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!

KVL easy question

  1. Sep 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Use voltage divider rule and KVL to find V1 and V2

    2. Relevant equations
    KVL, v1+v2+..=0
    V0*r1/(r2+r1)=V



    3. The attempt at a solution
    So I know how to use the voltage divider rule, my question is about the KVL of the loop below.

    I'm wondering about the 8 V source inside of the smaller loop, is the result of the voltage drop across the resistors, or if it's another voltage source entirely? Can someone explain in plain English what is happening as the voltage goes across each resistor? Here is what I think happens:
    As far as I know the 16 V source is sending voltage / current to the first resistor, which causes a voltage drop of 8 V. The 8 V is then sent to the other 3 remaining resistors, which have a voltage drop of 4.57 V.
     

    Attached Files:

    • kvl.jpg
      kvl.jpg
      File size:
      20.7 KB
      Views:
      72
  2. jcsd
  3. Sep 13, 2014 #2

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    I agree that it is not clear but it doesn't matter for this problem. It doesn't matter if the thing on the right is an 8V source or just a voltmeter indicating 8V. Either way node d is at 8V w.r.t nodes b/c.

    That's not the way I would describe it. It begs the question... How does the 16V source "know" how much current to send to produce an 8V drop across R?

    I would look at it like this...

    Node a is at 16V w.r.t nodes b/c.
    Node d is at 8V w.r.t node b/c.
    Applying KVL to the loop a->d->b/c->a shows that the voltage drop across R (eg V1) must be 8V.

    Just for info.. The current that flows from the 16V source through R is unknown because R is unknown. There is no way to calculate it from the info given.

    Then for V2...

    Node d is at 8V w.r.t node b/c regardless of how much current is flowing from the 16V source. So you can ignore/delete the 16V source and R from the drawing for this step. Just apply the potential divider rule to the 1,2 and 4 ohm resistor as if the thing on the right was an 8V source. Your answer V2 = 4.57V is correct.
     
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: KVL easy question
  1. Easy Question? (Replies: 2)

  2. An easy question (Replies: 0)

  3. KVL Question (Replies: 2)

Loading...