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Finding Vottage and Current Values in a Circuit

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Circuit3.png


    2. Relevant equations
    Kirchoff's Voltage Law - Sum of voltages = 0
    Kirchoff's Current Law - Net Sum of Currents going into a node = 0
    V = IR
    3. The attempt at a solution
    I am not sure if I am doing this correctly, could someone please check my work?

    Sum of I net at node B = - I2 - 8A + 5A = 0
    I2= -3A (3A but in the opposite direction from the arrow drawn)

    Sum of VABFH = -(5Ω*I2) - V1 = 0
    -5Ω*-3A - V1 = 0
    V1= 15 V

    Sum of VDEFG= -(I3*5Ω) + 50V - V1 = 0
    -5*I3 + 50V - 15V = 0
    I3 = 7A

    I am particularly concerned if I am treating the ampmeter or voltage meter right (Do I need to account for their inside resistance?) and whether or not I am using the correct signs.

    Thank you for your help.
    In case the image does not appear above:
    [PLAIN]http://img607.imageshack.us/img607/3518/circuitquestion6hw1.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 2, 2011 #2

    BruceW

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    It's been a while since I last did this, but it looks like you've done it all right. I don't think you need to account for their internal resistance, because no internal resistance values are given.
     
  4. Nov 2, 2011 #3

    phinds

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    I don't get this circuit at all. Are you saying that what appears to be a current source and a voltage source are in reality an ammeter and a voltmeter?

    Whether they are or not, the circuit doesn't make sense to me. If they are sources, then how is it that you have 10amps flowing the wrong way through a voltage source and if they are meters, how is it that you have 10 amps flowing through a voltage meter (which ideally is an open circuit)
     
  5. Nov 2, 2011 #4
    I am honestly not sure what they are supposed to be...this is a homework problem that my professor made up. I took them to be a Voltmeter since it wasn't displayed with a battery symbol.

    And I should mention, the blue arrows (only) are added in by me to help with the signs in the calculations.

    Are my calculations correct at all?
     
  6. Nov 2, 2011 #5

    phinds

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    Well, take a look at node D. You've got 8 amps going in and 17 amps (according to you) going out. You reckon that's going to work? Do you not know how to check your work on something this simple?

    EDIT: sorry ... that sounded a bit snippy. Obviously you haven't yet learned how to do/check the problems else you wouldn't be here asking.
     
    Last edited: Nov 2, 2011
  7. Nov 2, 2011 #6
    I have not learned how to go about solving this problems and this is for a class without a textbook so I really have nothing to refer to.
    How do you suggest I approach it?
     
  8. Nov 2, 2011 #7

    phinds

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    Well, if you don't know the fundamental laws of circuit analysis, I don't understand why you have been given the problem.

    Here's one that will get you most if not all of the way there. The sum of all the currents coming into a node has to be zero. That is, there has to be as much going out as there is coming in ("going out" is a negative "coming in").
     
  9. Nov 2, 2011 #8
    That is how I started the calculations earlier - by summing the currents at node B, but the solution doesn't make sense from what you pointed out.

    here's another try:
    Sum of currents at node D = 8A-10A-I3
    Solving for I3: I3= -2A (2A going INTO node D).

    Sum of Voltages DEFG (lower right loop) = 5Ohms*-2A - 50V - V1 = 0
    -10 - 50 + V1 = 0
    V1 = -40 V This looks like it's wrong(?)
     
  10. Nov 2, 2011 #9

    phinds

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    you got I3 right, V1 wrong (it's -60v)

    The whole circuit is nonsensical because the only way it works is if both R1 and R2 have NEGATIVE resistance values.

    I think you must have copied something wrong in stating the problem.
     
  11. Nov 2, 2011 #10
    The image is pasted straight from the homework handout, I haven't copied it myself.

    (V1 should have read -60, don't know how/why I ended up with -50, my mistake)

    So for the last part, to calculate I2:
    Sum on loop ABFH = -I2*5Ohms - V1 = 0
    I2 = 60/5 = 12
    But this doesn't make sense because a sum on node B gives (5-12-8) = 0 which is impossible.

    Again, I am stuck.
     
  12. Nov 2, 2011 #11

    phinds

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    Guy, you don't even seem to be able to do simple arithmetic. I'm not trying to be harsh with you here, but this is getting ridiculous.

    I2 = 5 - 8 = -3

    HOW can you not get that? What did I tell you about the sum of the currents into a node?
     
  13. Nov 3, 2011 #12

    BruceW

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    http://en.wikipedia.org/wiki/Current_source This webpage suggests that the symbols are a current source and a voltage source.

    Assuming this is true, we can calculate all the currents. And then we can calculate the sum of voltages around a loop, but then it doesn't come out right. The voltage source in the middle would have to be something different to agree with the rest of the circuit diagram.

    So I think something with the diagram is wrong...
     
  14. Nov 3, 2011 #13

    phinds

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    No, that's not true. I got perfectly good solution for all nodes and loops ASSUMING that the two resistors are, as I said, both NEGATIVE in resistance value. Read my post #9
     
  15. Nov 3, 2011 #14

    NascentOxygen

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    It's unlikely that an introductory class on circuit analysis would have a homework exercise where the solution involved negative resistances. (Except, perhaps, as a face-saving manoeuvre to cover the embarrassment of the person who set the exercise without analyzing it.)

    Probably nothing more than a simple drafting error that the teacher just didn't pick up.
     
  16. Nov 3, 2011 #15

    BruceW

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    Even if we did allow negative resistance, it still cannot be made to be consistent. Remember that there are 4 different closed loops, and the directed sum of the potential differences around any closed loop must equal zero. In this circuit, we cannot make this true. So for the circuit to be correct, it must disobey Kirchhoff's second law.
     
  17. Nov 3, 2011 #16

    phinds

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    OOPS ... you're right. In concentrating on making the currents come out right, I negleced to check the loop voltages.

    Besides, haveing a 10am reverse current going through a REAL 50v power supply would result in sparks, heat, melted wired, and general chaos.
     
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