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Transient analysis.

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data
    In the given circuit (voltage source is V) , we have to find initial and final values of current and also di(0+)/dt , di2)0+)/dt , d2i1(0+)/dt2 .
    Now , by intial and values of currents , does they mean i(0-) and i(0+) or the discharging or charging current??
    nd how to form the equations to solve it out??


    2. Relevant equations


    3. The attempt at a solution
    beforing closing the switch , the current across the inductor , say i(0-) = (R1+R2)i2 = i(0+)
    when we close the switch , capacitor will be shorted , and inductor will be open-circuited , so the KVL equation in the first loop will be V-i1R=0 by differentiating we get di1(0+)/dt = 0 similarly , d2i1(0+)/dt2 =0
    how we will solve for di2(0+)/dt and d2i2(0+)/dt2 then..??
     

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    Last edited by a moderator: Sep 28, 2014
  2. jcsd
  3. Sep 28, 2014 #2

    NascentOxygen

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    You don't differentiate your expression for V(0+) to get dV/dt (0+). There is no time information in the expression. You have to arrive at the derivatives by a different technique.
     
    Last edited: Sep 28, 2014
  4. Sep 28, 2014 #3
    Which technique..?? Please elaborate...
     
  5. Sep 28, 2014 #4

    NascentOxygen

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    How has this been done in worked examples in your class notes?
     
  6. Sep 29, 2014 #5
    If you are asking about class..please don't ask because there's nothing productive being taught..I am studying this topic from book..& in the book they are just using KVL.. & then differentiating..and using the corresponding conditions for inductor and capacitor.. but that is for simple circuits.. but I am not getting how to solve such circuits which I'hv attached..!!
     
  7. Sep 29, 2014 #6

    NascentOxygen

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    You can look at it like this....

    Immediately after the switch is closed, with as yet no current through the inductor and its branch, all current from the source is taking the path through the capacitor and R1. With this current charging the capacitor, you can determine what that capacitor's dV/dt would be with such a level of current.

    Bear in mind that in another circuit, for example, if there was initially current through the inductor before the switch is closed, then that same level of inductor current will continue at t=0+ so that needs to fit in with your concept of the inductor and it being "open circuited".
     
    Last edited: Sep 29, 2014
  8. Sep 30, 2014 #7
    Actually I worked on this problem and got the correct values... and worked on some more problems..and in of the problem..I just wanted to make it confirm whether my answer is correct or not..Can you please help out in getting me confirmed about it..so that I can move further..
    In the circuit (voltage source V0 and current i is flowing across capcitor) ) , we have to find i(0+) , di(0+)/dt , d2i(0+)/dt2.
    I got i(0+) =V0/2 , di(0+)/dt = 2V0 & d2i(0+)/dt2 = -3V0/2
     

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  9. Sep 30, 2014 #8

    NascentOxygen

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    Explain your thinking that led you to decide I(0+) will be Vo/2 .

    Connection to the source doesn't occur until t=0. Is that what we are told?
     
    Last edited: Sep 30, 2014
  10. Sep 30, 2014 #9

    NascentOxygen

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    What were your answers for the first circuit?
     
  11. Sep 30, 2014 #10
    for the first circuit I got di1 (0+)/dt = V/L-V/R12C and di2(0+)/dt=V/L..
    And in the second circuit....actually I did a mistake the current i(0+) is coming V0/L & I used the conditions at t0+ , inductor open & capacitor short-circuit,, so the current in 1st loop will be zero and in 2nd loop is also zero..
    then I wrote the KVL equations in the 2 loops & since these equ. hold in general so they'll hold for t=0+ also.. so at t=0+ I put the values and found it out..
    Should I write the equ. also??
     
  12. Sep 30, 2014 #11

    NascentOxygen

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    You need a pair of these ( ) around a denominator in that expression.

    That's di/dt(0+) = V0/L
     
    Last edited: Sep 30, 2014
  13. Sep 30, 2014 #12
    I am really sorry , I am doing blunders in typing , Yes it is di(0+)/dt = V0/L
     
  14. Sep 30, 2014 #13
    so Is that correct..??
     
  15. Oct 2, 2014 #14

    NascentOxygen

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    Even if there was no inductor, the capacitor current at t=0+ could not be as high as V0/2
     
  16. Oct 2, 2014 #15
    I guess You did'nt see the earlier posts.. I said that I did mistake in finding the currents in the 2 loops.. I am getting i(0+) = 0 & di(0+)/dt = V0/L & not V0/2...
     
  17. Oct 2, 2014 #16

    NascentOxygen

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    V0/L is di/dt for inductor current. This then halves as it divides equally between the two subsequent branches. I'd like to see how you are working these out.

    You are not attempting the second derivatives, d²i / dt² ?
     
  18. Oct 2, 2014 #17
    previously I'hv done many mistakes in posting my solutions.. and evverything got messed up ..Now I explain the whole thing...
    first of all at t=0+ , the inductor is open ckt & capacitor is shorted..so the current in 1st loop will be zero and in 2nd loop is also zero..
    then I wrote the KVL equations in the 2 loops & since these equ. hold in general so they'll hold for t=0+ also.. so at t=0+
    For 1st loop KVL equ. : V0-2i1R+iR -Ldi1/dt = 0 (I'hv assumed current in 1st loop as i1 )
    for t=0+ , V0-2i1(0+)R +i(0+)R=Ldi1(0+)/dt ______(1)
    di1(0+)/dt = V0 (R=1ohm & L=1henry )
    Again for 2nd loop KVL equ. : -2Rdi/dt -i/C +Rdi1/dt=0
    for t=0+ , -i(0+)/C+di1(0+)/dt =2di(0+)/dt _____(2)
    di(0+)/dt = V0/2
    for d2i(0+)/dt2 , I differentiated equ.2 & 1 once again...
    got d2i1(0+)/dt2 = -3V0/2 & d2i(0+) /dt2 = -V0
    Now You check what is wrong..??
     
  19. Oct 2, 2014 #18

    NascentOxygen

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    Those are the values I calculate, too. http://imageshack.com/a/img29/6853/xn4n.gif [Broken]
     
    Last edited by a moderator: May 7, 2017
  20. Oct 2, 2014 #19
    so now can we conclude that it is right..??? :rolleyes:
     
  21. Oct 2, 2014 #20

    NascentOxygen

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    Within a 98% confidence interval.
     
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