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!

Potential at point in circuit

  1. Jun 4, 2015 #1
    • Warned: URL/Image only problem statement
  2. jcsd
  3. Jun 4, 2015 #2

    cnh1995

    User Avatar
    Homework Helper

    Potential is assumed w.r.t some reference called 'ground' which is assumed to be at 0V. In your circuit, point x is at 0V and not the -ve terminal of the battery. So, if you say A is at 10V, that means -ve terminal of the battery is earthed (which is not). So, taking x as reference, A is at 7.5 V and the potential distribution from A to B is not 10V to 0V ..its actually 7.5V to -2.5V..
     
    Last edited: Jun 4, 2015
  4. Jun 4, 2015 #3
    Can you please tell me how to calculate that? Even a hint would do. Thanks
     
  5. Jun 4, 2015 #4

    cnh1995

    User Avatar
    Homework Helper

    • member has been warned that providing complete or almost complete solutions is forbidden at Physics Forums
    [ mentor note: content deleted ]
     
    Last edited by a moderator: Jun 4, 2015
  6. Jun 4, 2015 #5
    Why is placing the ground between the third and fourth capacitor any different than simply removing the fourth capacitor and unearthing the point X?
     
  7. Jun 4, 2015 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    @Kinta :

    Is this a question to help the OP, or are you asking this because you don't know the answer?

    IMG_20150604_140746.jpg
     
    Last edited: Jun 4, 2015
  8. Jun 4, 2015 #7
    I'm asking because I don't know. I should know, but I don't. I figured if I had difficulty understanding this concept, someone else (maybe even OP) might have similar troubles.
     
  9. Jun 4, 2015 #8

    cnh1995

    User Avatar
    Homework Helper

    My apologies..
     
  10. Jun 4, 2015 #9

    cnh1995

    User Avatar
    Homework Helper

    I think it will depend on where you assume the ground (0V) after X is unearthed. But I'm not getting your idea of removing the fourth capacitor..
     
  11. Jun 4, 2015 #10

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    If you simply remove the fourth capacitor, then the 10 Volt potential difference is split across only three capacitors instead of four.
     
  12. Jun 4, 2015 #11
    I understand that. What I don't understand is if this is any different than placing the ground at the point X in OP's picture.
     
  13. Jun 4, 2015 #12

    cnh1995

    User Avatar
    Homework Helper

    If you remove the 4th capacitor, automatically point x (ground) will be connected to the -ve terminal of the battery. So, A will now be at 10V instead of 7.5V..
     
  14. Jun 4, 2015 #13
    Once the ground is placed at point X, why does it matter that there is a capacitor between it and the -ve terminal? Aren't they both grounded?
     
  15. Jun 4, 2015 #14

    cnh1995

    User Avatar
    Homework Helper

    No they aren't..Ground is just a reference point..It can be anywhere in the circuit.. Battery neither knows nor cares where you place the ground. It will just establish a 10V pd between A and B. How to "see" the 10V path is decided by the ground. If ground is at B, then it will be
    10V(point A)---7.5V---5V---2.5V(point X)---0V(point B). In the problem, X is assumed to be 0V. So, the 10V path is
    7.5V(A)-5V-2.5V-0V(X)- -2.5V(B)..
     
  16. Jun 4, 2015 #15
    Ok. I think I've got it now. My confusion was stemming from a misunderstanding of what it means to have a grounded point in a circuit. Thanks. :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted