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A demonstration of inductive reactance

  1. Aug 27, 2010 #1
    I'm conducting a seminar at college on application of AC to passive components.
    To Demonstrate what inductive reactance is , i want to wind a coil , connect it to a DC source , find the resistance. Then i want to directly connect it the AC mains and find th effective resistance again and show that there is an extra reactive component that is preventing the AC mains from being shorted.

    Can you tell around how many turns and of what wire i will have to wind to get the necessary inductance so that the circuit breaker does not trip?
  2. jcsd
  3. Aug 27, 2010 #2
    It would help if you gave some figures.
    If you have 250V mains at 50Hz to get a current of about 5 Amps requires something around 300mH of inductance.

    If we assume an air coil of say 6 inch diameter, something around 1500 turns will give you the necessary inductance.

    That's quite a lot. If you use an iron or ferrite core it would be easy but the calculation then very strongly depends on the material. That would reduce the number of turns by a factor of a couple of hundred. - 10-20turns.

    The wire is pretty irrelevant - use copper, as thick as you can get it, to reduce resistance and heating losses.
  4. Aug 27, 2010 #3

    Andrew Mason

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    Not a good idea. You will have to put a load in line with the inductor - e.g. a light bulb. Using 110 VDC the light bulb in series with the inductor will give more light than 110 VAC in the same configuration. The current will be I = V/Z where:

    [tex]Z = \sqrt{R^2 + (2\pi f L)^2}[/tex]

    f is the frequency of the AC (0 for DC). R is the resistance in the circuit. L is determined by a number of factors: number of turns, diameter of coil, permeability of the core, length of coil

    An inductor of .5 H in series with a 100 watt light bulb double the impedance when using 60Hz AC (ie. the DC current will be double the AC current).

  5. Aug 27, 2010 #4
    Use the primary of a regular power transformer?
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