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One of the statements in my book:- "The magnitude of induced e.m.f in an LR circuit ( A circuit connect to a AC source having a resistor and a pure inductor in series) is more at the break of circuit (when the key is removed) than at the make of circuit (when the key is inserted).

Could someone explain me the reason behind this? I suppose the reason could be related to -π/2 phase difference in the circuit but I'm looking for definite answer.

Thank you.

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  • $\begingroup$ By key, do you mean a telegraph key (which is schematically a switch)'? It's considered good from to include a schematic or other diagram. $\endgroup$
    – Alfred Centauri
    Jun 16, 2018 at 12:07
  • $\begingroup$ I doubt if the phase difference has anything to do with this. Breaking the circuit of a series RL circuit with a DC source with a would produce a (very) large emf too. $\endgroup$
    – Alfred Centauri
    Jun 16, 2018 at 12:11
  • $\begingroup$ Yes, a switch in other words. The phase difference logic was just a guess to be honest. $\endgroup$
    – Dante
    Jun 16, 2018 at 16:31
  • $\begingroup$ "Breaking the circuit of a series RL circuit with a DC source with a would produce a (very) large emf too. " Yes, but how is it greater than the one produced when the switch is switched on? $\endgroup$
    – Dante
    Jun 16, 2018 at 17:06
  • $\begingroup$ user64829, assuming zero initial conditions, the inductor current is zero when the switch is closed and so the initial $\frac{di_L}{dt}$ is just $V_{DC}/L$. But, once there is non-zero inductor current, opening the (ideal) switch abruptly stops the current and then $\frac{di_L}{dt}$ is arbitrarily large. $\endgroup$
    – Alfred Centauri
    Jun 16, 2018 at 21:31

1 Answer 1

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A general idea here is that the emf induced into a circuit from an AC source cannot exceed the amplitude of the source voltage, but the emf generated by an inductor ($emf=-L\frac {di}{dt}$), when the current through the inductor is interrupted, could, theoretically, be very high.

An inductor may generate a high voltage when a circuit is broken, because it "wants" to maintain its current and the only way to make a current flow through a gap in the circuit is to apply a very high voltage to it.

For a given inductor and the type of a switch, the magnitude of the emf will depend on the magnitude of the current at the moment of switching: the higher the current, the greater the emf.

In this particular case, we have an AC voltage source and, therefore, an AC current flowing in the circuit. Therefore, the magnitude of the emf generated by the inductor will depend on the phase of the current at the moment the circuit is broken, e.g., emf will be at its maximum if the current was at its maximum, when the circuit was broken, and it will be close to zero if the current was was close to zero.

To summarize, in theory, the emf associated with breaking an AC circuit with an inductor could be higher than the emf associated making such circuit, but it could happen only if the current at the time of breaking is sufficiently high.

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