resistance between various points of resistor networks

Discussion in 'Homework Help' started by groundcontrol, Apr 22, 2016.

  1. groundcontrol

    Thread Starter New Member

    Sep 1, 2015
    24
    1
    Hi guys,
    I am stumped. I have tried inside out to solve the attached problem but I can not figure it out. It's driving me a little bit nuts as it looks like it couldn't be that hard... famous last words. I have attached several pics: The problem I can not solve is "Figure 6".

    I have been able to logically solve Fig1 thru Fig5. I have also put a few pics of my workings to the previous resistor networks in case my approach is incorrect but only now it is showing, and someone else can see it..

    The question is to calculate the resistance between points A and B, RAB.
    There's no mention of voltage at any stage, its just all about the resistance.

    I have included the answer to Fig 6 (80.5ohm). I have cross checked this with a simulation of the circuit, and that is the correct answer :) But how!!????


    I have ended up trying by changing the points of A and B to try and understand the circuit better, and to try and figure out why I cant solve this last one, what it is that's different about it that I do not understand.. but unfortunately I am still not able to figure out the correct way to solve it. I've attached one of those attempts too.


    I hope someone out there is up for shedding some light on this. I would really appreciate the help ! :)

    Thanks !!

    Question Fig 6.jpg
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    Re Figure 6. Temporarily remove the 100 ohm resistor. Can you calculate the resistance between A and B now? Since the 100 ohm is connected between A and B, what effect will it have on the resistance between A and B you calculated when it was not there?
     
    groundcontrol likes this.
  3. WBahn

    Moderator

    Mar 31, 2012
    17,726
    4,788
    You can slide any connection to anywhere else as long as it remains on the same node. With that in mind, redraw the circuit with the right hand connection of the 470 Ω moved down to the bottom to the left of the B label. Can you solve this circuit? If so, you are done, because it is the same circuit.
     
    groundcontrol likes this.
  4. grahamed

    Member

    Jul 23, 2012
    99
    11
    Hi

    470//330 =194

    194+200 = 414

    414//100 = 80.6
     
    groundcontrol likes this.
  5. WBahn

    Moderator

    Mar 31, 2012
    17,726
    4,788
    The idea is to guide the student along a path that lets them solve the problems on their own. Yet here, two approaches have been suggested and before the TS has had the opportunity to see if they are sufficient to let them now solve it, you feel the need to jump in and do their work for them. Why? Do you plan to do all their work for them throughout their entire career?
     
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  6. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,157
    if you redraw number six you end up with something like this

    redrawn.png

    or

    redrawn-1.png

    Does that look easier?
     
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  7. #12

    Expert

    Nov 30, 2010
    16,267
    6,778
    I'm with the, "redraw" crowd. Funny shapes trick the mind and fool the eyes. Some European schematics (Grundig brand) ran me crazy in my early days. I have had to re-draw many of them just to remove clutter or make a shape I could recognize. Get used to it. You will need it.
     
  8. groundcontrol

    Thread Starter New Member

    Sep 1, 2015
    24
    1
    Hi guys,
    First of all THANK YOU !! Nothing quite like the waive of realisation and learning something new :D

    Hello WBahn :) I did try your suggestion even though a more direct answer was given. I would love to say i could then figure it out as it makes me look better :p but even after moving the right leg of the 470 resistor to the left side of B, and even with making the 330 and the 470 clearly in parallel, i still was not able to solve it. I'm greatful to have done it though as it proved that the ohmmeter doesnt impact the configuration of the circuit kind of thing. And illustrated that if its the same point electrically that it doesn't mater where within that electrical connection you put the ohmmeter probe. You're right, I wasn't sure if you could you could move it like that :)

    I have to also say that grahameds answer really helped becasue he showed enough of the circuit analysis /maths for me to then to see what was actually going on, then have something correct to work backwards from. So it was a combo of your guidance and grahamed's workings as a way to go through and see the breakdown of the answer that made me able to fully understand whats going on. And learn :D Thank you to both of you !!

    My problem was not knowing what to do when there's a resistor in between points A and B. I can now understand that this put the two lots of resistors in parallel with each other. I have attacehd a pic of what I've learnt. THANK YOU BOTH HEAPS !!! :D

    Solved.JPG
     
  9. groundcontrol

    Thread Starter New Member

    Sep 1, 2015
    24
    1
    You're right, this is where the hole in my understanding is. So its in parallel with the rest of it. That's ho you do it :D Thanks for taking the time !!
     
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