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I take a 12 volt battery, take a strong electromagnet, and fire it up.

I then touch the two terminals that are currently connected, one terminal of the battery, and one terminal of the electromagnet.

When I disconnect the terminal of the electromagnet to one terminal of the battery, while still placing my thumb over the terminal of the electromagnet and the battery, I receive a little shock.

I know that the inductor releases a higher voltage when disconnected, but why?

•The magnetic field is induced with an inductor with X amount of turns when the battery is connected.

• The battery is now released, and the magnetic field closes down, and induces a current on the inductor with X amount of turns.

Heres where I am confused- both X's are the same amount of turns, therefore a 1 to 1 ratio. Why are different voltages present with the same voltage source, a 12 volt battery?

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1 Answer 1

Changing currents in an inductor result in changes of magnetic flux in the inductor. Then, since the flux changed, the induced voltage changes. That is why

\$v = L{di \over dt} \$

Thus, when you disconnect the inductor, the current changes from whatever it was to 0 in a very short amount of time. The resulting \${di \over dt} \$ becomes very large making the induced voltage very large.

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