- #1

ehrenfest

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## Homework Statement

Read the image. I was always confused about this. Is Kirchhoff's loop rule here:

[tex]-L\frac{dI}{dt}+IR = 0 [/tex]

or

[tex] L\frac{dI}{dt}+IR = 0 [/tex]

. Please explain your answer carefully.

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- Thread starter ehrenfest
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- #1

ehrenfest

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Read the image. I was always confused about this. Is Kirchhoff's loop rule here:

[tex]-L\frac{dI}{dt}+IR = 0 [/tex]

or

[tex] L\frac{dI}{dt}+IR = 0 [/tex]

. Please explain your answer carefully.

- #2

Astronuc

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http://farside.ph.utexas.edu/teaching/316/lectures/node104.html

Also,

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indcur.html - look at the polarity on the inductor.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indtra.html#c1

- #3

ehrenfest

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Well what they do in the second link is use the given fact that the current is increasing to decide what the sign must be. But the same equation should hold whether the current is increasing or decreasing. How do you decide which of my equations to use without knowing whether the current is increasing or decreasing through the inductor?

BTW, the do we know the current is decreasing here? Why?

I don't mean to be rude, but can you help me with this specific example?

BTW, the do we know the current is decreasing here? Why?

I don't mean to be rude, but can you help me with this specific example?

Last edited:

- #4

malawi_glenn

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explain YOUR thoughts carefully, then someone might help you.

- #5

ehrenfest

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- #6

Astronuc

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However there is a current and stored energy in the inductor. There is also a voltage drop across the resistor, and only across the inductor when current is changing.

If the battery is by passed, i.e. short-circuited, what then happens to the current?

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