Inductor operation confusing

Discussion in 'General Electronics Chat' started by elecidiot, Jun 29, 2010.

  1. elecidiot

    Thread Starter New Member

    Jun 29, 2010
    In the ebook i found an example of inductor regarding the connection of an induction to a circuit having a source of 6 V to light a neon lamp of 70V. When switch is opened the inducot produces a high voltage since current suddenly goes to zero. Thats ok. But when closing the switch also the current should rise instantaneously na? But it has been told in the ebook that the rate of rise of current during the closing of the switch is not enough to generate a high voltage. How is it so?

    Link to the doubtful section of the ebook:
    Last edited: Jun 29, 2010
  2. retched

    AAC Fanatic!

    Dec 5, 2009
    The current "remaining" in the circuit when the switch opens causes the surge or spike.

    When closing the circuit, it draws up slowly. Instead of a large spike, you have large continuous current, once powered up, it equalizes.

    Notice how the lights dim when the A/C kicks on in a house?
    That is the motors coil in the A/C drawing HUGE current from the home supply. The resistance of the circuit changes as the unit energizes. So a slow change, no spike.

    When you turn the unit off, its like driving a speeding car towards a wall.

    You have a lot of energy (momentum) and it cant stop immediately, hence the voltage spike.

    The energy in the coil surges while discharging.

    The larger the voltage change in an inductor, the larger the spike.
    So with the 70v neon light, the inductor is changing from 70v to 0v when the circuit is opened.
    elecidiot likes this.
  3. hgmjr

    Retired Moderator

    Jan 28, 2005
    It is a bit difficult to explain the behaviour of an inductor to a first timer. The article you cited in your initial post is a good introduction. The very first equation in the article is key in describing the behaviour of an "ideal" inductor when a constant "ideal" voltage source is applies across it. The current builds up at a constant rate (di/dt) immediately upon the application of the constant voltage. This is the "dual" of the behaviour of an ideal capacitor when a constant current is applied to it. The only difference is that in the case of a capacitor, it is the voltage across the capacitor that builds up at a constant rate.

    As for the behaviour of an inductor in which a constant current is flowing when the circuit is opened up instantly, the current stored in the inductor cannot change instantaneously. The voltage that is almost instantly developed briefly across the inductor is dictated by the current that was in the inductor times the open-circuited resistance across the inductor (Ohm's Law in action). The reason this voltage can rise to such high values is that the open circuit resistance can be quite high thus the resulting voltage can be quite high as well.

    elecidiot likes this.
  4. elecidiot

    Thread Starter New Member

    Jun 29, 2010
    Now i clearly understand the operation of the inductor. Ur reply was exactly pin point and dispelled my dobuts in a second. I am glad to be in this forum. Again thanks a million
  5. elecidiot

    Thread Starter New Member

    Jun 29, 2010
    Thanks for ur kind reply. U and hgmgr blasted off my doubts. Thank u sooooo much