Where to start ?! (Simple circuit)

Discussion in 'Homework Help' started by Dàrk Proféssor, Oct 2, 2016.

  1. Dàrk Proféssor

    Thread Starter New Member

    Oct 1, 2016
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    I simply don't know how start solving this question

    [​IMG]


    The idea that I am struggling with a such a simple question is so annoying.
    I know I need to use Kirchhoff's law and my be do some source transformation. But where do I start ?

    Thank you,
     
  2. takao21203

    Distinguished Member

    Apr 28, 2012
    3,577
    463
    Added confusion on purpose

    Its not clear if the Amp meter like shown is measuring 5 Amps.
    Or if it is supposed to be the current through Rn

    In which case you should consult Ohms law and Kirchhoff rule all over again.

    In physics they enjoy to mount special case which can be solved some way normally not be solveable, and not all required parameters given or explained. By a special case, you could substitute a parameter.

    If you look matrix transformations you get the idea.

    When the A meter is measuring 5 Amps its difficult to say because it has very low resistance and its not known even. So I assume its the current through Rn.

    You simply calculate the supposed total resistance for 5 Amps.

    If the Amp meter is supposed to be parallel, I dont know how to find solution.
     
  3. AlbertHall

    Well-Known Member

    Jun 4, 2014
    1,951
    387
    Is the 5A not a current source? The red writing shows 5A flowing from that branch.
     
  4. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,159
    The 5 ampere is a current source. An ampere meter would have an A in the center .... Not an arrow.
     
  5. wayneh

    Expert

    Sep 9, 2010
    12,126
    3,048
    Just take your time and start a list of things you know, based on Ohm's law, Kirchoff's law and so on. You want to get enough independent equations to solve for the unknowns. As you know, you need at least one equation per unknown.

    Once you have the minimum, then work the math.

    If you don't have enough and have exhausted all equations, you cannot get a numerical solution and will have to accept answers in terms of one remaining unknown.
     
  6. #12

    Expert

    Nov 30, 2010
    16,298
    6,809
    It's also allowed to redraw the circuit so it looks familiar.
    Doing that can reduce intentional confusion, like takao mentioned.
     
  7. takao21203

    Distinguished Member

    Apr 28, 2012
    3,577
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    If it's a current source I don't see how the right and left currents should have a relationship. You could substitute values for Rn and see how much this makes sense.

    Maybe there is equilibrium condition where the induced voltage from both sides is the same.

    Maybe a formula can be found.

    Is this the complete description did you find similar questions presented?
     
  8. takao21203

    Distinguished Member

    Apr 28, 2012
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    Hmm flipping 90 degree. First you calculate the current caused by the two resistors then derive the Rn

    Or is it?
     
  9. WBahn

    Moderator

    Mar 31, 2012
    17,748
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    Can you find Va (the voltage at 'a')?

    Can you find Vb (the voltage at 'b')?

    If so, the Vab is simply (Va - Vb).

    So this breaks the problem into two smaller problems -- find Va and find Vb. Treat them as separate problems.

    You are given the voltage Vn. Can you find i1 knowing just this fact combined with the 20 V source and the two resistors between it and the top-right node?

    If so, can you use the value of i1 in order to find Vb?

    If you know i1 and you know you have 5 A from the current source, can you find i2?

    If you have i2 and you have Vn, can you find Rn?
     
    anhnha likes this.
  10. RBR1317

    Active Member

    Nov 13, 2010
    232
    48
    If the voltage at node 'n' is given, what would happen if you forced node 'n' to be that voltage?

    It would make it rather easy to find the voltage at node 'b' which would then make it simple to find the current flowing in the 2Ω resistor.

    Node-Equation-E2.png
     
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