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Circuit with Two Batteries and Six Resistors

  1. Oct 12, 2014 #1
    1. The problem statement, all variables and given/known data
    A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1 = 18 V and V2 = 12 V. The positive terminals are indicated with a + sign, The values for the resistors are: R1 = R5 = 40 Ω, R2 = R6 = 156 Ω R3 = 69 Ω, and R4 = 113 Ω. The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows.


    The First question asks me to find the voltage at R4
    h10_twoloop.png

    2. Relevant equations
    V = IR

    3. The attempt at a solution
    I wasn't seeing any other way to find v4 than by making a loop for all three sections of the circuit
    For the first loop I chose a clockwise orientation
    -i3R3 - V1 +i1R1 = 0

    For my second loop I chose another clockwise orientation
    i2R2 - V2 - i6R6 + V1 + i3R3 = 0

    and for the final loop I also choose a clockwise orientation
    V2 -i5R5 - i4R4 = 0

    Looking at the junction next to R2 I get one more equation
    i2 - i3 = i1

    Looking at junction B I get yet another, fifth, equation
    i4 - i5 = i6

    my problem is going to arise every time I have to label currents and come up with equations at junctions or loops. I understand how to create a system of equations using these but I will sometimes get solutions to the matrix with all my currents being zero. Also, i'm sometimes creating systems where I have Nx(N+2) rows and columns and I find that creates problems as well, I usually miss an equation. Any advice on how to fix these problems would be greatly appreciated, thank you.
     
    Last edited: Oct 12, 2014
  2. jcsd
  3. Oct 12, 2014 #2

    ehild

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    Indicate i6 in the figure.

    Note that i5 and i4 are the same, as r5 and r4 are connected in series.

    Also note that the voltage across the series resultant of r4 and r5 is equal to V2, so it is very easy to answer the first question.

    ehild
     
  4. Oct 12, 2014 #3
    okay so i found R45 and calculated the voltage through it to be 8.86V
     
  5. Oct 12, 2014 #4

    ehild

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    You mean the voltage across R4 is 8.86 V.

    ehild
     
  6. Oct 12, 2014 #5
    So I can't use an equivalent circuit using R45 as opposed to R4 and R5 for the rest of my questions?
     
  7. Oct 12, 2014 #6

    NascentOxygen

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    Probably can. What are the remaining questions? That 1st one was very easy (a trick question to catch anyone napping).
     
  8. Oct 12, 2014 #7

    ehild

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    You can, but the voltage across R45 (which is R4+R5) is V2.
    And you do not need to use R45 for the further calculations. It does not change anything if you omit it.
    What you need, is the relation between I1, I3, I6 at node a.

    ehild
     
  9. Oct 12, 2014 #8
    okay i'm still confused about labeling currents on my diagram.

    what do I call the currents coming out of batteries #1 and #2? It should matter because it will affect the equations at the nodes which will effect my system
     
  10. Oct 12, 2014 #9
    I don't understand why you have I3 at node a, wouldn't that cancel the current coming out of R3 how can the battery produce current when the current always cancels over R3. If those currents do cancel, then I1 = I2 at the node to the left of R2.
     
  11. Oct 12, 2014 #10

    ehild

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    The battery produces emf, that is, voltage. The same current flows through each element between two nodes. So I3 flows through R3, then through the battery, and enters into the node a. Name the currents between the nodes.

    ehild
     
  12. Oct 12, 2014 #11
    there's I3 in between the nodes through V1
    I labeled current going through V2 I4
     
    Last edited: Oct 12, 2014
  13. Oct 12, 2014 #12
    i1 + i6 = i3 at node a
     
  14. Oct 12, 2014 #13

    ehild

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    I1 and I3 both flow into the node a. If I6 flows into the node, too, I1+I3 +I6 =0.

    ehild
     
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