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Determine iL(t) of the circuit

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data
    The switch is initially open, closes at t = 0, and then reopens at t = 0.5. Find iL(t).
    swcu9c.png

    2. Relevant equations
    3. The attempt at a solution

    i really have no idea how to do this. i have a series RLC circuit at t=0-, t=0.5 and the circuit with the 18V source at t=0, t=0.5-.

    I got iL(0)=iL(0-)=0
    Vc(0)=Vc(0-)=0

    iL(0.5)=iL(0.5-)=18V/6ohm= 3A
    Vc(0.5)=Vc(0.5-)=5ohm*3A= 15V
    iL'(0.5)=vL(0.5)/L=15/2 = 7.5 A/s

    a = R/2L (series)= 7ohm/4 = 1.75
    w0 = LC^(-1/2)=1/sqrt(2/17)=2.92

    underdamped because a<w0.

    so far is this right? and also i have no idea how to find a single equation iL(t) for t=0-, t=1.5-, and also t=infinity. what do i do??
     
  2. jcsd
  3. Apr 9, 2013 #2

    berkeman

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    Staff: Mentor

    You are correct that the currents and voltages in the right-hand circuit (past the switch) are zero for t=0-.

    But to solve for the behavior after the switch closes at t=0, you need to write the KCL equation for the currents leaving the top right node.

    What is the relationship between voltage and current for an inductor? How about for a capacitor? You will use those relations when writing the KCL equation.
     
  4. Apr 9, 2013 #3
    this is the hints our instructor gave, didn't think i would need it. i would like to try it the loop method bcause it seems easier to understand. but i am still having problems, on step #4. so far i have:

    1. I1+2(I2)+Vc=18
    2. I2=iL= C*Vc' = Vc'/17
    3. I1 = 18-Vc-(2*Vc')/17
    4. How do i find Vc"? for the left loop i get:
    I1+5(I1-I2)=18
    when i substitute I1 and iL I get:
    18-Vc-(2*Vc')/17 + 5(iL)=18
    -Vc-(2*Vc')/17 + 5(iL)=0
     
    Last edited: Apr 9, 2013
  5. Apr 9, 2013 #4

    berkeman

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    Staff: Mentor

    KVL equations use voltages across components as you go around each loop. It looks like you are writing a mix of currents and voltages above?
     
  6. Apr 9, 2013 #5
    i am sorry i got confused. left out the 1ohm i multiplied I1 by, so that would make it voltage.
    1. 1ohm(I1)+2ohm(I2)+Vc=18
    2. I2=iC= C*Vc' = Vc'/17
    3. 1ohm(I1) = 18-Vc-(2*Vc')/17

    so it is now a voltage equation. how would i find I1 alone in terms of Vc'?
     
  7. Apr 9, 2013 #6

    berkeman

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    Staff: Mentor

    You need to write both KVL equations before starting to solve for t=0+.

    And you will only have the right-hand loop obviously after the switch is opened, with some initial conditions determined by how the inductor current and capacitor voltage ramp up for that first 0.5 seconds...
     
  8. Apr 9, 2013 #7
    alright, so KVL equations for outer and left loop, and leave out the right loop for now?

    1. for outer loop: 1(I1)+2(I2)+Vc=18
    2. using equation I=C*dv/dt, since I2 is same as current through capacitor i got I2=iC= C*Vc' = Vc'/17

    also for left loop, the equation before substituting for I1/IL is: 1(I1)+5(I1-I2)=18
    also noting that IL=I1-I2


    i think this is right so far, but the problem is i don't know what to do from here. specifically how to do step #3, can't find a way to relate I1 to Vc'
     
  9. Apr 10, 2013 #8
    OK, suppose i try node method and write KCL for top node:
    -I1+IL+IC=0

    in terms of Vc, -I1=(Vc-Vs)/(R1+R3) = (Vc-18)/3
    also, iC= C(dv/dt)

    so:
    Vc/3 + IL + C(d(Vc)/dt) = 6

    Since Vc is equal to voltage across 5ohm plus the V across inductor:
    Vc = L(d(IL)/dt) + IL(5)

    if i substitute this into the equation:

    1/3[L(d(IL)/dt) + 5(IL)] + IL + C(d/dt)[L(d(IL)/dt) + 5(IL)] = 6

    following the method that my book uses:
    IL'' + [L+(R1*R3*C)/(R1*L*C)]IL' + [(R1+R3)/(R1*L*C)]IL = Vs/(R1*L*C)

    where R1=3, R3=5, in this case. also L=2 and forgot to mention that C actually simplifies to 1/17 = 0.05882F so:


    a = (L+R1*R3*C)/(R1*L*C) = (2+(15/17)) / (6/17) = 8.1666
    b = (R1+R3)/(R1*L*C) = 8/(6/17) = 22.666
    c = Vs/(R1*L*C) = 18/(6/17)= 51

    A= a/2 = 4.08
    w0 = sqrt(b) = 4.76

    A<w0 so underdamped.
     
    Last edited: Apr 10, 2013
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