1. Limited time only! Sign up for a free 30min personal 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!

Potential at a point in a circuit

  1. Sep 19, 2013 #1
    ImageUploadedByTapatalk1379582898.613208.jpg

    Why is the potential at X , denoted by V, not

    V/8 = 4.8/(4.8+7.6) ? Why is numerator of resistance part 7.6 and not 4.8 by potential divider principle?
     
  2. jcsd
  3. Sep 19, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi vadevalor! :smile:

    conventionally, the current flows from the positive terminal (long line) to the negative terminal (short line)

    (the electrons, of course go the other way, towards the positive!)

    so the short line is at zero potential, and you have to start from the short line to get to X to find the potential at X, ie the potential difference from zero

    (see eg "Electric Potential Diagrams" at http://www.physicsclassroom.com/class/circuits/u9l1c.cfm)

    ie you go through the 7.6Ω, so 7.6 is in the numerator :wink:
     
  4. Sep 19, 2013 #3
    The values 4.8V and 1.6V shown on the circuit diagram are misleading. Should not the unit be Ω instead of V?

    The equivalent circuit is a voltage supply of 8V connected to 4.8Ω in series with 7.6Ω as shown correctly in your diagram.

    Hence 8V is the voltage on (4.8 + 7.6)Ω.

    And the voltage on 4.8Ω can be found be using direct proportion.
     
  5. Sep 19, 2013 #4
    Thank you all for your inputs! I realised its the direction after thinking for so long. And yea it should be ohm instead of v.

    I have another question regarding potential could you guys help me? Answer is A but i dont understand the solution as there are 'conflicting battery direction' and i dont know how to start

    ImageUploadedByTapatalk1379591023.772679.jpg

    Thanks!!
    - curious lost soul
     
  6. Sep 19, 2013 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi vadevalor! :smile:
    start by replacing the three batteries by one 3V battery …

    what is the potential at Y? and where is that measured from?

    and now adjust that to start from X instead :wink:
     
  7. Sep 19, 2013 #6
    Should i put the 3v battery to the right or left of X and should i go from + terminal to -terminal from X to Y like current flow or electron flow?

    Smiles :D
     
  8. Sep 19, 2013 #7

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi vadevalor! :smile:
    you must always always ALWAYS use current flow not electron flow!

    X is somewhere in the middle of the 3V battery, isn't it? :wink:
     
  9. Sep 19, 2013 #8
    ImageUploadedByTapatalk1379593167.784346.jpg

    Think i got it! Yes! So its 2-1.5= 0.5v following current direction. But when do i know to use current or electron flow direction? Above for potential divider i was told to look at electron flow
     
  10. Sep 19, 2013 #9

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    yes, i like your diagram …

    you've split the batteries into two: a 2V and a 2 minus 1 = 1V battery going the same way,

    so it's easy to see whether to add or subtract :smile:
    i don't understand that :redface:

    it's always current flow​

    what exactly did they say? :confused:
     
  11. Sep 19, 2013 #10
    Its the third paragraph from your first answer :) starting from the short line (electron flow direction)
     
  12. Sep 19, 2013 #11

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    ah, i see, you're interpreting the flow direction as the direction for measuring the potential

    no, the flow direction (= the current direction) is the direction the charge is moving (which we assume is a positive charge, so the opposite of the electron flow direction)

    since the charge moves from high to low potential, the potential decreases as the charge moves, so the potential difference is measured in the opposite direction:

    ie we use a positive charge (ie not an electron), and so it will flow outside the battery from the long line (+ve) to the short line (-ve), and the potential will be highest at the long line, and decrease to zero at the short line :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted