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Applying Kirchoff's Law in an R-L circuit with parallel resistors

  • Thread starter amolv06
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http://img296.imageshack.us/img296/6961/circuithk7.jpg [Broken]

Assume that the switch has been closed a long time.

1.) Find the current through the inductor before the switch is opened.
2.) Find the voltage in the right resistor before the switch is opened.
3.) Find the current in the right resistor right after the switch is closed.

My question here is how do you treat both resistors? My hunch tells me that the right resistor is irrelevant, and that loop should be treated as superfluous, but I'm not sure. If anyone could answer that question, I would greatly appreciate it. Thanks.
 
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Answers and Replies

454
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I think 3.) should end in "After the switch is opened".

For a DC current, an inductor is just a wire. This makes the exact value of the right resistor indeed irrelevant.

for 3.) I can't say that the right loop is irrelevant. What happens to the current in the inductor right after you close the switch? where will this current now go?
 
46
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Thanks, I completely mistyped that. 3 should read opened rather than closed.

Just to clarify, the answer to #1 then should look like the following?

[tex] \frac{V}{R} = I[/tex] where I is 55/150?
 
Last edited:
46
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Thanks, I completely mistyped that. 3 should read opened rather than closed.

Just to clarify, the answer to #1 should be:

[tex]I=V/R[/tex] where I = 55/150?
 
454
0
Thanks, I completely mistyped that. 3 should read opened rather than closed.

Just to clarify, the answer to #1 should be:

[tex]I=V/R[/tex] where I = 55/150?
Yes, that is the answer
 

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