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Determining resistor polarity

  1. Mar 4, 2016 #1
    I guess there's a simple rule, but I can't find it anywhere and it seems I was supposed to know this. I'll appreciate it if someone explains me how it works and also if (and how) is related to the way you consider positive and negative voltages when you apply Kirchoff's second law as shown in the picture below.
    Last edited: Mar 4, 2016
  2. jcsd
  3. Mar 4, 2016 #2


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    Looks like homework. If so, please post in a homework forum. And fill in the template: now it is unclear what is given, and what is asked
  4. Mar 4, 2016 #3
    No, it's not homework. It's just a figure, supposed to explain how Kirchoff's second law applies to a circuit. I just can't understand it.
  5. Mar 4, 2016 #4
    There are many "methods" of how we can apply KVL around the loop. And in your circuit the author gives a "-" sign when he goes from positive to negative (voltage drop) or he gives a "plus" sign when he encounters a voltage rise (from negative to positive).
    As for resistor polarity, to determine the voltage polarity of a resistor you need first defined the current direction through each resistor with arrows. Once the arrow is defined, then the polarity is defined (current flow from positive to negative ).
    http://www.mwftr.com/netw1/04 pol rev.pdf
    http://www.ittc.ku.edu/~jstiles/312/handouts/312_Introduction_package.pdf (page 12 )
  6. Mar 4, 2016 #5
    Thanks for the sources. I now know how you determine resistor's polarity. But I still struggle to comprehend the way polarity was determined on resistors R3 and R4 on my circuit. Why is the arrow on top right loop this way? I thought it shows the current flow.
    Can you or anyone point out the procedure I should follow if the circuit had no symbols at all? Here; I erased everything.
  7. Mar 4, 2016 #6
    Simply assume current direction first. Next mark resistor polarity based on assumed current direction and write the KVL.
  8. Mar 4, 2016 #7
    Let's see. I assume that current direction is from + to - for both sources. And that's how I'd mark the polarities base on the current direction from each source. But it's not correct. I also have no idea where and how the two currents "interfere".
  9. Mar 4, 2016 #8
  10. Mar 4, 2016 #9


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    Firstly you should name the currents I1, I2, I3 etc.

    Then I note you have a duplicate on your drawing... eg the current through R6 is the same as R5 so you only need one name for that current

    Then regarding..

    Take the current through R3 for example. This can be expressed as equation the sum (or subtraction) of two other currents by applying KCL at one end (eg the current through R4 and R6). Have a go after naming the currents.
  11. Mar 4, 2016 #10


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    Say you call the upper loop current: I3
    and the lower loop currents: . . . I1 , I2

    then your last equation as for the I3 loop should be:

    -E2 - ( R3+R5+R6 ) * I3 + R3 * I2 = 0

    because the voltage drop across R3 is due to the sum of currents ( I3 - I2 ) passing R3.

    Substituting e.g. V5 by I3*R5, you will get 3 equations as the 3 loop current as variables. ( You can never solve 6 voltages by means of 3 equations ).

    Now solve the loop currents. Afterwards you can determine the voltages using ohm's law. E.g. V2 = ( I1 - I2 ) * R2, according to your choise of loop current directions.
  12. Mar 4, 2016 #11
    The direction of the assigned current is arbitrary. Once the circuit is solved it may turn out that the current is negative implying you guessed wrong, but that's fine.

    You do need to be consistent though. In your OP diagram the blue circles that look to be current-ish end up flowing through some resistors in both directions. (R2, R3) Pick a single direction.
  13. Mar 5, 2016 #12


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    I agree with Jeff. You assign arbitrary polarities, and the answers will be positive or negative based on that polarity. You can't guess them all correctly to begin with.
  14. Mar 6, 2016 #13


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    Even more than that - you should realise that it really doesn't matter and you will (should) get the right answer whatever you choose.
  15. Mar 6, 2016 #14

    jim hardy

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    To Sollicitans -

    There's an exercise a beginner should do if he is uneasy about the method.
    Work the same problem three times.
    First time write your polarity signs , let's just say all +'s to left (or up)
    Second time write them all right or down
    third time according to your best guess , which by now oughta be pretty good.
    If you don't get same answers each time you are making mistakes of algebra not EE.

    That will demonstrate the value of rigorous step by step algebra and of neatness on your worksheet. Keep your lines horizontal and your columns aligned.

    I did that in high school electronics class . One needs to have that subliminal confidence in the method.
    Kirchoff is very reliable. It's our own sloppy math that frustrates us.
    Develop rigor in your methods early on.

    My two cents, and that's probably overpriced...

    old jim
  16. Mar 6, 2016 #15
    I noticed that you probably think I know what equations to use or how to apply the law. Truth is, I don't even know where you start from - not only is this case.
    Anyway, I didn't have enough time this weekend to follow all of your instructions, but I read them all. Also, English is not my main language, so I need more time to assimilate what I read, especially on those huge guides you sent me.
    It'd be a lie if I said you helped me figure out what's happening on that circuit, but you helped me figure out a few other things - potentially useful for circuits in general.
    I think the best thing I can do is start reading a book on basic electronics analysis, and I usually trust the "For Dummies" series.
  17. Mar 6, 2016 #16

    jim hardy

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    okay we all started from there

    I think Hesch did that for you in post 10
    here it is again, perhaps a little more cookbook

    What: Arrive at the equations for KVL
    1. Assume directions of polarities and current flow.
    2. Imagine yourself a tiny particle of current and walk the loops, writing down the first polarity sign you encounter for each voltage .
    3. Solve the resulting algebraic equations.

    Step 1:

    i removed some more stuff from your drawing.

    Note all +'s are left or up
    all currents circulate CW
    that's just assumed
    any that were assumed wrong will have negative value at the end of (proper) calculations.
    If you're sloppy you'll miss a sign. That's why it is necessary for a klutz like me to work them over and over until i get same answer 3 times in a row.

    Okay, step 2:

    inject yourself in the circuit at some convenient spot, maybe lower left corner and walk the circuit. I'll just follow I1.
    Writing down the voltages i encounter, using first sign, i get
    -E1 + I1R1 + (I1-I3)R2 = 0
    Why did i write (I1-I3)R2 ? Because I1 and I3flow opposite directions so (their difference ) X R2 is the voltage. Were I3 also flowing down it'd be their sum.

    Next inject yourself in lower right corner and walk I3's path
    -(I1-I3)R2 + (I3-I2)R3 +I3R4 = 0

    and top right corner walking I2's path
    +E2 -(I3-I2)R3 -I2R5 + I2R6 = 0

    step 3:

    If you know the resistor values and E1 & E2
    then you have 3 equations and 3 unknowns
    -E1 + I1R1 + (I1-I3)R2 = 0
    -(I1-I3)R2 + (I3-I2)R3 +I3R4 = 0
    +E2 -(I3-I2)R3 -I2R5 + I2R6 = 0
    so it's 9th grade algebra from there.

    we used to say "plug and chug"..

    i hope i did that right - been fifty years now .....

    Would somebody younger check me above for flipped signs ?


    old jim
    Last edited: Mar 7, 2016
  18. Mar 7, 2016 #17
    Turns out the kind of instructions I needed were the cookbook ones!
    Also, turns out I was rushing a bit. Today we started working on simple circuits, explaining everything from scratch. I probably panicked thinking that there is stuff I should know and I just don't.
    Thank you, Jim and everyone for your precious time.
  19. Jan 27, 2017 #18
    Does the assignment of +'s and -'s have to do with the flow of current? I tried to solve the problem by arbitrarily assigning +'s and -'s and did not get the same answer after plugging in the same resistor and voltage source values.
  20. Jan 27, 2017 #19

    jim hardy

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    No, initial assignment is completely arbitrary, up to you. As you develop skill you get good at predicting how they're gonna work out .

    That's arithmetic mistakes not flawed EE theory.

    See post 16. I arbitrarily put all +'s left or up . You should work it several times as suggested in post 14.
    When you can crawl to the correct answer irrespective of initial polarity assumptions you are nearly ready to try walking.

    old jim
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