1. Oct 30, 2011 #1

   pyroknife

   

   I attached the link to the problem. Shouldn't the Vlmax be 330V? I tried entering that and it wasn't accepting my answer.

    

   pyroknife, Oct 30, 2011
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3. Oct 30, 2011 #2

   sandy.bridge

   

   Where is the link?

    

   sandy.bridge, Oct 30, 2011
4. Oct 30, 2011 #3

   pyroknife

   

   oh oops sorry lol! no wonder why i didn't get any replies.

    

   Attached Files:

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   pyroknife, Oct 30, 2011
5. Oct 30, 2011 #4

   sandy.bridge

   

   When dealing with ac sources you want to be dealing with the rms voltage.
   [tex]V_{peak}=\sqrt{2}V_{rms}[/tex]

    

   sandy.bridge, Oct 30, 2011
6. Oct 30, 2011 #5

   pyroknife

   

   Well thats the 2nd part of part b where it asks for the rms Voltage. The 1st part is asking for the peak. I thought the peak was defined by emf=emfpeak * cos(wt+weird symbol) and in this case wouldn't emf peak just be 330V?

    

   pyroknife, Oct 30, 2011
7. Oct 30, 2011 #6

   gneill

   

   Staff: Mentor

   You have a series circuit (resistor and inductor), so it's a voltage divider. The peak voltage across either component must be less than the voltage supply's peak. You'll have to deal with the reactance of the inductor (or its impedance if you're familiar with that).

    

   gneill, Oct 30, 2011

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