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Finding current in the 25 ohm resistor

  1. Sep 27, 2016 #1
    1. The problem statement, all variables and given/known data
    The problem is to find the current in the 25 ohm resistor (left side)
    http://imgur.com/a/luKCW

    upload_2016-9-28_8-5-42.png [Image inserted by moderator]

    2. Relevant equations
    KCL
    KVL
    ohms law (v=IR)

    3. The attempt at a solution
    My attempt at the solution was to add both power sources together (-5v(-5v because the source is - +) and 15v). Then using ohms law, I could find current by dividing voltage (10) by resistance (25) to get .4A. However, this is wrong. I was told that I am not allowed to add the voltages together because there is a resistor between them (10 ohm resistor between the 2 power sources). The final answer is -.4A but I am completely clueless how to come up with this answer if I cannot add the sources together. I have found I2(top right of picture) to be -1.5A and i3 (bottom right of picture) to be -.5A but i'm not sure if those will be helpful in finding the current in just the 25 ohm side.
     
  2. jcsd
  3. Sep 28, 2016 #2

    phinds

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    Whoever told you that does not know what they are talking about. The voltage across the resistor in question is exactly defined by the sum of the two voltage sources and the rest of the circuit is irrelevant to that resistor.
     
  4. Sep 28, 2016 #3
    This was my professor which is why I am very confused. She said a lot of students make this mistake but because there is a resistor, we are not able to add the voltages together. She said I have to find the current in the rest of the circuit, again, which is very confusing because why would finding the current be helpful in finding the current in question (just I in this case).

    If I am able to add the voltages as you say, can you please explain why the answer is -.4A rather than +.4A (which is what I got initially). I'm just very confused as to what I am supposed to do, because she told me the way I am doing it is wrong but I can't think of another way to do it. Sorry if it seems like I'm not trying or I am fishing for answers but I'm truly trying to figure how to do it.

    Quick edit: Just to be clear, the resistor she told me that is in way of the power sources is this one: http://imgur.com/a/m6zUC
     
  5. Sep 28, 2016 #4

    phinds

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    I cannot imagine why she does not have exactly the point of view I expressed. I repeat, the rest of the circuit is irrelevant as far as the 25 ohm resistor is concerned. It may as well not exist. The resistor "between" the two power sources has an effect on the REST of the circuit but the 25 ohm resistor can't even tell whether it is there or not. You could remove any or all of the rest of the circuit (other than the two voltage sources) and the 25 ohm resistor would see no change.

    As to the sign of the result, please pay attention to the directions of the voltage relative to the current arrow for I1
     
  6. Sep 28, 2016 #5
    Ok, thank you very much for the help. There's a bit of a language barrier so she may have not been able to explain herself the way she wanted to. Again, thank you for all of the help. I've been stuck on this for a little while and didn't know what to do.
     
  7. Sep 28, 2016 #6

    Simon Bridge

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    Kirkoffs voltage law applied to the left-most loop, traversed clockwise from the bottom left corner, I get:
    -25i + 5 - 15 = 0 means i = -10V/25##\Omega## = -0.4A

    To get +0.4A you must have got the voltage directions switched somewhere - I suspect you got the voltage across the resistor in the wrong direction wrt the labelled current even though it was correct for the physical reality. Remember that current is a vector.

    The way to get it right all the time is to sketch arrows on the circuit diagram across each voltage.
    Voltages across voltage sources go from negative to positive, the voltages across resistors go opposite the direction of the current through them.
    When you apply the loop law, you make a "vector sum" of the voltages: so when you traverse the loop, a voltage is positive if you travel in the direction of the arrow and negative if you travel opposite the arrow.

    note: a negative value for the current just means the direction of the current arrow drawn on the diagram turned out to be incorrect.
     
  8. Sep 28, 2016 #7

    gneill

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    Your professor may be thinking of the definition of series connected components. Since the junction between the two voltage sources is not shared by the sources alone (it's shared with the 10 Ω resistor), strictly speaking they are not in series.

    However, as these are voltage sources with fixed potential differences regardless of what currents flow through either of them, the total potential difference from one end to the other as you perform a "KVL walk" through them is also fixed.
     
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