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An LR circuit contains a battery hooked up to an inductor and a resistor in series. A switch effectively removes the battery from the circuit. The switch is initially closed. Immedeiately after the switch is opened, the voltage at location B is...

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equal to, higher than, or lower than the voltage at location A?
higher than the voltage at location A because inductor opposes the change.

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Other answers:

@ramkrishna is right about this in the most commonest sense. No offence@ramakrishna. But if you look at it from the point of view of pure physics, its wrong. Because an inductor is just a piece of wire. There cannot be a "potential difference" between the ends of a piece of metallic wire, remember!!! If you are confused, I suggest you read one of Prof. Walter Lewin's most excellent lecture supplements here:
inductor acts like a piece of wire when connected to battery. i'm sorry, i didn't make it clear in the drawing, the battery disconnects from the circuit so now the inductor does not act as just a piece of wire. it keeps the current flowing for time t.
i don't understand how to get potential difference without any given values. V_R=-IR and V_L=-LdI/dt
even then it is a piece of wire
but there is potential difference...?
no no... you've got it all wrong. This is all a misconception. that is why I posted the link. Read carefully and find out for yourselves!!!
a current carrying loop has a net magnetic flux. and whenever there is change in magnetic flux the loop will oppose that. (Lenz's law)
all of you who think there is a "potential difference" between the ends of an inductor should seriously watch this video completely: PLEASE FOR THE SAKE OF CORRECT UNDERSTANDING
thank you i will be watching it soon but must study so i have no time right this moment. thx~

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