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Kirchhoff's Loop Rule Conceptual Question

  1. Feb 22, 2008 #1
    1. The problem statement, all variables and given/known data

    The circuit shown below consists of four different resistors and a battery. You don't know the strength of the battery or the value any of the four resistances.

    [​IMG]

    Select the expressions that will be equal to the voltage of the battery in the circuit, where [tex]V_{A}[/tex], for example, is the potential drop across resistor A.

    [tex]V_{A} + V_{B}[/tex]
    [tex]V_{B} + V_{C}[/tex]
    [tex]V_{B} + V_{C}[/tex]
    [tex]V_{A} + V_{B}+ V_{C}[/tex]
    [tex]V_{A} + V_{B} + V_{C} + V_{D}[/tex]
    [tex]V_{A} + V_{D}[/tex]
    [tex]V_{D} [/tex]

    2. Relevant equations

    Kirchhoff's voltage rule for closed circuit loops

    3. The attempt at a solution

    So I know Kirchhoff's voltage rule for closed circuit loops states that the voltage in a closed circuit is 0.

    I think that the answer would be[tex]V_{A} + V_{B} + V_{C} + V_{D}[/tex].

    am I right?
     
  2. jcsd
  3. Feb 22, 2008 #2
    Think about what you're looking for, you want to know the voltage of the battery. If the battery were, say, 5 V, that means you could say the top part of that circuit, all the lines immediately connected to the + terminal of the battery, were at 5 volts, whereas all the lines immediately connected to the negative terminal were at ground

    Also I'll tell you I think there are two answers, however I think only one is present as a choice, and it's not the one you picked
     
  4. Feb 22, 2008 #3
    Well, according to Kirchhoff's voltage rule for closed circuit loops, V_B and V_C should be 0, so if that's true then it should be V_A+V_D since the voltage has been split into the two paths
     
  5. Feb 22, 2008 #4
    Voltage doesn't split like that. If Vb and Vc were not there, then the voltage drop across Va and Vd would be the same(that'd driven by the voltage source), and your answer

    By saying that Vb and Vc = 0, well, you're basically saying they're just more wire, not resistors, and that Va is hooked to the plus terminal of the battery, as is Vd, and both other ends connect directly to the negative terminal, and your answer would be either Va or Vd
     
  6. Feb 22, 2008 #5
    Lets start over here. You said you know that Kirchoff's voltage law (KVL) states that the sum of all voltages in a closed loop is equal to zero. Simply apply this law to one of the closed loops in your circuit. In the circuit you have drawn above there are 3 loops that go to ground, 2 of these 3 loops can directly provide you with the voltage you seek namely the voltage across the battery, let's call it V_bat.

    I will apply KVL to one of the 3 loops as an example in a clockwise direction:

    -V_bat+V_A+V_B=0

    rearranging terms....

    V_bat=V_A+V_B

    I hope this helps you, there is also another solution to this problem that I will leave for you to figure out. The same priciples apply.
     
  7. Feb 26, 2008 #6
    Yes, got the answer. Thank you both.
     
  8. Mar 25, 2011 #7
    I don't understand this, I thought A,B,C, & D summed together would equal the voltage in the battery because there are parallel components in the figure, so the voltage splits unequally between A and D and splits equally between B & C.
     
  9. Mar 25, 2011 #8

    gneill

    User Avatar

    Staff: Mentor

    A loop is a single, closed, non bifurcating, non self-crossing path through the circuit. Can you show us how you would draw the loop A,B,C,D?

    A corollary of the loop law is that if you pick any two distinct points (call them 1 and 2) along the path of the loop (that is, they are not the same node), then the loop is divided in to two distinct paths from one point to the other, and the sum of the voltages for both paths from point 1 to point 2 must be equal.

    In the case of this circuit, if you pick the two points to be at the two battery leads, then there are precisely three paths through the circuit that can form the "other path" of the closed loop. You should be able to identify them.
     
  10. Mar 25, 2011 #9
    Ok, I got it now. Thank you :)
     
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