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How do we know if a resistor is added or subtracted?

  1. Dec 14, 2016 #1
    1. The problem statement, all variables and given/known data
    I was reading in my physics book about loop currents. However, some resistors are added to the result (R*I) whereas some resistors are subtracted from the result -(R*I). The book uses the phrase "uphill" and "downhill" to describe this, and, indeed, the so called "uphill" resistors where the values are added instead of subtracted the chosen current was headed north as it passed through them, but I doubt that can be the rule.

    My answer, however, is WHY does this work? Is it not possible to construct a circuit where the resistor could be on either side of the loop (like a circuit with only one loop, a battery and two resistors?)

    So how do we know whether to add or subtract resistors?

    2. Relevant equations
    (N/A)

    3. The attempt at a solution
    (see above.)
     
  2. jcsd
  3. Dec 14, 2016 #2

    cnh1995

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    Homework Helper

    You don't add or subtract resistors. It's the potential changes that are added or subtracted. Have you studied Kirchhoff's voltage law? The term "uphill" indicates rise in potential and "downhill" indicates drop in potential. In any circuit loop, the sum of all the potential changes is ZERO.
     
  4. Dec 14, 2016 #3
    KVL I understand (or at least I think I do. However, I can't seem to identify whether the voltage drops or rises accross a resistor. (Sorry that I may have worded that poorly, I've been up for 22 hours.)
     
  5. Dec 14, 2016 #4

    gneill

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    Staff: Mentor

    When you are analyzing a circuit using KVL the first thing you do is assign currents through all the components, usually indicated by arrows showing their assumed directions (whether those assumed directions turn out to be correct or not doesn't matter, the math will take care of sorting that out). When you do your "KVL walk" around a loop, the rule is that if you walk though a component in the same direction as its designated current, then a potential drop occurs. If you walk through the component against the flow of the current, then a potential rise occurs. To extend your book's analogy, it's like following a stream where the running water represents the current flow. If you walk against the flow you are moving uphill. If you walk with the flow, you are moving downhill.
     
  6. Dec 14, 2016 #5

    CWatters

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    Science Advisor
    Homework Helper

    k_squared... Perhaps try posting an example circuit, apply the method gneill describes and we can see where you are going wrong if at all.
     
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