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Kirchoff's Loop Rule-Direction of Current?

  1. Apr 15, 2013 #1
    Kirchoff's Loop Rule--Direction of Current?

    I came across this question, and it really confused me-- how does the current decide which direction it goes?
    The pictures are attatched below.
    The first picture is of the original questions. It gives two scenarios, and asks you what the value of the current "I" is for each scenario.
    I didn't understand how the resistors were connected in both pictures.
    The second picture was an alternative method of connection for circuit 2 in the first picture, but how are the two cases were equivalent? Where did the "I" branch on the first picture go?
    The third picture is the calculated current for all branches in circuit two.
    Mathematically, everything make sense, but I would've never figured it out without the answer key.

    So could someone please help me with
    1) Why does I=0 in the first picture circuit 1?
    2) What decides the direction of current? (Higher, lower voltage drop? but I couldn't figure out which was which)
    3) Which branch in the third picture is equal to branch "I" in circuit 2?

    Thank you so much for your help!
     

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    Last edited: Apr 15, 2013
  2. jcsd
  3. Apr 16, 2013 #2
    Well, if you don't know which direction of current flows to where, you can "blindly" apply kirchhoff's loop rule. but whatever convention you use, do remember to stick with it.

    if you get a negative I (representing the I in circuit 1), it means that the current will flow in the negative sense or in the direction opposite to your assigned direction. if I turns out positive, congratulations! you have guessed the correct direction, but it's no biggie anyway.

    i'm not sure what picture 2 is trying to show, so i'll omit it in the discussion. thing is, not all circuits can be reduced nicely into clear parallel and series, but ideas for parallel and series can be applied to sections of such circuits.

    the annoying wire carrying the current I does one annoying thing which annoyingly annoys you: making potential the same. here, i'll use a comical way to describe this, which strictly speaking isn't correct because... you'll see.

    referring to picture 3: After the 0.5A current passing through the 10 ohm resistor on the left, it encounters a cross road. it can either take the direct route to face the 20 ohms or the easier path of 10 ohms provided by the said annoying wire. because the 20ohm path seems more difficult, most current decided to go through the path made by the annoying wire...

    but a more accurate picture would be to see what the wire and junctions do; the make potential the same. this is observed in parallel circuits. say, the middle annoying wire carrying current I isn't there, then potential has to be the same across the 2 parallel configurations.

    but now that the wire is there, the top can be said to be parallel to each other, and the bottom 2 resistors can be said to be parallel to each other (which could be what picture 2 is trying to say). if, for instance, no current flows through the wire (no 0.25A in picture 3), we see that there is a violation: the 0.5A flowing through the 10Ω on the left goes straight down ignoring the alternative path into the 20Ω. Applying V=IR we get V = 15V. Similarly for the left, 0.25A thru the 20Ω will go into the 10Ω, giving us V=10 V.

    to make this consistent, current flows through the annoying wire carrying current I, making the top and bottom 10V.

    if you follow the arguments you should be able to see why the circuit 1 in the first picture has I=0.
    as a small (but tedious) exercise you can use kirchhoff's loop rule to find the current. (hint: 3 loops, one consisting on EMF, 2 others loop in the small rectangle divided by the wire I)

    Related: Wheatstone bridge.
     
  4. Apr 16, 2013 #3

    CWatters

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    Look at the symetry of the circuit. What does that suggest ?

    The horizontal link in the middle means that the voltage on each side is the same. Lets call that link "Node B". Looking at the bottom half of the circuit you have a 20Ω and a 10Ω connected between B and 0V. So whatever the voltage on node B is the current through the 10Ω will be twice that through the 20Ω. Similarly for the top half of the circuit. If you write I and 2I on the four resistors it should be obvious what has to flow in the horizontal branch (or at least it's direction).

    More generally NaOH is correct. A way to solve circuits is to mark arbitary direction arrows showing the direction you assume the current is going (and add voltage arrows that are consistent with your assumption for the current!). Write simultaneous equatons using KVL and KCL. Solve them. If a variable turns out to be negative then your assumption about the direction of current flow was wrong. You don't need to do anything. For example do not go back and change the arrows on your drawing otherwise an examiner will not be able to follow your working out.

    It is very very easy to make an error with the sign, either while marking up the drawing or when writing an equation using KVL or when solving the equations.

    The same one. The horizontal marked ".25ΩA"

    Regarding picture 2 Fig 14.11....

    I'm not sure why they included Figure 14.11 (picture 2) as thats confusing. Fig 14.11 is roughly equivalent to Circuit 2 in Fig 14.8 but there isn't a branch that is equivalent to the horizontal branch I. There is however an equivalent node.
     
  5. Apr 26, 2013 #4
    so, when the current from the 10ohm resister on the left first, it's more likely to go towards the second 10 ohm easier path, but when current goes from the 20ohm resistor on the top right, won't a small portion of current still have to go down to the 20 ohm resistor on the bottom right? So why won't the two currents clash since they're moving in opposite directions? or do they just cancel each other out?

    Yes, but picture 3 is the answer key, so I have no way of knowing the value of current flowing through any of the resistors...

    only the first picture is part of the question, pic2 and 3 are part of the answer key, just to clarify.
    thank you for the detailed explanation though.
     
  6. Apr 26, 2013 #5
    haha, did you just quote the answer key? That was exactly what I got from the explanation, and exactly what I didn't understand. the circuit does have symmetry, but...I'm thinking, that after the current passes through the top left resistor, it faces two equal paths...so the current may as well divide in half and go both ways. The current that comes from the top right resistor does the same thing. So in branch I, two equal currents in opposite directions meet. Does this mean they cancel out, or am I overthinking everything?

    so basically all empty branches can be compressed into nodes...is that correct? It would make calculations a lot easier, because if i squeeze branch I into a dot and then stretch it vertically, it'd be much easier for me to analyze the curcuit since it would be exactly like picture 2.
     
  7. Apr 28, 2013 #6

    CWatters

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    You could look at it that way. Another way would be to realise that by symmetry the voltage at each end of the wire linking left to right is the same. What I mean is the voltage will be the same even if the wire was removed. So there is nothing "pushing" current down that wire.

    Imagine if branch "I" had a resistor in it. Would that change anything?

    That is correct. Lines on circuits represent ideal conductors which can be treated as nodes.

    Yes that's exactly what you can do. In circuit 3 the approach would be to..

    1) squeese branch I into a node like circuit 2
    2) calculate the voltages and currents flowing in each resistor
    3) expand the node again and apply KCL to calculate the current I.

    As an exercise have a look at the resistor cube problem. How would you calculate the resistance between points A and B. All resistors have identical value R. Hint - It's a good example of applying symmetry to solving a problem..

    http://www.rfcafe.com/references/electrical/images/resistor-cube-kirt-1.gif
     
  8. May 3, 2013 #7
    I suppose it wouldn't change anything...right?

    but in a way...if, "there isn't anything 'pushing' the current down the wire", then the wire might as well not be there at all. So...instead of colapsing the nodes together, wouldn't taking the branch out entirely be another option?
    Of course, if i do that, i get two 10ohm and 20ohm resistors connected in parallel with two other 10, 20ohm resistors, which gives me a different equivalent resistor as circuit 2.
    So...where did I go wrong?
    (also, current does pass through nodes, and if i do "collapse the two nodes together", I'm taking for granted the fact that there is current flowing through I branch, which in fact isn't true. so how... )

    wow. ok. Symmetry.

    I tried. I really did. but...I didn't get anywhere. is the answer 3 ohms?...3-demensional problems...and I'm already getting confused with 2-d.
     
    Last edited: May 3, 2013
  9. May 4, 2013 #8

    CWatters

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    Correct.

    The first circuit 1 in Fig 14.8 is symetrical. What I mean is if you removed the horizontal link the voltages and current wouldn't change.

    In the following "/" means divide and "//" means "in parallel with".

    As drawn with the link you have

    (10//10) + (20//20)
    = 5 + 10
    = 15

    Without the link you have

    (10 + 20)//(10 + 20)
    = 30//30
    = 15

    Which is the same. So I have shown that in Circuit 1 of Fig 14.8 you can remove the link or change it to a high value resistor, either way it has no real effect on the circuit. Likewise if you have a circuit with several nodes at the same voltage you could connect them together with a wire and it would not effect the circuit. However it might make it a lot easier to analyse. So with that in mind perhaps take another look at the 3D cube of resistors problem. By symetry.. which nodes are at the same voltage? There might be more than one group of nodes (eg a few nodes all at voltage X and a few nodes all at voltage Y). You can add more than one wire but you can only connect together nodes that would be at the same voltage if there wasn't a wire.

    Now Circuit 2 in your figure 14.8 is a totally different circuit to Circuit 1 in 14.8. If you removed the link in circuit 2 the voltages on the nodes at each end would change. So you can't do that. You have to analyse it with the link in place.

    However there is still a certain symmetry to the circuit which you can see if you write an equation for the voltage on the I node...

    Using the potential divider rule I get...

    Vi = Vbat * (20//10) / {(20//10)+(20//10)}

    =Vbat * (20//10) / 2 * (20//10)

    = Vbat/2

    Perhaps not surprising given the symmetry?

    Now you can write equations for the current in each resistor. For example the current in the lower 20R would be Vi/20 and you have an equation for Vi. Likewise for the other currents which should calculate out as shown in the third image.

    To work out the current I you can apply KCL to a node at one end of the wire link.
     
  10. May 5, 2013 #9
    Hey! this suddenly makes a whole lot of sense! So...I gave another shot at the problem, and I got [itex]\frac{5}{6}[/itex]ohms. would that be closer to the answer?
    and so...if i have two points in the circuit at the same potential, I'd be able to connect the nodes, or take away the branch inbeween, without changing the total resistance, but I would be changing the way the resistors are connected (parallel-in series),is that right?

    Right! my mistake, I got so caught up in circuit 2 that everything I saw started blending and meshing together into the same question.

    Thank you so much for going out of your way to explain this problem to me so thoroughly, and even extending the problem along the way! This is the first time I've been able to really talk about physics with someone and not have them scoff at my lack of comprehension.
     
    Last edited: May 5, 2013
  11. May 5, 2013 #10

    CWatters

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    Correct. The circuit simplifies to

    3 in parallel + 6 in parallel + 3 in parallel
    = 1/3 + 1/6 + 1/3
    = 5/6

    As with all circuits you can also solve it by writing KVL and KCL equations and solving them but the potential for making a trivial error is high. There are 6 equations and 6 unknowns but the symetry helps a bit. They do it that way in here..

    http://qedinsight.files.wordpress.com/2011/06/resistor_cube_problem.pdf

    Did you see this circuit in another thread that looks tricky but turns out to be easy once you see it a different way. Again all are 1 Ohm.
     

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  12. May 5, 2013 #11

    CWatters

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    It's possible this has already be covered in your book but in case not it's worth a look at this wiki page..

    http://en.wikipedia.org/wiki/Wheatstone_bridge

    It explains one application/advantage of that original circuit (circuit 1 in 14.8).
     
  13. May 5, 2013 #12
    Wow...that first method was sure tedious...I'm glad you suggested symmetry before I did the problem.

    so...in this question, the total resistance is 1 ohm...I think...yeah.
    This is a cool way of looking at circuits, thanks so much for helping me clear up the confusion!
     
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