Kirchhoff's loop rule problem

  • Thread starter ehrenfest
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  • #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.

Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #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?
 
Last edited:
  • #4
malawi_glenn
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please show YOUR attempt to solve this first.

explain YOUR thoughts carefully, then someone might help you.
 
  • #5
ehrenfest
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say we go clockwise around the smaller loop starting from S and record the voltage changes as we go around. How do we get the sign of the first voltage change?
 
  • #6
Astronuc
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The battery establishes an initial potential. When the switch has been set for a long time, it means that it has reached a steady-state situation, i.e. the current is constant, which means di(t)/dt = ?

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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