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Finding potential difference between two points in a ladder circuit

  1. Jan 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the potential difference ##V_A– V_B## for the circuit shown in the figure. Capture.PNG

    2. Relevant equations
    Kirchhoff's laws.
    3. The attempt at a solution
    If we assign potentials to junctions starting by setting ##A=0## Current ##i_1## flow through each vertical resistances and ##i_2 ## through slant resistances. B has potential ##i_1+2## and there exist a relation ## 1+i_2+i_1=0## between ##i_1## and ##i_2##. I need one more equation. In which loop should I apply KVL. Is there a better way to approach this ?
     
  2. jcsd
  3. Jan 22, 2016 #2

    TSny

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    Welcome to PF!

    You might try working with some junctions (KCL).
     
    Last edited: Jan 22, 2016
  4. Jan 23, 2016 #3
    How exactly ? Could you please elaborate ?
     
  5. Jan 23, 2016 #4

    gneill

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    Staff: Mentor

    Choose A as the reference node. You have two supernodes, one associated with B and the other A (the reference node). Write a node equation for supernode B.
     
  6. Jan 23, 2016 #5
    Have I proceeded correctly ?
     
  7. Jan 23, 2016 #6

    TSny

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    Consider the 4 emf's along the bottom. In terms of ##i_1## and ##i_2##, how much current is in the emf next to point A? Then how much current is in the next emf as you move to the left along the bottom of the circuit. Keep going for all 4 emf's along the bottom. This should get you another relation between the two currents.

    Your relation ##1+i_1+i_2 = 0## looks correct for a particular choice of current directions.

    I don't know what a supernode is, but gneill's suggestion might get you to the answer quicker.
     
  8. Jan 23, 2016 #7

    cnh1995

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    KCL will surely give you the answer. But you need not consider all the 8 sources.
    Replace the vertical 1Ω below point B with a parallel combination of two 2Ω resistors and cut the circuit into two halves at point B such that each 2Ω resistor is on either half. Now, you have two circuits with 4 sources in each. Use the one containing both A and B(i.e. discard the left circuit). It will be easier. This is possible because of the symmetry of the circuit w.r.t. point B.
     
    Last edited: Jan 23, 2016
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