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Determine Unknown Elements and Node Voltages (ECE 210)

  1. Aug 28, 2016 #1
    1. The problem statement, all variables and given/known data
    Screen Shot 2016-08-28 at 12.51.14 PM.png

    2. Relevant equations
    V=IR

    3. The attempt at a solution
    Hello. I am working on this problem and I just wanted someone to check it out if I did it correctly. I was kinda confused about some of the nodes and why we even needed to find voltages at those points because in a lot of cases those nodes seemed to be equal to Va, Vb, Vc, etc... Like node 1 (V1) looks like it's just equal to Va. Same with Vb = V2 and Vc = V3. Is that correct? Also, does the rest of this seem correct?

    New Doc 15.jpg
    New Doc 15_2.jpg
     
  2. jcsd
  3. Aug 28, 2016 #2

    gneill

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    Staff: Mentor

    Take a close look at ##v_c## and how it's defined on the circuit diagram. Which way is the current flowing through the 4 Ω resistor?
     
  4. Aug 28, 2016 #3
    The current is flowing to the left. I forgot to mention that there was a typo made by the professor in that diagram. The signs should be flipped for Vc. The diagram in the picture of the solution should be correct.
     
  5. Aug 28, 2016 #4

    gneill

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    Staff: Mentor

    Okay, that makes a difference.

    Your work looks okay except for 2(C). They're asking for the voltage rise from node 2 to node 3, and since that direction is against the flow of the current, will the potential increase or decrease?

    I think they're just checking your understanding of potential in a circuit with respect to a reference node. The potential at a node may be the same as the potential drop across a component connected to that node if that component has one "leg" connected to the reference node. If there's no such connection then the potential across the component does not define the potential at the node.
     
  6. Aug 28, 2016 #5
    Decrease right? It should be negative. I think a voltage drop (+ to -) is positive, while a voltage rise (- to +) is negative. So that should be -12V for Vc (or node 2 to node 3) right? Then the sum of the voltages should be 0.

    Ok that makes more sense now.
     
  7. Aug 28, 2016 #6

    gneill

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    Staff: Mentor

    Ah. Don't confuse the common practice of summing potential drops as positive values while writing KVL loop equations with what's actually taking place; That practice is a matter of convenience.

    When you traverse a resistor in the same direction as the current there is a potential drop (a negative change in potential). Likewise, if you traverse a resistor against the flow of current there is a potential rise (potential increase). We tend to sum potential drops as positive values because we typically write loop equations by following around the loop in the same direction as the assumed loop current, and it is inconvenient to write +(-x) for every term that's a potential drop. Instead we sum the drops as positive values and place the sum of the changes due to voltage source rises on the other side of the equals sign, thus enforcing a change of sign that keeps things consistent mathematically.
     
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