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Why does Back EMF produce so much voltage?

  1. Dec 4, 2015 #1
    I apologize if this question is a bit confusing, but I am studying electrical engineering and what keeps baffling me is why when you run current through an inductor then place an open in the circuit how come the Back EMF current is able to cause a spark across the open. I get that the inductor stores the electric energy in its magnetic field but how does that back current for a brief time constant not care about resistance and create a high potential difference or voltage to keep that current constant for a brief moment in time.
    Last edited: Dec 4, 2015
  2. jcsd
  3. Dec 4, 2015 #2
    Hi electric jake:

    I am not sure which of two questions you are asking:
    1. Why is there a high voltage across the open space in the circuit?
    2. Why does a high voltage cause a spark across an open space?

    1. When a voltage is maintained across a inductance, a constant magnetic field (like an electromagnet) in established and remains until the circuit is opened. When the circuit is opened, the magnetic field collapses, and as the collapsing field lines cross the loops of the coil create a voltage which remains (not necessarily constant) until the field has completely collapsed. The effect is similar to that of an electric engine in which a coil is moved through an electric field.

    2. A sufficiently large voltage across a open air gap can remove electrons from the gas molecules, leaving ionized gas molecules. The positive side of the voltage attracts the electrons, and the negative side attracts the ions. The moving electrons and ions is what you see as a spark.

    Hope this helps.

  4. Dec 4, 2015 #3
    I see so the collapsing magnetic field induces a voltage in the coil when the field lines cross the loops in the wire so then the voltage spikes keeping the back current equal to the current ran through the coil initially for a brief period. That helped immensely and sorry about the the second question was basically restating the first question asking how the inductor causes enough voltage to reach the breakdown voltage of air. Thank you very much.
  5. Dec 4, 2015 #4


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    Just thought it might help to look into Montgolfier’s hydraulic ram. Only an analogy like water in a pipe circuits, but it shows how low pressure energy can generate high pressure pulses.
  6. Dec 4, 2015 #5


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    The equation for the voltage on an inductor is..

    V = L di/dt

    When you open the switch the current goes from some value to zero. So di/dt is very large.
  7. Dec 4, 2015 #6


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    In open circuit, the resistance is very large .So rate of decay of current di/dt is very high, inducing a large back emf Ldi/dt. If you short a fully energized inductor, there will be very slow decaying of current and hence, very small voltage will be induced as di/dt is small.
  8. Dec 5, 2015 #7


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    The current through any resistance will always be V/R. It will not be a constant value here and neither will the induced Voltage. If there is a finite high resistance, rather than an open circuit, the resulting waveform will have a time constant. This link shows how to calculate the time constant and dI/dt will give you the voltage at any time in the process. The total resistance in the circuit (source plus load) is what should be used and the start current is defined by the source volts/source resistance.
    If an arc is formed then all bets are off and a much more complicated analysis is needed. The initial, striking voltage will be set by the separation and shape of the contacts and it will reduce drastically once the arc has struck.
    If the separation / insulation between the ends of the coil is high enough then the self Capacitance of the coil and its inductance can form a resonant circuit and there will be 'ringing', which will gradually decay as energy is dissipated by the coil's resistance (or some other electrical path happens to take the current).
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