inductors

Discussion in 'Homework Help' started by lemon, Feb 8, 2010.

  1. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    A circuit comprises an inductor of 700mH of negligible resistance connected in series with a 15Ω resistor and a 24V d.c. source. Calculate:

    a) i) the initial rate of change of current ii)the time constant (T) for the circuit
    b) the value of the current after 1T seconds, and ii) the time to develop maximum current.


    The attempt at a solution
    The inductor has negligible resistance so we can use V=IR to calculate the current through the circuit = 24/15 = 1.6A
    V-iR/L = 24-(1.6x15)/(700x10-3)=0

    So, I got zero. So, I got nothing, right?
    What am I doing wrong?
    I suck at this! :confused:
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
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    Write down your try step by step and indicate for which question is each part.
     
  3. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    I could't work out in the editor how to show something raised to a power. Hope you can understand this easily.

    a)i) So, the initial rate of change of current is when di/dt=V/L
    Which is 24V/(700x10^-3)=34.3A/s (3s.f.)

    ii) the time constant (T) for the circuit
    This is not given in the same way as the time constant for a series connected CR.
    It is, instead, circuit T=L/R
    Therefore T=(700x10^-3)/15Ω=0.0467s or 0.05s (2s.f.)

    b) the value of the current after 1T seconds
    Well there are two equations here - one for growth of current and one for decay.
    I think it is growth because the question says 'initial rate'
    Growth=I(1-e(-t/T))=34.3(1-e^(-0.05/0.05))=21.68A (2s.f.)
    But - Currents will be approximately 63% or their final values after 1T when growing.
    Imax=V/R=24/15=1.6A
    So, at 1T, I=0.63x1.6=1.008A
    These figures don't agree

    ii) the time to develop maximum current. When t≥5T.
    Imax=V/R=24/15=1.6A
     
  4. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,535
    Another way is the same way calculators do it, 2.2E4 is 2.2X10^4.

    There is another way, I'm trying to figure it out, using LaTeX.

    2.2X10^4

    Use {tex][/tex}with the number between them, and brackets [] around tex and /tex.

    A how to guide for math on this forum...

    http://forum.allaboutcircuits.com/showthread.php?t=21380
     
    Last edited: Feb 13, 2010
  5. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    You have to put I=Imax in the equation and not the rate of change.
     
  6. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    To find the time rearrange i=Imax(1-e(-t/T))

    as to get t=-T*ln(-i/Imax+1)

    when i=Imax, ln(-1+1)=ln(0)=infinity

    Thus, the time to reach Imax is infinite. This the reason it is said that Imax is reached after 5T.
     
  7. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    t=-T*ln(-i/Imax+1)

    Well. I had two values of i (not sure which one was correct)

    t=-0.05*ln(-1.008/1.6+1)=i get mathematical error

    or

    t=-0.05*ln(-21.68/1.6+1)=i get mathematical error
     
  8. mik3

    Senior Member

    Feb 4, 2008
    4,846
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    In the first case you shouldn't get a mathematical error however the value for i is wrong. It should be i=Imax.

    In the second case you will get a mathematical error because you have a negative number in the ln() which is not allowed.
     
  9. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    if i=Imax then i=1.6A

    t=-0.05*ln(-1.6/1.6+1)

    but i'm confused cause they both have a minus sign in the ln(-1.008) and ln(-21.68)

    but which one of these values is also correct from the earlier workings?

    b) the value of the current after 1T seconds
    Well there are two equations here - one for growth of current and one for decay.
    I think it is growth because the question says 'initial rate'
    Growth=I(1-e(-t/T))=34.3(1-e^(-0.05/0.05))=21.68A (2s.f.)
    But - Currents will be approximately 63% or their final values after 1T when growing.
    Imax=V/R=24/15=1.6A
    So, at 1T, I=0.63x1.6=1.008A
    These figures don't agree
     
  10. mik3

    Senior Member

    Feb 4, 2008
    4,846
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    ln(-1.008/1.6+1)=ln(0.37)

    ln(-21.68/1.6+1)=ln(-12.55) this is wrong because 21.68 is wrong

    The correct value of current after T is 1.008.
     
  11. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    t=-0.05*ln(1.008/1.6+1)=0.05 (2d.p.)

    I should not put the minus sign for i=Imax?
     
  12. mik3

    Senior Member

    Feb 4, 2008
    4,846
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    You have to put the minus for i unless you will get a negative time value which is not valid.
     
  13. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    ln(-1.008/1.6+1)=ln(0.37)

    but -1.008/2.6=-0.387...

    and you don't have a minus, ln(0.37)
    :confused:
     
  14. mik3

    Senior Member

    Feb 4, 2008
    4,846
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    First, you have to make the division and then the addition, simple maths.

    It is ln[(-i/Imax)+1]
     
  15. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    right. -1.008/1.6 + 1=0.37

    ln(0.37)=-0.9942....
    -1s (2s.f.):)
     
  16. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    hold on!
    Your earlier post:

    To find the time rearrange i=Imax(1-e(-t/T))

    as to get t=-T*ln(-i/Imax+1)

    when i=Imax, ln(-1+1)=ln(0)=infinity

    Thus, the time to reach Imax is infinite. This the reason it is said that Imax is reached after 5T.

    So, the answer to this part of the question is infinity?
    It can't be -1s. It doesn't make any sense for the current to be maximum at -1s, the circuit isn't connected before time t=0.
     
  17. mik3

    Senior Member

    Feb 4, 2008
    4,846
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    If you calculate the time mathematically, then (t) is infinity. However, if you assume that (i) reaches Imax after 5T, then t=5*0.05=0.25.
     
  18. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    ahh! I see. So the answer to my question is:
    (i) reaches Imax after 5T, then t=5*0.05=0.25.

    But, you have been showing me how to calculate the time mathematically, then (t) is infinity?

    But why is my answer -1s?
     
  19. mik3

    Senior Member

    Feb 4, 2008
    4,846
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    Yes, the exact time is infinity however we assume that after 5T it reaches Imax because (i) is very close to Imax.

    How you get -1?
     
  20. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    How you get -1?

    It's a good question. I must have eaten a strange fruit this morning. I will use 0.25s then. Once again mik3, thanks for all your great support today. It is very appreciated.
     
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