1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

RL Parallel Circuit

  1. Feb 27, 2014 #1
    ex6.jpg

    1. The problem statement, all variables and given/known data

    Find the open circuit voltage VL(0) across the 5ohm resistor



    2. Relevant equations

    Z1*Z2/Z1+Z2

    V(Z2*Z1+Z2)



    3. The attempt at a solution

    I thought for the open circuit voltage that due to infinite resistance of OC that the voltage across j35 inductor would be the equivalent of VL(0).

    So i'd use the voltage divider equation across the 2 inductors in the left hand loop.

    j35/(j35+j0.3)

    Solving this using by multiplying the numerator and denominator by the conjugate of j35+j0.3

    j35/(j35+j0.3)

    Am I on the right track?

    Thanks
     
    Last edited: Feb 27, 2014
  2. jcsd
  3. Feb 27, 2014 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    If you have interpreted the question correctly, then you have started it the right way.

    But I'm not convinced that the question is "Find the open circuit voltage VL(0) across the 5ohm resistor". Is that your wording?

    Can you quote the instructions exactly?
     
  4. Feb 27, 2014 #3

    gneill

    User Avatar

    Staff: Mentor

    Yup. Should be fine so long as the inductors are not coupled by mutual inductance.
     
  5. Feb 27, 2014 #4
    Sorry, it's two seperate questions, first find the open circuit voltage VL(0) where there's an OC instead of the resistor.Then secondly find the voltage over the resistor.

    Sorry, misquoted the potential divider equation, it should have been

    V(Z2/Z1+Z2)
     
  6. Feb 27, 2014 #5
    Am I right in thinking that as it's just 2 inductors then the voltage across will be in phase with each other, hence

    v*(j35/j35+j0.3) the j will cancel leaving 100*(35/35.3) =99.2v? For the OC voltage?

    Thanks
     
  7. Feb 27, 2014 #6

    NascentOxygen

    User Avatar

    Staff: Mentor

    okay

    I'd not hesitate to mark that wrong.
     
  8. Feb 27, 2014 #7
    Sorry I'll try again

    VL(0)=V*(Z2/(Z1+Z2))

    =100*(j35/(j35+j0.3))

    =100*(35/35.3)

    VL(0)=99.15v
     
  9. Feb 27, 2014 #8

    NascentOxygen

    User Avatar

    Staff: Mentor

    Looks okay. That's the easy part out of the way ... :smile:
     
  10. Feb 27, 2014 #9
    Thanks, so for VL would loop analysis work? Or should I be looking at using Z1*Z2/(Z1+Z2) to simplify the circuit?

    Thanks
     
  11. Feb 27, 2014 #10

    NascentOxygen

    User Avatar

    Staff: Mentor

    Any valid method will work. Choose one, or be daring and try a few different ways!
     
  12. Feb 27, 2014 #11
    The aim was to get us practising with complex impedences.So my plan is to find overall impedence using

    Z1*Z2/Z1+Z2

    then add this to the 0.3 inductor closest to the voltage supply.This gave me 1.28j+4.82 which gives me the wrong VL compared to the answer provided (98.45v)
     
  13. Feb 27, 2014 #12

    NascentOxygen

    User Avatar

    Staff: Mentor

    There seems to be a problem with the parentheses keys on your keyboard....?

    You won't get the answer in one step. There are a couple of 'potential divider' steps to consider when determining VL.

    If you don't provide your working, it is difficult to say whether you are going about it the right way or not.
     
  14. Feb 28, 2014 #13
    Sorry again :)

    Thanks, I think I understand now, it's just the complex arithmitic I think I slipped up on.It'll be a struggle to type it all up on here, so I'll stick with it and look for a solution.

    Thanks again for your assistance.
     
  15. Feb 28, 2014 #14

    NascentOxygen

    User Avatar

    Staff: Mentor

    You can scan a neatly handwritten page and attach the jpeg to your post here using the paperclip icon.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: RL Parallel Circuit
  1. RL circuit (Replies: 7)

Loading...