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RL circuits

  1. Mar 31, 2003 #1
    :smile: hello all
    it is interesting to ask geniuses of physics a small question
    the question says in LR circuits
    when you close the switch the induced emf is a maximum value
    then it starts to decrease
    why does this happen from the physics point of view?
  2. jcsd
  3. Mar 31, 2003 #2
    I am no genius. In fact, I’m not a physics fiend either. You can see you are really going to be out of luck with me,

    Lenz’s law;
    An induced effect is always such as to oppose the cause that produced it.

    The instant the current begins to increase in magnitude, there will be an opposing effect trying to limit the change. This is why the call them ‘chokes’. It takes a period of time (determined by the coil and the circuit resistance) before the inductor stops opposing. The ability of a coil to oppose any change in current is a measure of it self-inductance… blah blah blah.

    Anyway, that’s more of what happens then why it happens, sorry…
  4. Mar 31, 2003 #3
    Better than I can do, Boulderhead. Anyway, I believe the "back EMF" (induced voltage) is produced by the creation of a magnetic field in and around the coil. In an L-R circuit, current eventually dissipates through the resistor, decreasing the B-field and with it the induced voltage. Can anyone remember if there is a characteristic time associated with the voltage decay in an L-R circuit? Maybe L/R?

    My university E-M professor and I had personal disagreements, and it shows.
  5. Mar 31, 2003 #4


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    I believe the question you're asking deals with an inductance, a resistor, switch, and a battery all connected in series. When the switch is closed, current begins to flow.

    The answer, as Loren Booda said, deals with the inductance. When you run current through an inductor, a magnetic field is built up through the coil. The magnetic field stores energy.

    Faraday's law essentially says that a changing magnetic field through a conductor induces a voltage in the conductor. Futhermore, the faster the magnetic field changes, the larger is the induced voltage.

    When the switch is first closed, the magnetic field is changing the fastest. It is built up in an (inverse) exponential fashion -- the field strength rises rapidly at first, and slows down.

    So when the switch is first closed, the inductor's magnetic field is changing very rapidly, so the induced voltage across the coil is very large. As the circuit "settles," the magnetic field increases more slowly, so the induced voltage across the coil decreases.

    After a period of time (which can range from microseconds to seconds, depending on the size of the inductor), the circuit is stable -- the field through the inductor is constant, which means there is zero voltage across it. In a steady-state condition, the inductor acts as a short circuit. It is sometimes said that inductors are shorts to DC.

    - Warren
  6. Apr 1, 2003 #5
    Not quite Warren, The time to fully charge the circuit also depends on the size of the resister or resistance in the circuit, the RL time, resistance X inductance. Also in such a circuit there would always be some current flow until the battery wears down to zero or the switch opened. With current flow there would always be some electromagnetic field about the inductor. Inductors oppose a change in current flow. When the switch is first closed back emf opposes current flow until it stablizes. When the switch is opened and the field collapses it generates emf in the inductor tending to keep the current flowing creating the "inductive kick" always associated with an inductive circuit.
  7. Apr 1, 2003 #6


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    Correct. Perhaps I should have elaborated that point.
    This is also correct. I did not include the opposite scenario in which the switch is opened again (or the battery fails) at some later time, since that was not in the domain of the question. Instead, I just described how the circuit behaved on its way to steady-state.

    - Warren
  8. Apr 1, 2003 #7
    back emf

    Just found out via email the real question was WHY is back emf maximum at first when the switch is closed.
    For any of you who are interested - When current first begins to flow in the coil it creates a rapidly expanding magnetic field in the inductor. It is the moving magnetic lines of force cutting across the conductor that literally generates the the voltage that is opposite in polarity to the applied voltage and called back emf. Once again it it the changing (moving) magnetic field that causes back emf or as the field collapses, the kick, usually much higher in voltage than the applied voltage because once the circuit is open the field collapses much faster than it built up due to counter emf. Once the voltage and current stabalize the magnetic field no longer changes though it still exists and has no more effect on a DC circuit until the switch is opened.
    Referring to another tread that is some what related, capacitance is just the opposite. In the first instant it acts as a short circuit to the applied voltage until it charges up then blocks any DC component of the applied votage but passing AC thus acting as a filter or as an AC coupler.
    Last edited: Apr 1, 2003
  9. Apr 1, 2003 #8
    I still say the characteristic time for an L-R circuit is ~R/L since R/L, not RL, is in units of time.
  10. Apr 2, 2003 #9
    You may be right. It has been so long I would have to look it up. The principle however is the same. It takes time and during the first few circuit time constants a lot is happening.
  11. Apr 2, 2003 #10
    The "time constant", t, for an R-L circuit is in fact:
    t = L/R
    This is a measure of how fast the current builds towards its final value. Specifically at time t after throwing the switch the circuit will have reached (1 - 1/e), or about 63% of it's final current value.
    So that in time 2t, the current will have reached about 86%, at 5t it's about 99%, at 10t about 99.995%.
  12. Apr 2, 2003 #11
    thanks j-Man now I or we don't have to look it up.
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