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Voltage problem

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    23roas9.png

    Three resistors (R1 = 3 ohm, R2 = 4 ohm, R3 = 5 ohm) is arranged like the picture above.
    The voltage at point 1 is 24 Volt and the voltage at point 2 is 10 Volt.
    The currents that go in R1 and R2 are 2 Amperes.
    The voltage at point 3 is ... Volt

    A. 10
    B. 12
    C. 16
    D. 18
    E. 24



    2. Relevant equations
    V = I R

    3. The attempt at a solution

    Umm...
    I think there are weird things, such as I think that the problem does not obey the Kirchhoff 1 Law.
    If the current 1 is 2 ampere into the junction, so I think current 2 and current 3 will share. (The problem says that The current 2 is 2 ampere which means that current 3 is 0 ampere (huh?) )

    Voltage at R1 is I1 * R1 = 2 * 3 = 6 Volt
    Voltage at R2 is I2 * R2 = 2 * 4 = 8 Volt
    Voltage at R3 is I3* R3 = 0 * 5 = 0 Volt ( I get the current is zero because the Kirchhoff 1 law which states that current into junction = current leave junction)

    But, then I think that the voltage at point 2 and 3 must be the same because it's parallel connected, right?
    So, I think it's 10 Volt.
    But, really, I'm still confused.
    Please help
     
  2. jcsd
  3. Mar 27, 2016 #2

    gneill

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    Staff: Mentor

    There are no parallel or series connections shown: No resistors share connections at both ends, so there are no parallel connections; The only junction shown has three connections, so that rules out series connections.

    What can you say about the potential at the junction?
     
  4. Mar 27, 2016 #3
    Potential at the junction?
    Hmm... I have no idea...
    Maybe it is the same as the voltage in R1 = I1 * R1 = 6 Volt?
     
  5. Mar 27, 2016 #4

    gneill

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    Staff: Mentor

    When a current flows through a resistor there is a potential drop across the resistor. If the potential at point 1 is 24 V, then what is the potential at the junction? DO a "KVL walk" from point 1 to point 0.
     
  6. Mar 27, 2016 #5
    Oh yeah, I get the idea...
    The voltage at the junction is 24 - 6 = 18 Volt
    Then?
     
  7. Mar 27, 2016 #6

    gneill

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    Staff: Mentor

    Confirm that the 2 A flowing through R2 satisfies KVL for point 2. If it does, what can you conclude about the current through R3?
     
  8. Mar 27, 2016 #7
    So V2 is 2 * 4 = 8 Volt
    It means that V3 is 18 - 8 = 10 Volt (I'm not sure about this since V3 is at another branch)

    The current is V3/R3 = 10/5 = 2 ampere ??

    But, I really doubt it. Since by KCL, it should be zero ampere since I1 = I2 + I3 so I3 = I1-I2 = 2-2 = 0 ampere..
    I'm confused
     
  9. Mar 27, 2016 #8

    gneill

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    Staff: Mentor

    No. It means that point 2 is at 10 V: 18V - 8V = 10V , which is what was given in the problem statement, so it's confirmed that the currents and potentials given for points 1 and 2 are good, and that the center junction is at 18 V.
    And I3 is zero, which is a perfectly good value. What does that make the potential drop across R3? Then what's the potential at point 3 (KVL walk from the junction to point 3).
     
  10. Mar 27, 2016 #9
    Since the voltage at the resistor is zero, it must be 18 volt at point 3, right?
     
  11. Mar 27, 2016 #10

    gneill

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    Staff: Mentor

    That's right.

    So to sum up, from the given information and KCL you know that the current through the R3 path is zero. By a KVL walk from one of the known potentials (we used point 1 at 24 V) you know the potential at the junction is 18 V. Then the potential at point 3 is determined by a KVL walk from the junction to point 3. Since that current is zero there's no drop, so it's also at 18 V.
     
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