# Why is the magnitude of induced e.m.f more in an LR circuit at the removal of key?

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.

• 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. – Alfred Centauri Jun 16 '18 at 12:07
• 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. – Alfred Centauri Jun 16 '18 at 12:11
• Yes, a switch in other words. The phase difference logic was just a guess to be honest. – Dante Jun 16 '18 at 16:31
• "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? – Dante Jun 16 '18 at 17:06
• 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. – Alfred Centauri Jun 16 '18 at 21:31

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.