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[Circuit] Potential Difference question

  1. May 4, 2010 #1
    1. The problem statement, all variables and given/known data
    http://www.sewanee.edu/physics/PHYSICS102/Ch28.pdf [Broken]

    It is number 18 in the link.

    3. The attempt at a solution
    I found the three currents but i don't know how to find the Potential Difference between two points. what formula am i supposed to use?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 4, 2010 #2


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    Well, you know the current through each resistor. What's the voltage drop (i.e. potential difference) across the terminals of resistor, with resistance R and current I? Once you have those, add things up appropriately to find Va relative to Vb.

    (Hint: There are 3 ways to solve the problem. Each method should give you the same results. :wink:)
  4. May 4, 2010 #3
    still confused. do you mean i just multiply

    (Current1)*(20 Ohms) = Vb
    (Current 2) *(10 ohms) = Va
    and then Va - Vb?
  5. May 4, 2010 #4


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    I don't think so.... Although it would be more clear if I knew how you defined "Current 1" and "Current 2."

    But what I can say with certainty is this.

    (Current going through a particular resistor)*(Resistance of that particular resistor) = Voltage drop across the terminals of that particular resistor.

    Once you have all the voltage drops of all the resistors, and since you already know the voltages of the supplies (batteries), you can sum the voltages across any given path from point b to point a, which gives you Va - Vb. There are three different paths, but each should give you the same answer, if you calculated the currents correctly.

    Something to consider. Electric potential (i.e. voltage) is always a difference between two points. Sometimes in electrostatics, infinity is chosen to be one of the two points, but that is merely a convention. Usually in circuit design, the ground (earth) node is chosen as the reference point. But that too is also merely a convention.

    In other words, it doesn't make any sense to define a Va on its own, unless you've already established a reference voltage. Otherwise, the best you can do is express Va relative to some other voltage. Which the end result is Va - Vsomething.

    If it helps you conceptualize this, place a Ground symbol on point b. That way, by your own definition, Vb is assumed to be zero volts (because Vb - Vb = 0). Now solve for Va. By your definition of placing the ground signal on point b, "Va" is really just shorthand for Va - Vb.
    Last edited: May 4, 2010
  6. May 4, 2010 #5
    thank you for the explanation
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