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How to determine the sign of currents in complex circuits?

  1. Feb 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A combination circuit powered by a 6.0 V battery is shown.
    Complex Circuits.png

    What is the total current through this circuit?
    I don't know how to determine where the signs go, for example if the right side of a resistor is positive or the left side is positive
    2. Relevant equations
    What is the total or equivalent resistance of this circuit.

    3. The attempt at a solution
    I found the total resistance and I tried to find the currents using Kirchhoff's law
    I1=I2+I3.
    For the first loop, I found the equation: -6+220*I1+2100*I2=0
    For the second loop, I found: 4500*I3-2100*I2
    Here is my work:
    -6 + 220(I2 + I3) + 2100*I2
    -6 + 220*I2 + 220*I3+2100*I2
    220*I3=-2320*I2+6
    I3=-10.54*I2+0.027
    This answer to I3 was plugged into the second loop equation:
    4300*(-10.54*I2+0.027)
    0=47430*I2 + 122.7
    122.7/47430=I2=2.6*10^-3 A or 0.00259
    I highly doubt that 2.6*10^-3A is the current for I2 but I don't know what I did wrong. I think it has to do with the fact that I don't know where each sign goes where (+/-). Any advice is appreciated!
     
  2. jcsd
  3. Feb 20, 2016 #2

    Merlin3189

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    First: why do you think I2 is not 2.6 * 10-3 A?
    It seems a plausible answer to me.

    Your arithmetic is a bit strange to me, so I'm not sure I can follow your method. For eg. you say "This answer to I3 was plugged into the second loop equation:" but you have not written an equation for the second loop.
    The first loop has an equation, but for the second loop you just give an expression.

    I wonder if you can use a simpler method.
    Can you calculate the resistance of R2 and R3 in parallel ?
    Then can you calculate the total series resistance in the circuit?
    If you can, then you can calculate the current I1 and the voltage across every resistor. Then you can calculate the current through each resistor.
     
  4. Feb 20, 2016 #3

    Merlin3189

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    <Quote>
    I1=I2+I3. Yes
    For the first loop, I found the equation: -6+220*I1+2100*I2=0 Yes
    For the second loop, I found: 4500*I3-2100*I2 = ?
    Here is my work:
    -6 + 220(I2 + I3) + 2100*I2 = ?
    -6 + 220*I2 + 220*I3+2100*I2 =? (ok. follows from above)
    220*I3=-2320*I2+6 suddenly the = sign appears! ok.
    I3=-10.54*I2+0.027 ok, follows from above. I'd keep 4sf during calculations.
    This answer to I3 was plugged into the second loop equation: loop expression = ?
    4300*(-10.54*I2+0.027) this looks a bit like I3 substituted into part of the second loop expression, but where does 4300 come from?
    0=47430*I2 + 122.7 Now you've lost me completely. An = sign has appeared again! And where does 47430 come from?
    122.7/47430=I2=2.6*10^-3 A or 0.00259 This seems to follow from previous line.
    I highly doubt that 2.6*10^-3A is the current for I2 but I don't know what I did wrong. I think it has to do with the fact that I don't know where each sign goes where (+/-). Any advice is appreciated!</Quote>
    Well, I can't follow the arithmetic, but I can't see any reason to think that 0.00259 could not be the answer. (I don't think it is, but that's just arithmetic.)

    Edit : I have now followed your stated method and it does give the same result as the simple method.

    It just seems that you've made some arithmetic errors, which is hardly surprising with your setting out.
     
    Last edited: Feb 20, 2016
  5. Feb 20, 2016 #4
    I apologize for the errors. I adjusted the errors found above to make it easier to follow:
    I1=I2+I3.
    For the first loop, I found the equation: -6+220*I1+2100*I2=0
    For the second loop, I found: 4500*I3-2100*I2=0
    Here is my work:
    -6 + 220(I2 + I3) + 2100*I2=0
    -6 + 220*I2 + 220*I3+2100*I2=0
    220*I3=-2320*I2+6
    I3=-10.54*I2+0.027
    This answer to I3 was plugged into the second loop equation:
    4500*(-10.54*I2+0.027)-2100*I2=0
    Using Distributive property and subtracting 2100*I2
    0=-47430*I2+121.5-2100*I2
    0=121.5-49530*I2
    49530*I2=121.5
    I2=121.5/49530
    I2=0.002453 or 2.45*10^-3
     
    Last edited: Feb 20, 2016
  6. Feb 20, 2016 #5

    Merlin3189

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    Ok. I agree as far as your I3 and the substitution into 2nd loop equation.
    4500( -10.54 I2 + 0.02727) -2100 I2 (except again you omit the = 0 )
    But I still can't get your next line. I can only suggest you work it step by step without skipping. You just need to be careful.
    "Using distributive property and subtracting" seems to show the correct intention, (though I wouldn't think of it that way.)

    I disagree with 47430. 121.5 is near enough, but notice you are going from 0.027 with only 2sf to 121.5 which claims to have 4sf. It would be better to keep 4sf all through your working. (I get 122.7 here.)

    Your use of positive and negative in deriving the loop equations seems fine. (I'll have to leave it to others to tell you what the rule is these days - I learned it 50 years ago and rules seem to have changed in modern physics teaching.) Your error seems to be in the arithmetic.
     
  7. Feb 20, 2016 #6
    I added the zero to the equation and calculated it step by step to get a new answer of 0.002453
     
  8. Feb 21, 2016 #7

    Merlin3189

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    First loop -6+220*I1+2100*I2 = 0
    second loop 4500*I3 - 2100*I2 = 0

    -6 + 220(I2 + I3) + 2100*I2 = 0
    -6 + 220*I2 + 220*I3 + 2100*I2 =0
    220*I3 = -2320*I2 +6
    I3 = -10.54*I2 +0.02727 (4sf)
    This answer to I3 was plugged into the second loop equation:
    4500*(-10.54*I2+0.02727) -2100*I2 = 0
    Using Distributive property
    4500*(-10.54*I2) + 4500*(+0.02727) -2100*I2 = 0
    -47430 * I2 + 122.7 -2100 * I2 = 0
    Now for me, isolating the I2 terms is easiest,
    +122.7 = +2100 * I2 +47430 * I2
    but continuing your route,
    0=-47430*I2 +122.7 -2100*I2

    0 = 122.7 -49530*I2
    49530*I2 = 122.7
    I2 = 122.7/49530
    I = 2.477 * 10-3 = 2.48 * 10-3 (to 3sf)
    I2=0.002453 or 2.45*10^-3 just slightly different due to arithmetic accuracy.


    Now, coming back to your thinking that 2.6mA must be wrong,
    firstly I think it is so close that there is no reason to doubt it,
    secondly you can make a rough estimate just looking at your diagram:
    if you ignore R3 for the moment,
    the total resistance is a little over 2k (actually 2320),
    so current I2 is less than 6V / 2k or 3mA (2.6mA a bit more carefully)
    Putting R3 back in reduces I2 slightly ,
    so a bit less than 2.6mA could well be around 2.5mA
     
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