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Circuit Question

  1. Nov 20, 2007 #1
    View attachment circuit.bmp

    Hello everyone,

    I am trying to learn about electrical circuits for a physics class.

    I have attached a drawing of a circuit related to my question. The battery is 9 volts and each resistor has a 3 ohm resistance.

    I need to figure out the resistance through the middle resistor.

    The answer is stating it is 3 amps to the left, but I need help understanding this.

    Is this circuit considered a parallel curcuit?

    If so, the voltage should be shared between branches correct?

    Looking at the battery, the positive charge is on the left and the negative charge is on the right correct?

    Does the current start flowing from the positive charge? I thought I read that somewhere.

    Under these assumptions, I think the current is flowing from the battery left and upwards until it gets to the first branch. There I think there is a voltage drop (split in two) so it looks like 6 volts are going across the left resistor. Then, I would assume the voltage is split in two between the first and second resistor at that branch, so I am thinking 3 volts are across resistor two.

    Voltage = Current X Resistance

    So if the resistance is 3 ohms and the voltage is 3 volts, then wouldnt that make the current 1 ampere across the 2nd resistor? By my assumptions, it looks like current would be flowing to the right as well and the answer says to the left.

    I am really confused about this problem. I have put some serious thought into it. So I would appreciate if anybody can help me with this.

    Thanks in advance.

    Sincerely,
    Travis Walters
    admin@codebuyers.com
     
  2. jcsd
  3. Nov 20, 2007 #2

    stewartcs

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    Where's the drawing?
     
  4. Nov 20, 2007 #3
    Hey there,

    I attached it to the beginning of my post just now.

    I'd really appreciate any assistance with this. Thanks again!

    Sincerely,
    Travis Walters
     
  5. Nov 20, 2007 #4

    stewartcs

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    Is this a trick question or a typo?

    Try redrawing the circuit. It will be a lot clearer after that.

    Yes.

    Depends on which school of thought you buy into. Conventional flow theory (taught in most schools) says that it flow from the poisitive to negative. However, Electron flow theory says just the opposite (the US Navy teaches this).
     
  6. Nov 20, 2007 #5
    Hey there,

    Thats exactly what the curcuit looks like out of the text.

    The answer sheet has been known to have some wrong answers on it.

    I guess I will forward this question to the professor.

    I will let you know what comes out of it.

    Thanks again.

    Sincerely,
    Travis Walters
     
  7. Nov 20, 2007 #6

    Integral

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    If you study the drawing a bit you see that the only thing tricky is how it is drawn. Redraw, untangling the resistors, it is really very simple. (Note the relationship between each resistor and the battery.

    As for direction of the current, that will depend on the conventions used in your class. Are your current carriers postitive or negitive?
     
  8. Nov 20, 2007 #7
    Hey there,

    I found something in the class that says according to convention, the direction of electric current is defined as the direction positive charge is transported.

    I am pretty sure current is starting from the positive side of the battery.

    However, I am not sure what you mean by redrawing it. I have been looking at this drawing forever and I do not see the trick yet. Could you elaborate a little on how you would redraw it?

    Thanks again.

    Sincerely,
    Travis Walters
     
  9. Nov 20, 2007 #8

    Integral

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    Trace the path from the end of each resistor to the battery. What is in the path? Rearrange the resistors to show how they are connected.
     
  10. Nov 22, 2007 #9
    I have attached a re-draw of the circuit. Hope this helps. I solved it using three KVL loop equations one for each loop and ended up with three equations and three unknowns. I got 3A to the left or down in the redrawn circuit.
     

    Attached Files:

  11. Nov 24, 2007 #10
    Hey there,

    A fellow student responded that power follows the path with the least resistance.

    However, from my thinking, there are three paths that have equal resistance. So my new question is how is power shared among these paths? Is power shared equally if there are multiple paths of equal resistance? Or perhaps power travels the path with the shortest distance?

    The maximum current through the middle resistor would be 3 amps to the left. However, if power is shared equally among all resistors, would the power be 1 amp?

    Furthermore, if power takes the path with the shortest distance, then I think the first resistor in my initial drawing would get all the power.

    Please clarify these points so I can totally understand the concept behind this.

    Any help will be greatly appreciated. Thanks again.

    Sincerely,
    Travis Walters
     
  12. Nov 25, 2007 #11
    Please see the attached file. I have redraw and describe in there. They are all parallel; the Rtotal=1ohm.
     

    Attached Files:

  13. Nov 26, 2007 #12

    stewartcs

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    Current follows the path of least resistance.

    If you look at the redrawn circuit provided by ferncpi you'll see that there are three paths of equal resistance. This essentially means the same current will flow through each resistor (technically there will be a relatively small amount of extra resistance in the wires since they are of different length but this normally ignored since it is so small in relation to the other components).

    The current through each resistor is 3 amps, which is already shared. The total current in the circuit is 9 amps which was found using the Req and voltage...

    Itotal = V/Req

    It doesn't take the shortest distance, it take the path of least resistance. But like we've already discovered, the paths have equal resistance.

    Hope this helps.
     
  14. Dec 1, 2007 #13
    Hello again,

    First, you guys have been a really big help so far and I am very thankful :)

    Let me expand on this question a bit and get into some theory.

    Theoritically, lets say the resistance in the 2nd and 3rd resistors are higher than the 1st resistor. Does that mean ALL the current will go through the first resistor and none will go through the 2nd and 3rd resistor?

    I see that stewartcs stated that current follows the path with least resistance. I am just wondering if this means most of the current or all of it.

    If it is all of it, why does it do that?

    Thanks in advance for those who respond.

    Sincerely,
    Travis Walters
     
  15. Dec 1, 2007 #14

    Ouabache

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    Actually the current going through each resistor will be proportional to resistance. Here, the "water flowing through a pipe" analogy is a useful way to think about resistive circuits. The smallest resistor = largest diameter pipe, lets say the 2nd and 3rd resistors are of equal resistance but greater than the 1st. Both 2nd & 3rd resistors = smaller diameter pipe (because the 'resistance to flow' is greater, with decreasing diameter of conduit) When you turn on the source of water, the rate of flow (current) will be greatest through the large pipe (1st resistor) and less through each of the smaller pipes (2nd & 3rd resistors).

    I suspect he meant "more" current will flow through the path of least resistance.
     
  16. Dec 1, 2007 #15
    Hey there,

    Thanks a bunch to everyone who contributed!!!

    It really helped me understand circuitry a bit better.

    Now.. onto magnetism :)

    Sincerely,
    Travis Walters
     
  17. Dec 1, 2007 #16

    stewartcs

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    Yes I did. I just assumed that was implied.
     
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