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Circuit with inductor; find current for t>0

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/a/img832/1651/homeworkprobsg32.jpg [Broken]

    At t= 0, the switch is opened. Calculate the current i(t) for t>0

    2. Relevant equations

    V(t) = L* dI/dt

    V = IR

    I = V/R

    voltage division, current division,

    KCL, KVL

    3. The attempt at a solution

    Not so experienced with inductors but, at the beginning with the switch connecting:

    voltage division:

    V across right branch: (12Ω/14Ω)*36V

    V = 30.85V

    so this must be the total voltage across the branch with the 6Ω and 2H

    Voltage across 2Ω = 36V - 30.85V

    = 5.142V

    I across 2Ω resistor = 5.142V/2Ω

    = 2.571A, and

    2.571A enters the parallel branches

    so this is the total current combined through the parallel branches

    I know now that the voltage across an inductor is 0 if the current does not change

    so with that and current division:

    I = (12/18)(2.571A)

    = 1.714A through the inductor branch with the switch connected at the beginning

    and then the current through the other branch is 0.857A at the beginning.

    My problem I think is calculus related again since V(t) = L* dI/dt

    is the only equation I know for an inductor, I try to set up a problem going like

    v(t) = 2H*dI(dt) + 6Ω(1.714A)

    but not sure

    and couldn't the current be 0 if the switch is opened?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 6, 2013 #2

    rude man

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    Defining current i counterclockwise, sum the voltages around this loop = 0. That will give you a diff. eq. to solve.

    Solve the differential equation with the appropriate initial condition on i. What is i just after the switch is thrown? Hint: current thru an inductor cannot change instantaneously so what was the current thru L just before the switch was thrown?
  4. Mar 6, 2013 #3
    Well, one more thing to know I guess.

    Assuming everything I did was right before, then I'd say the current through L before throwing the switch is


    Can't see any mathematical mistake I did before... don't I already have the (initial) voltages around the loop? Which loop do you mean?
  5. Mar 6, 2013 #4

    rude man

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    Incorrect. Show me how you computed the inductor current before the switch was opened.
    There is only one loop after the switch opens.

    Remind me - what is your loop equation around that loop?
  6. Mar 6, 2013 #5
    Well first I did voltage division between the parallel branches and the 2Ω

    and i got 30.85V as the voltage across the 8Ω and 4Ω in series (while completely ignoring the 6Ω resistor, unless you can't do it this way?)

    Okay maybe this is wrong, and probably need to start all over again with this.

    I know little about inductors; the 6Ω is in series with the 2H inductor but can you say that the 6Ω is in parallel with the branch containing the 4Ω and 8Ω resistors?

    (After that I used Ohm's Law to find the current through the 2Ω, and then current division).

    And for the loop after the switch opens, would it be

    V = IR

    with Req = 18Ω ?

    Then at the same time V = L * dI/dt so would you set them equal by any chance?
  7. Mar 6, 2013 #6

    rude man

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    The inductor is a short circuit until the switch is thrown open. So recompute the initial current thru the inductor.

    (There's no voltage across L before the switch is opened. Why? Because V = L di/dt for an inductor, and since we waited a long time before throwing the switch open, di/dt = 0 everywhere. So any L looks like a short circuit if its di/dt = 0. A short circuit is defined by V = 0 across it).
  8. Mar 7, 2013 #7
    Sorry for getting back to this late.

    I did it over and got V= 24V across the branch with the inductor with the switch still closed.

    Then got I0 = 4A across the inductor

    If the current isn't allowed to instantaneously change through an inductor, what about the voltage?

    Because if it does, then the voltage when the switch opens would be 4A*18Ω = 72V, right?

    How would I go about setting up / solving the differential equation now?

    I'm thinking you also still need a formula like V(t) = -V0*e((-t/RC), but there's no capacitance?
  9. Mar 7, 2013 #8

    rude man

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    I said to sum the voltages around the loop. When you do that you get a diff. eq. that looks like L di/dt + (R1 + R2 + ...)i = 0.
    Solve the equation with the initial condition on i.

    Do you know how to solve a 1st order constant-coefficient diff. eq.?

    I'm thinking you also still need a formula like V(t) = -V0*e((-t/RC), but there's no capacitance?[/QUOTE]

    With a capacitor the time constant is RC. With an inductor the time constant is L/R.
    Expect to get a solution that looks like i = i_0 exp(-Rt/L).
  10. Mar 8, 2013 #9
    Sorry for getting back to this late again..

    So for the voltage in the loop like you said I got:

    (dI/dt)(2H) + 18Ω*I = 0

    Then with it as a differential equation I got

    dI/I = -18Ω*(dt)/2H

    then integrating:

    ln(I) = (-18t/2)(Ω/H) + C

    And the initial condition was if t = 0 then I = 4


    C = 1.386 then

    ln (I) = -9t*(Ω/H) + 1.386

    Forgot how e and natural log work when you raise them and then try to combine them so:

    I = e^(-9t) + e^(1.386).

    I think because e^(1.386) = 4 then

    I = 4e^(-9t)

    because this is the right answer now except it's negative.

    Thanks for the help, by the way.
    Last edited: Mar 8, 2013
  11. Mar 8, 2013 #10

    rude man

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    OK, stop there because you haven't done the exponentiation correctly.

    Also, before I proceed: fix symbols to your components right at the outset instead of carrying inductance and resistance etc. values thru your equations. I will now declare R = 6 + 4 + 8 ohms = 18 ohms, and L = 2 H.

    So go back to your last correct equation which was L di/dt + R i = 0
    and you solved that OK by saying ln i = -Rt/L + c1, c1 = constant.

    Now you need to exponentiate correctly. Hint: e(a+b) = eaeb.
  12. Mar 9, 2013 #11
    Okay, thanks.

    So it was

    ln (I) = -9t*(Ω/H) + 1.386

    I take it you exponentiate EVERYTHING then at the same time? So

    ln (I) = -9t*(Ω/H) + 1.386


    e^(ln I) = e^(-9t + 1.386)

    I = -4e^(-9t).

    (so the negative sign comes down then right?)
  13. Mar 9, 2013 #12

    rude man

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    "Comes down then"?

    Look at the diagram. Which way is the current going to flow? And that makes the sign of the current as shown on the diagram + or -?

    Hint - your answer is correct, including the sign.
  14. Mar 9, 2013 #13
    Okay, so simply because it flows the opposite way then, right?
  15. Mar 9, 2013 #14

    rude man

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