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RLC circuit problem

  1. May 23, 2009 #1
    1. (See attachment) Q1 a, Determine I1,I2and I3 after 5 time constants
    b, when the switch is first closed ie t=0
    c,when the switch re opens




    2. I total=I1-(i1+I2)

    i=I[1-e --RL/L




    3. i havnt practiced any circuits with resitance in series and parallel just looking for a way forwerd on how to write the equations down for each one
    with the circuit initially closed i imagine its basic kirchoffs laws to calculate I1,2 and 3 but these will be very small as almost all of the supply voltage is going to the inductor but after 5 time constants at 63.2% steady state the current values will increase


    any help appreciated

    regards Ian
     

    Attached Files:

  2. jcsd
  3. May 25, 2009 #2
  4. May 25, 2009 #3
    Been awhile since circuits for me. I'm going to assume that you are working with simple differential equations and not something more advanced like Laplace transforms, etc. I think I would sum up the voltage of each loop. The sum of voltage around each loop is set to zero. That will give you 2 equations for I1, I2, I3. The third equation is I1=I2 + I3, the current entering the top node has to equal the two currents leaving it. One of these three equations will be a first order differential equation (DE) in I3. Use the other two equations to eliminate I1 and I2 from this DE and then solve the DE. You'll need an initial condition to solve for the constant of integration. The propery of inductors to oppose sudden changes in current should lead you to the proper initial condition to use for I3 at time zero. Once that is solved for I3, you can use the other two equations to get I1 and I2. The start plugging in your times of zero to get part b, e^(+/-5) to get part a. I assume that part c is meant that after a long time the switch is opened so that the currents have reached steady state.
     
  5. May 25, 2009 #4
    Thankyou very much for the replies
    Uskalu

    As i am only learning, the method i used to calculate the currents for I1 I2 I3
    was a very long method giving similar results ie

    I2=I1x(R1/R1+R3)
    I2=.037(100/300)=12.3mA (.0123A) etc

    now the part i dont understand on your working out
    is the equation you used to calc the voltage across R2, R2=200x30/800
    i dont know where the 30 and 800 come in (as i am only a novice at this)
    i got 7.5v as well doing it the long way

    also the time constant

    when the switch is closed is the time constant

    tor=L/R now is the R always 200ohm in this equation for each one of the I1,I2 and I3

    as i get it ro be 5mS

    your equation is eg I3=.025x.819 (not sure of the .819).

    when the switch is closed can it be drawn as (5v with two 100ohm resistors in series and the inductor?)

    Please forgive my questioning its just my lack of knowledge in this


    Best Regards Ian
     
  6. May 25, 2009 #5
    I got a different value for the currents at 5 time constants.

    I1=0.0375-0.0125*e^(-5)=0.03741
    I2=0.0125*(1+e^(-5))=0.01258
    I3=0.025*(1-e^(-5))=0.02517

    5 time constants is a long time. So the current values should be very close to the final values after the switch has been shut a long time, when the inductor is acting like a short.
     
  7. May 25, 2009 #6
    I got time constant of 5 ms too. However, you don't really need to know it. In general, the exponential can be written as
    exp(-t/tau)
    where tau is the time constant. You are told that time is 5 time constants or t=5*tau. So the exponential terms are
    exp(-5*tau/tau) = exp(-5).

    Since the I1 and I2 currents can be written in terms of I3 by simple algebra equations, then the same exp(t/tau) will appear in all three current formulas.
     
  8. May 25, 2009 #7
    goto this.
    http://knowhowstuff.blogspot.com/2009/05/tension-problem.html [Broken]

    i updated it


    "when the switch is closed can it be drawn as (5v with two 100ohm resistors in series and the inductor?)"


    initially when time=0 current through inductor is zero.
     
    Last edited by a moderator: May 4, 2017
  9. May 25, 2009 #8
    Thank you both for you swift and knowledgable replies
    that has made a lot of sense now.
    I think it has finally sunk in

    Many Thanks

    ian
     
  10. May 26, 2009 #9
    One last question, part 4 of this question asks to plt all three currents on a time axis wavefrom. starting from 0mS to 80mS however the switch opens at t=50mS.

    As I3 is the current through the inductor i have plotted that fine but as part of the other questions
    we found that when the switch is first closed t=0 i1 and I2 = .025A so when i start plotting my graph will i start from 25mA for i1 and i2 and then use the formula i=I[1-e-rt/L]

    Basically i can do the inductor circuit plotting fine its whats happening in the rest of the circuit from t=0 onwards and t=50 decaying as switch opens



    Many thanks Ian
     
  11. May 26, 2009 #10
    You should find the currents I1 and I2, in terms of I3 using the circuit laws. You already have I3, so plug that into I1 and I2. However, you can cheat some to get the current equations. Solve the currents when the switch just closes and the inductor is open circuit. You already have those currents. Now solve the circuit for when the switch is closed a long time, the inductor is a short. That gives you the starting currents and the ending currents. The plots for each current should go from the initial value to the final value in an exponential fashion in which you already know the time constant of 5 ms.
     
  12. May 26, 2009 #11
    Right i see so my graph will start at 0 from t=0 (as inductor is open circuit)with i3 rising exponentially to 99.3% of max steady state current until switch opens and then decays exponentially.

    And at T=0 (instant switch closes I1=37.5mA and i2 = 12.5mA

    after a steady state and the switch reopens
    I3 will equal I1 + I1 and will be 25mA

    will i1 drop straight to 0A
    and i2 jump instantly to 25mA?

    or do i keep the steady state figure and use the decay exponential equations for them

    Thanks again for your help, i tought i had it sussed just try ing to get my head around what happens in the rest of the circuit.

    Best Regards Ian
     
  13. May 27, 2009 #12
    Yes, for I3 this is what will happen.

    Nope. Those are the final values after the switch has been closed a long time.

    I think you meant I3 = I1 + I2, and not that's not correct. You need to make sure to keep the signs right. Both I2 and I3 are leaving the junction. I1 is entering it. So you have to write it as I1 = I2 + I3 (unless you change the picture). With I1=0, you have I2 = - I3.

    In ideal cuircuit laws, there is nothing to prevent I1 from suddently changing (step function) from one value to another, like to zero. I1 does not pass through any inductors which prevent that. Same for I2. I2 is allowed to suddenly change. I3 can't change suddenly because of the inductor. The I3 current just before the switch is opened and just after have to be the same.

    Answered above. However, don't forget that the change in circuit configuration when the switch is opened will lead to a different time constant when the switch is opened.
     
  14. May 28, 2009 #13
    Thanks for all the help Chris Im starting to understand it now, the only time i get to work through it is when im working nights and my brain usually gets puddled quickly.

    one final question

    as the time constant for the ciruit when the switch is closed is tor=L/R 1/200=.005s

    how do i get the open circuit time constant im not sure what figures to plug in as the switch is open no current will flow through R2 and the only circuit flow will be reverse emf for the 5 times constants as the the inductor decays.


    initially i thought both open circuit and closed had the same value?
    but now i realise fom you help that it probably wont, im not sure how to work the open circuit time constant out

    Sorry if im being simple its just that when i first started working it out and completing it from your replies i went back through my work to check what i had and realised i hadnt initially understood what was being asked in the question.


    Regards Ian
     
  15. May 28, 2009 #14
    That's the answer. No current flow through R2. So R2 can't effect the circuit operation. However, there is a current through both R1 and R3, so ....

    (your have only a single circuit loop with L1, and the resisters in series)
     
  16. May 31, 2009 #15
    Thanks for all your help Chris i think i have it sussed now

    couldnt have done it without you explanations


    Regards Ian
     
  17. Mar 17, 2011 #16
    How did you solve for part c. When the circuit reopened? Specifically for I2.
     
  18. Mar 17, 2011 #17
    Welcome first time poster. Most of the answer has been given above in the discussion. To reiterate, when the switch is opened, no charge can flow in the left hand side of the circuit. There is not a complete path for them to follow. So I1 goes to zero. This means that resister R2 no longer has any effect on the circuit. Suggest you redraw the circuit with just the right hand side since left hand side has no effect after the switch is opened.

    Since what is left is a single loop, I2 has to be the same as I3. No charge can just appear from nowhere at that top node point to make I3 and different from I2. Because the arrows are drawn in opposite directions, I2 will be of opposite sign of I3 but same magnitude.

    Can you solve the circuit that is left, two resisters in series with the inductor? The two things you need to know is 1) This circuit will have a different time constant than part a and b, because R2 not longer matters. 2) Since I3 is going through the inductor it can't change instantaneously after the switch is opened. It has to have the same current as before the switch is opened. That becomes your new initial condition for solving this second circuit.
     
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