Solve for L (inductance)

Discussion in 'General Electronics Chat' started by Overclocked2300, Oct 3, 2005.

  1. Overclocked2300

    Thread Starter Senior Member

    Apr 24, 2005
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    J=1/2L I^2

    Where,

    J=Joules
    L= Inductance
    I=Current through the inductor

    So To solve for L I did:

    J=1/2L I^2
    -I^2 -I^2

    so I^2-J=1/2L

    Then: I^2-J=1/2L
    ----
    1/2

    Finally
    I^2-J=L
    ------
    .5

    I come up with a negative Number...
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Hi Overclocked2300,

    In looking at your problem, it appears as though you subtracted i^2 from both sides of the equation. To isolate the value L you would need to divide both sides of the equation by i^2 and then multiply both sides of the equation by 2.

    Performing these two operations would yield:

    L = (2*J)/(i^2)

    hgmjr
     
  3. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
    OT: In the school we just talked about education in usa. Teacher told us it is not on very high level.
    So: Do you (all) consider this equation and solving L to be complicated, or it´s just a bad example? (I mean no offense :)

    Kubeek
     
  4. Overclocked2300

    Thread Starter Senior Member

    Apr 24, 2005
    124
    0
    It wasnt complicated, I went wrong some where just wanted to know where.

    Now on with inductive reactance.

    (taken directly from the ARRL handbook)

    X=V/I

    V=XI

    I=V/X

    But

    XL=2pi F L

    So, if I knew the voltage (AC) and the current, I could find XL Correct? But would the XL found using V/I be the same as XL =2pi F L?

    So Then once I have XL I could rewrite the formula to find L so that:

    L= XL/2pi F

    EDIT: I have also read on how to measure inductors, but I get weird values. The circuit consists of an unkown inductor in series with a known resistor, R. Voltage is 18vRMS, but the resistor takes up most of the voltage and I measure a very small voltage across the inductor (0.05V). Shouldnt the voltage be greater? Or does it matter on which polaritys I measure the voltage of the inductor with? Btw I used a 1K resistor. Im thinkings of making it smaller.

    Ahem, Isnt that prejedice to say about americans? Granted some people here are idiots, but others are above those idiots
     
  5. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Hi Overclocked2300,

    We are all learning and we all make errors occasionally in the process. The key is to learn from our errors and end up smarter for the experience.

    In your example of the series resistor and inductor in which the resistor has almost all of the voltage appearing across it, for a constant value of R in your circuit the amount of voltage appearing across the inductor is a functioin of the inductance as well as the frequency of your AC input voltage. Since the impedance of the inductor as you have already shown is XL=2 *pi* f *L, you can see that the impedance increases with an increase in frequency. The voltage being dropped across the inductor in your experiment is low because either

    A.) your inductor is low in value
    B.) your frequency is very low

    or

    C.) both A. and B. are true.

    hgmjr
     
  6. Overclocked2300

    Thread Starter Senior Member

    Apr 24, 2005
    124
    0
    I am using 60Hz (US standard) I could of course hook up a rectifier and use 120Hz

    So Could I still use that method of finding the inductance of the coil?
     
  7. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    The method should work fine. The only problem is that the accuracy will be a bit off due to the small voltage drop across the inductor. It might benefit you to decrease the value of your series resistor to get the voltage drop across the inductor to a couple of volts if possible. Of course you may need to use a higher wattage resistor as you decrease the resistor value.

    Good Luck,
    hgmjr.
     
  8. Overclocked2300

    Thread Starter Senior Member

    Apr 24, 2005
    124
    0
    Hmm I have a Vair-AC (variable transformer), but its output is 2.5 Amps, Im guessing I couldnt use that unless I want to be fried...

    I do how ever have 250ohm 5W, and 2Ohm 25 or 50W resistors on a heatsink..

    The resistance (DC) of the coil is .5 Ohms.
     
  9. Overclocked2300

    Thread Starter Senior Member

    Apr 24, 2005
    124
    0
    Hey I just thhough of another way to measure inductance!

    I have a 50W amp hooked up to some speakers, and Ive found a program on the net that will output a sine wave (or triangle, or square wave) at a constant amplitude.

    What if I connect the unknown inductor on the output side of the amp (to act like a passive filter) and then adjust the freqency of the signal generator, and every 100Hz Measure voltage and create a graph?

    It should work since inductors only allow a certain frequency to pass Correct? So all I would have to do is find that F.
     
  10. Firestorm

    Senior Member

    Jan 24, 2005
    353
    0
    Like everywhere in the world, the intelligence of people vary as do their learning capacities and everything else. If you compare the EC kids from 1 area to the validictorians of another area, of course the validictorians will look smarter.

    US schools vary and some states are on a very high level. Look at the colleges such as MIT and Harvard. Your teacher wasn't totally wrong but it all just depends on where you look to withdraw an opinion. Just curious, what country are you from and what level of education do you have? thx l8er

    -fire
     
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