shorts

Discussion in 'General Electronics Chat' started by pilotnmech, Feb 26, 2005.

  1. pilotnmech

    Thread Starter Member

    Feb 26, 2005
    27
    0
    This may seem like a stupid question for someone who's been in electronics for a long time, but I need to understand this. How do you know what the effect of a short will be on a resistor? I am using Thevenin's theorem to figure out load voltage and current, and I have a problem where I've removed the load on the right side (verticallly drawn resistor) and am trying to figure out total resistance. To check the resistance I am starting from the top right side and I have a point where there is one vertically drawn resistor, a horizontally drawn resistor, then another vertically drawn resistor, then the voltage source has been replaced by a short (which means I have 2 parallel,and a series resistor, and then a short in parallel with the parallel resistors.) The book says that the resistor closest to the short is shorted out, but the other parallel resistor isn't. How can that be? How do the electrons know that there is less resistance at the other end? I thought some electrons will go through any resistance. Shorts make sense when there is a direct path back to the source, but not when there are resistances in between. The resistors are R1-10 ohms (first vert. resis. on the left), R2-20 ohms (horiz. resis.), R3-20 ohms (vert. resis. on the right) Thanks for your help.
     
  2. haditya

    Senior Member

    Jan 19, 2004
    220
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    can u post a schem(with the original source and everything)...draw it in paint and save it as a jpeg file....

    moreover everthing and anything takes the path of least resistance...so if we assume than conducting wire hav zero resistance the work any electrons has to do to get across the short(which we regard as a perfectly conducting wire) is zero and hence electrons like any a component in other system in the universe will prefer that path...

    however in prctice wire do hav a small(very small ) reistance...and when a short is made across a resistor the currnet is divided such that more than 99% of the current flows thru the short and current of a few nano-amps might flow thru the resistor
    the math also explains the phenomema i=v/r;r=0 then i =inf.
     
  3. n9xv

    Senior Member

    Jan 18, 2005
    329
    1
    Yea try to draw it it paint/paintbrush. It should save/upload directly as the .bmp.

    I remember thevinins theorem, and I think I know whats puzzling you. But would be clearer if you could draw. You dont have to be fancy about it, just draw lines (representing wires) into and out of the label i.e. ----- R1 ---- R2 ----- etc.

    Just pointing something out;

    Generally speaking a very small resistance would mean a large current flow, something much above nano or pico amps.
     
  4. haditya

    Senior Member

    Jan 19, 2004
    220
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    oops!! sorry thats a mega blooper on my part...i ll hav that post edited...thx n9xv
     
  5. n9xv

    Senior Member

    Jan 18, 2005
    329
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    No sweat. I think I knew what you really meant. The mind works faster than the keyboard :) .
     
  6. pilotnmech

    Thread Starter Member

    Feb 26, 2005
    27
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    After I wrote my post last night I was explaining it to my wife and I noticed the answer on that problem (because the electrons had two equal resistors to go through first), but how would it work if they weren't equal, and there was a short at the other end? I'll put in a drawing if I can figure out how.
     
  7. pebe

    AAC Fanatic!

    Oct 11, 2004
    628
    3
    Are you describing two resistors in series making a potential divider across supply rails, with a resistor coming from their junction?
     
  8. pilotnmech

    Thread Starter Member

    Feb 26, 2005
    27
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    Well, I understand how it works if the electrons are coming in at the top and facing two equal 20 ohm resistors, but what if the electrons are coming in the bottom and they go through the 20 first, then the 10, and then the short. How do you know exactly what's shorted? It will have to go through R2 to get shorted, but if the electrons take the path of least resistance, it seems like R1 and R3 would be shorted. (See attachment). Sorry I'm stupid.
     
  9. pebe

    AAC Fanatic!

    Oct 11, 2004
    628
    3
    You are connecting the LH side of R2 to the bottom of R3, so R2 and R3 are in parallel and their combined resistance viewed from the RH terminals is 10ohms.

    The formula is 1/Rcombined = 1/R2 + 1/R3

    But in your original posting you mentioned Thevenin'd theorum. How is that involved?
     
  10. pilotnmech

    Thread Starter Member

    Feb 26, 2005
    27
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    I had removed the load and shorted where the battery was.
     
  11. n9xv

    Senior Member

    Jan 18, 2005
    329
    1
    The Thevinin resistance at the open end (accross R3) is indeed 10-Ohms. Just think of the shorted resistor as being invisable. Electrically it does not exist. In thevinins Theorem you mentally "short" The supply and compute the resistance backwards toward the Thevinized or "loaded" end. Which is what you apparently have done. As opposed to the standard method of computing resistance at the point farthest from the supply at the last point where current splits. Not a bad drawing of resistors by the way!
     
  12. haditya

    Senior Member

    Jan 19, 2004
    220
    0
    a short can be defined as a connection (between 2 or more points in a circuit) such that the 2 points have the same electric potential

    now back to the circuit...no matter how the electrons enter either from top or from bottom they ll by-pass the 10 ohm resistor..this is beacause the potential difference across the 10 ohm reistor is 0 V....both ends of the reistor are at the same potential owing to the short and hence their difference is zero.

    so what actually happens is that all current is divided into 2 and one part of it passes thru 20 ohms and the other part passes thru the short and goes thru the other 20 ohm resistor..

    so as n9xv puts it that 10 ohm reistor in the circuit show is as good as non existent(from an electrical point of view)...

    now if the resistors r2 and r3 were unequal the electrons would divide such that the potential difference across the the two is the same...ie i*r in one branch is equal to the product i*r in the other branch....therfore greater current would flow thru the lesser resistive branch..

    now potential difference is defined as the work done in moving a unit charge it from one point in a circuit to another...so greater no of charges can be moved across a smaller resistance in unit time
     
  13. pilotnmech

    Thread Starter Member

    Feb 26, 2005
    27
    0
    So what you're saying is that the electrons may start to go through the 10 ohm resistor, but moving at 186,000 mi. per sec. they figure out it's easier to go around it so the potential difference (voltage drop ) becomes nothing, and if the 20 ohm resistor weren't there R3 would be shorted too? I'm trying to understand this from an electrons point of view. Why wouldn't R3 be bypassed also? The path of least resistance is through the short. Please don't be upset with me.
     
  14. n9xv

    Senior Member

    Jan 18, 2005
    329
    1
    Hey, no question is a stupid one. Fearing to ask is to reject knowledge, or something like that. :) .

    From the currents point of view, the bottom of R2 is connected to the bottom of R3. If you were to take out the "slack" and redraw the circuit (minus R1 due to it being shorted) you can easily see that R2 is in parellel with R3 as per the provided drawing. As far as the path of least resistance thing goes, try the water analogy. Imagine water flowing (current) through a pipe. Then at some point the pipe splits off into two parellel branches and then recombines again at the other end. We'll call the branches the "left" side and the "right" side. If the left side had a big nasty hair ball (resistance) in it and the right side was perfectly clear of obstructions (wire short), the water would easily flow through the unobstructed pipe (shorted path).

    The current is not doing anything "intelligent" in the above example. What happens is this;

    The electrons can only flow through the path provided. Some may flow through the resistance in inverse proportion to the resistance. The electrons dont "turn around" and go the other way, they simply continue to flow in any possible path that will allow them to pass. In the above example using the water analogy, replace the nasty hair ball (resistor) with an adjustable water valve (potentiometer). If the valve is shut off, then current flows right up to the valve but cant go past it.
     
  15. haditya

    Senior Member

    Jan 19, 2004
    220
    0
    yes pilot...if r2 werent there and replaced by a wire instead....r3 wud be bypassed too and the entire circuit would function as if there were no resistance.
    remember a short is implies that the points it connects have the same potential.
    and for current to flow there must be a non zero potential difference and finite resistance(by finite i mean not infinite)
     
  16. mozikluv

    AAC Fanatic!

    Jan 22, 2004
    1,437
    1
    hi

    i believe what we have here is an attenuator in a pi-pad configuration. have also attached a schem. hope it would help in the analysis :)
     
  17. Victor Beard

    Member

    Mar 20, 2005
    11
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    How much voltage does the circuit have.
     
  18. Victor Beard

    Member

    Mar 20, 2005
    11
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    How much voltage does the circuit have,what,s the load value in resistance.
     
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