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Series Resistive Circuit

  1. Feb 9, 2014 #1
    1. The problem statement, all variables and given/known data
    http://puu.sh/6PZNJ.png [Broken]

    2. Relevant equations

    This is supposed to be a pretty easy problem, but can i assume the resistors are in series? The other loops are throwing me off.


    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 9, 2014 #2

    gneill

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    What have you done to try to resolve whether the resistors are in series or not?
     
  4. Feb 9, 2014 #3
    Well in the other two loops there are no circuit elements, and there is only one current source. Should i instead go through this with nodal analysis?
     
  5. Feb 9, 2014 #4

    gneill

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    The circuit can be simplified. It might be best to do that before anything else.

    How many distinct nodes are there in the circuit?
     
  6. Feb 9, 2014 #5
    I believe I'm seeing 4 nodes, 1 at the left and right, and 2 in the middle.
     
  7. Feb 9, 2014 #6

    gneill

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    Nope. Remember that a node consists of a contiguous conducting path. Everywhere that path goes is part of the same node.

    Take some highlighting markers and highlight the contiguous conducting paths (wires). How many different colors do you need to use?
     
  8. Feb 9, 2014 #7
    Oh starting from the Current source, I see 2 different paths.
     
  9. Feb 9, 2014 #8

    gneill

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    Right. So there are two distinct nodes.

    Now try this: Call the two nodes 'a' and 'b', corresponding to the terminals that are already labeled. Now, everywhere wires meet on a given node place the same label. Then, start a new sketch, placing two dots called a and b to begin with (they will represent your two nodes). Now, for each component in the original circuit, place it on the new sketch with the same node connections. So, for example, the 20 Ω resistor connects to both node a and node b in the original. Place it onto your new sketch preserving those associations. Do the same for all the other components, too. How does it look now?
     
  10. Feb 9, 2014 #9
    I think i see where this is going. They are all connected in the same way. Aren't they all short circuited?
     
  11. Feb 9, 2014 #10

    gneill

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    Yes, they're all connected in the same way. No, they are not short circuited. To be short circuited would imply a wire connection between a and b. But that's not possible if you identified them as distinct nodes:

    attachment.php?attachmentid=66451&stc=1&d=1391978765.gif

    If a and b were shorted, they'd both be the same color.
     

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  12. Feb 9, 2014 #11

    NascentOxygen

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    I see 3 separate paths, 3 different paths that current can get from a to b. That says there are 3 parallel paths.
     
  13. Feb 9, 2014 #12

    gneill

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    The 'paths' being referred to were for the contiguous wires representing the distinct nodes. So these 'paths' stop when ever they encounter a component.

    You're perfectly correct though that there are 3 paths that current can follow, passing through components on the way from a to b.
     
  14. Feb 9, 2014 #13
    (Vb - Va)/40 + (Va - Vb)/20 + (Vb - Va)/40 = 0

    Va = (.003)(40) = .12 V

    Did i set this up right?
     
  15. Feb 9, 2014 #14

    gneill

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    No, your current directions seem to be a bit confused. Which node (of the two that were identified) are you taking to be your reference node for potentials? Also, you haven't accounted for the current source.

    Did you figure out the simpler way to look at the circuit via the re-drawing of it? I don't suppose you could post your sketch?
     
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