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Circuit problem Kirchkoff's law

  1. Jun 2, 2016 #1
    1. The problem statement, all variables and given/known data
    a) With the switch open, what is the current I1 ? (A positive sign means that current
    flows in the direction of the arrow.)

    b) Once the switch is closed in the circuit, what is the power dissipated in the 47 Ω
    resistor?

    c) How much current passes through the closed switch?

    d) With the switch closed, what is the voltage difference, VA-VB ?

    2. Relevant equations
    junction rule: I3 = I2 + I1
    Loop rule(s)

    3. The attempt at a solution
    I have attempted all 4 parts, please check my solutions? (parts a to c should be correct, but d not too sure.)

    For a) Uses the same picture as part b

    I believe that because there is a open switch, then the whole circuit became a series.
    1st step: I got the Req value = 5ohms + 22ohms + 47ohms + 15ohms = 89ohms.
    2nd step: I got the Veq value = 10.0V + (-6.0V) = 4.0v
    3rd step: I did I = V/R = 4.0V/89ohms = 0.0449 A = I1.

    For b) I have very badly drawn out all of the current directions(the switch is also closed) the top loop and the bottom loop would be separate with a junction at the middle piece on either end.

    Untitled.png

    First, I got that I2 + I1 = I3.

    Then, I solved for the top loop, starting at the negative terminal side of the battery with 10V moving clockwise.
    Top loop(clockwise): +10V1 - 15*I1 - 47*I1 + (0V since i'm assuming no Voltage drop across switch) = 0
    Bottom loop(counter clockwise): -5*I2 + 6V2 - 22*I2 + 0V = 0

    For the top loop: I isolated I1 and got 10/62 = 0.161 = I1
    For the bottom loop: I got 6/27 = 0.222A = I2

    Power dissipated in 47 ohm resistor I am assuming would be I1^2 * R = (0.161)^2 * 47 = 1.22 Watts?

    Part c)
    I'm not totally sure how I do this one, but I think that the current passing through the closed switch is just I3? which is I3 = I1 + I2 = 0.161 + 0.222 = 0.383A.

    Part d)
    For this part.. I am not quite sure.
    I was thinking of like, VA - VB means we have to get from VB to VA in the circuit.
    So... like VA - VB = +6V2 - 22*I2 + 47*I1? (going counter clockwise up the loop from the 6V battery.)
    which would give me VA - VB = 6 - 22(0.222)+47(0.161) = 8.68 V.
     
    Last edited: Jun 3, 2016
  2. jcsd
  3. Jun 3, 2016 #2

    haruspex

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    That all looks right.
     
  4. Jun 3, 2016 #3

    ehild

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    Your work is correct.
     
  5. Jun 5, 2016 #4
    anyone know how I would do part d.
     
  6. Jun 5, 2016 #5

    gneill

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    Staff: Mentor

    You can do a "KVL walk" from node A to node B, summing the potential changes along the way. You'll want to have values for both I1 and I2 to use when your walk passes through resistors carrying those currents.
     
  7. Jun 5, 2016 #6
    Exactly as you did it above or you could use I1* 47+I2*5 or -I1*15 +10V+I2*5 which should all give you the same result.
    I.e. no matter which path you use between A and B the voltage difference should be the same.
     
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