How to calculate Inductive Impedance???

Discussion in 'Homework Help' started by JDR04, Nov 28, 2012.

  1. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
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    Hello everyone. Could somebody please help me on how to calculate inductive impedance. I've attached my calculations for your perusal.
    I've taken the calculation as far as I can but have gone wrong somewhere and cannot establish where.

    My tutor says my answer should be in the megaohm region but I cannot for the life of me figure that out.

    Your help will be greatly appreciated, as always, thanks- JDR04
     
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  2. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    The imaginary part will be equivalent to iωL, where ω is equal to 2πf (2*π*frequency) and L is the inductance in Henrys. The i is the notation for the imaginary number. The real part will be any resistance in the circuit (I can't view the document at the moment, but I'm hoping to cover all the bases here).

    I hope this helps!
    Regards,
    Matt
     
  3. bonshie

    New Member

    Nov 21, 2012
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    0
    Hi JDR04 I had a look with your calculation.And you got the correct formula for inductive impedance calculations. I tried to calculate it bymyself and I got magAohm answer on it.Just do your calculation again and be careful with your calculator on putting those numbers.
    Cheers!
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    (7.29E12 + 3.549456E10) does not equal 5.802295954E10

    Furthermore, sqrt(5.802295954E10) does not equal 6.576.569046

    You've got some plain old arithmetic errors.
     
  5. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
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    Thanks Bonshie, I've never done this before so I'm kind of struggling with it. I see what you mean by being extra careful when using the calculator.

    Check my new calculations on The Electrician post. Comments are always appreciated. Thanks again JDR04
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    Youa've got one practice that you need to stop.

    You are expressing 7.29x10^12 as 7.29^12. This is wrong. This says 7.29 multiplied by itself 12 times (or 11, depending on how you count).

    For instance, you have 7.29^12 + 188400^2. How is someone supposed to know that in the first term you mean 7.29E12 but in the second you really do mean the square of 188400 and not 188400E2?

    Next, you need to work on your estimating ability so that you can ask if the answer makes sense.

    If I take sqrt(1000^2 + 10^2), I know that the result HAS to be greater than the larger of the two values, hence the answer HAS to be greater than 1000. Furthermore, if one value is significantly greater than the other value, I know that the answer is going to be very close to the larger value.

    So when you have

    Z = sqrt( (2.7E6)^2 + (0.1884E6)^2)

    You know the answer has to be greater than 2.7Mohm and that, in fact, it will be very close to this value.
     
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  7. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Thanks The Electrician, your help is appreciated.

    I've recalculated what you pointed out was wrong. Would appreciate your comments.

    7.29E12 + 3.549456E10 = 2.252871695E10

    so, square root of 22528716950.0 = 150095.6926

    Am I getting any better? Where do I go from here? Thanks a lot -JDR04
     
  8. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
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    Hi WBahn aned thanks for the help.

    As I mentioned before I've never done these equations before in my life so it's a steep learning curve for me. At times frustrating but very worthwhile in the end.

    I'll go through what you have pointed out in the morning and digest it...slowly.

    Really appreciate your input.....thanks again - JDR04
     
  9. JDR04

    Thread Starter Active Member

    May 5, 2011
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    Thanks DerStom8 (Matt), do appreciate your input but its way,way above me at the moment.

    I'm sure it will make good reading for the more experienced members of this forum. Take care -JDR04
     
  10. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Allow me to clarify.

    An impedance has both a real part (resistance) and an imaginary part (reactance). You can have either inductive or capacitive reactance.

    An inductive reactance is shown by a positive iX, where iX is the reactance.

    Capacitive reactance is shown by a negative iX, where iX is the reactance.

    An impedance is made up of two parts--the resistance and the reactance. The basic formula for impedance is

    z = R + iX

    Where 'z' is the impedance, R is the resistance in the circuit, and iX is the reactance. If you see the formula with R - iX, it means the reactance is more capacitive than inductive. If you see it with R + iX, it means it's more inductive than capacitive.

    If you don't understand this right away (and I wouldn't blame you if you didn't :p) then try reading it again. Eventually you will understand.

    From what I've seen though, most of your problem was just arithmetic and notation, so I'll just leave this here for future reference.

    Good luck,
    Matt
     
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  11. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    No.

    Write your numbers without using scientific notation; do you know what scientific notation is?

    Here's your addition without using powers of 10:

    Code ( (Unknown Language)):
    1. 7290000000000
    2.   35494560000 +
    3. -------------------
    4.  
    Carry out the addition and tell what you get.
     
  12. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
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    No your not getting better. Your making basic errors.

    3.549456E10 equals .03549456E12

    Now add the two numbers and continue on with your calculations.

    7.29E12 + .03549456E12

    You need to exercise due dilligence when working with the values.
     
  13. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
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    Hi The Electrician, thanks again.

    I get 7.32549456E12 which I figure to be 7325494560000

    Am I improving yet??
     
  14. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    That is correct. Are you using a scientific calculator? Your calculator should have done that addition correctly assuming you entered the numbers correctly. What do you get when you take the square root?

    I think you need some practice working with numbers in scientific notation.
     
  15. bonshie

    New Member

    Nov 21, 2012
    4
    0
    Hi JDR04, I had a look on the electrician post.And just like what I have said just be careful on using the scientific calculator.Try to use the ( open & close ) parenthesis when you add value or numbers like this.
     
  16. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,805
    I can't figure out how you are possibly getting this result. You are adding a huge number to a very large number and getting a result that is smaller than either of the two numbers you are adding together. Does that make sense?

    7.29E12 = 7.29 x 10^12 = 729 x 10^10
    3.549456E10 = 3.549 x 10^10

    729 x 10^10 + 3.549 x 10^10
    = (729 + 3.539) x 10^10
     
  17. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
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    Thanks The Electrician, if I square root 7325494560000 I get 2706565.085.

    So that will be 2.7M That looks about right ?? I hope.

    Thanks JDR04
     
  18. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    It's important that you remember that the exponents MUST be the same before you add the main number together. I think that's part of your problem. In simple terms, let's say you're adding 4.5*10^4 (4.5E4) to 3.2*10^5 (3.2E5). You MUST make sure the E#s are the same before adding 4.5 and 3.2. Therefore, 3.2E5 ==> 32E4, so instead of adding 4.5 + 3.2, you now have to add 4.5 + 32, and the answer is in terms of E4.
     
  19. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Thanks again WBahn, your points are valid, I'm getting there.

    The way you laid out the calculations has helped me a lot. Your comment on my ability to estimate is correct and I am going to work on it. Thanks for that.

    I got 5 questions out of 19 wrong, all becauase of my poor maths understanding. I've been going through the others as I learn from you guys on this one and hopefully I'll get to grips with them.

    I'll post them later today (now 06H00, where I am), been up the whole night working on these questions.

    I appreciate the time and effort you give me on my studies. JDR04
     
  20. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Hi DerStrom8, thanks for your input as well. I NOW see what you mean when I look at WBahn's post 1907.

    Thanks a lot -JDR04
     
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