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Homework Help: Simple circuit question

  1. Sep 4, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm going to try drawing the circuit. Numbers are added to it for reference. I apologize for the formatting. I'm not sure how to do white space. Edit: for VB the + is on top.

    VA and VB are switches.

    I need to figure out the voltage across VA and VBB when they are:
    1. open, open
    2. open, closed
    3. closed, open
    4. closed, closed

    1-------(+ VA -)-------(+ 8 V -)-------2--------------3
    12Ω____________________________15Ω_______(+ VB -)

    2. Relevant equations
    V = IR
    Kirchhoff's Laws

    3. The attempt at a solution

    1. When they are both open there isn't a circuit. So, VA = 0 V and VB = 0 V.
    2. Again, no circuit.
    3. There is a circuit. It has a total resistance of 27 Ω (I don't know if this matters). VA = -8 V and VB = 0 V.
    4. I'm not sure about this one. There are two loops, but loop 1256 has a total resistance of 27 Ω and loop 1346 has a total resistance of 12 Ω. So 1256 gets 27/39 = 9/13 of the current and voltage and 1346 gets 12/39 = 4/13 of the current and voltage? So VA = -8*9/13 V and VB = -8*4/13 V?
    Last edited: Sep 4, 2012
  2. jcsd
  3. Sep 5, 2012 #2


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    Staff: Mentor

    Circuit or no circuit, does not give the answer. If a voltmeter (or any device) were to be connected between the 2 terminals of the switch in question, would that form a path for current to flow through the meter? If so, what voltage would it register? That's the question you must address.
  4. Sep 5, 2012 #3


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    Staff: Mentor

    Code (Text):
    If you wish to construct ASCII diagrams, enclose those lines
    between [b][[/b][b]code][/b] and [b][/code[/b][b]][/b] statements to display in a
    fixed-width font.
  5. Sep 5, 2012 #4


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    Science Advisor
    Homework Helper
    Gold Member

    You need a circuit for current to flow. You don't need a circuit for a voltage to be present. For example a 9V battery still produces 9V even when it's disconnected from the circuit.
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