[itex]\Omega[/itex] resistor) effectively cancels out the two parallel 3k[itex]\Omega[/itex] resistors. So you're left with basically a single-loop circuit, so Ifinal would be equal to the current through the 3k[itex]\Omega[/itex] resistor in series with the voltage source, so Ifinal=0.001A.

    [itex]\tau[/itex] = [itex]\frac{L}{R}[/itex] = [itex]\frac{1 H}{3000 \Omega}[/itex] = 3.33 x 10-4

    That gives you equation:
    i(t) = 0.001 - 0.001[itex]e^{-t/3.33 x 10^{-4}}[/itex]

    Solving the equation at t=2ms give you i(2ms) = 0.998mA.

    Am I correct so far?

    For part b, wouldn't it take t=∞ amount of time since the current never really reaches 0.001A? The professor seems to expect a definite answer though...

    For part c, is this a trick question? If I've done everything right, the inductor current never reaches 2mA.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 30, 2013 #2

    gneill

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    Staff: Mentor

    R2 and R3 will influence the value of the time constant. Try turning the voltage source and resistor network into its Thevenin equivalent, with the inductor serving as the 'load'.

    Your query about part c has merit.
     
  4. Mar 30, 2013 #3
    (oops... ignore this post. I'll fix it soon.)
     
  5. Mar 30, 2013 #4
    I'd get:

    http://img145.imageshack.us/img145/5180/screenshot20130330at324.png [Broken]

    right?

    So the above equation is right, but now [itex]\tau[/itex] = 0.001.

    So i(2ms) = 0.865mA.
     
    Last edited by a moderator: May 6, 2017
  6. Mar 30, 2013 #5

    gneill

    User Avatar

    Staff: Mentor

    Yup. That looks good.
     
  7. Mar 30, 2013 #6
    Ok great. But what about part b? The limiting current in the inductor is 1mA, but it never really reaches that current, so t=infinity.
     
  8. Mar 30, 2013 #7

    gneill

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    Staff: Mentor

    True. (Although as an engineering rule of thumb, it's common practice to assume that all the excitement is essentially over after five time constants have elapsed)
     
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