Need help with resistor network from hell

• Need help with resistor network from hell
  Posted by modalenjr on 19 Nov, 2015 23:26
• Does anyone have a math formula to solve this one? (see attachment)

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• #1 Reply
  Posted by fivefish on 19 Nov, 2015 23:31
• R4 || R2 || R5 are in parallel. Compute equiv. resistance of these 3 parallel resistors, let's call it R2'.
  
  1/R2' = 1/R2 + 1/R4 + 1/R5 .... solve for R2'
  
  Then do R1 + R2' + R3
• #2 Reply
  Posted by T3sl4co1l on 19 Nov, 2015 23:31
• Quote from: modalenjr on 19 Nov, 2015 23:26

  Does anyone have a math formula to solve this one? (see attachment)
  

  
  23.3... kohm
  
  Spot the resistors in parallel and use the parallel formula:
  1/Req = 1/R1 + 1/R2 + ... + 1/Rn
  
  Then spot the resistors in series and use the addition formula:
  Req = R1 + R2 + ... + Rn
  
  Tim
• #3 Reply
  Posted by modalenjr on 19 Nov, 2015 23:43
• Sorry i draw it wrong

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• #4 Reply
  Posted by Andy Watson on 19 Nov, 2015 23:50
• Easy! Remove R2. Work-out the resulting network.
  
  Now put R2 back ... if you want - it makes no difference.
  
  
• #5 Reply
  Posted by modalenjr on 19 Nov, 2015 23:56
• Quote from: Andy Watson on 19 Nov, 2015 23:50

  Easy! Remove R2. Work-out the resulting network.
  
  Now put R2 back ... if you want - it makes no difference.
  

  
  Don't mind 10K, Every resistor can have a different value.
• #6 Reply
  Posted by wraper on 20 Nov, 2015 00:02
• For easier understanding the same thing drawn differently:
  

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• #7 Reply
  Posted by helius on 20 Nov, 2015 00:03
• In this case the key is that both sides of R2 are equipotential nodes. So no current will ever flow through R2.
  It's clearer if drawn this way:
  
       /--R1--+--R4--\
  >----+      R2      +---->
       \__R5__+__R3__/
  
• #8 Reply
  Posted by Andy Watson on 20 Nov, 2015 00:16
• Quote from: modalenjr on 19 Nov, 2015 23:56

  Don't mind 10K, Every resistor can have a different value.
  

  You didn't mention that!
  In that case: Label the ends of R2 as points A and B. Split the network at A and B such that R1, R2 and R5 form an isolated network. Use a delta-star transform on R1, R2 and R5. Now reconnect the star-network to the points A and B.
• #9 Reply
  Posted by T3sl4co1l on 20 Nov, 2015 02:00
• For reducing circuits of this shape, you need nodal analysis (the sledge hammer), or if you prefer a graphical approach, apply delta-wye transformations as mentioned by
  

  Quote from: Andy Watson on 20 Nov, 2015 00:16

  In that case: Label the ends of R2 as points A and B. Split the network at A and B such that R1, R2 and R5 form an isolated network. Use a delta-star transform on R1, R2 and R5. Now reconnect the star-network to the points A and B.
  

  
  
  Application of this will lead to equations with a lot of R / (R1 + R2 + R3), which after putting things together, probably looks a lot like what blueskull has. (I don't particularly care to check the result for correctness.. )
  
  Tim
• #10 Reply
  Posted by Syntax_Error on 20 Nov, 2015 03:03
• Quote from: blueskull on 20 Nov, 2015 02:07

  Dave will have a video on nodal analysis, mesh analysis and superposition theorem soon.
  

  
  In the meantime, you can look at the lectures here:
  
  http://www.digilentinc.com/Classroom/RealAnalog/
  
  Click show all and watch chapter 3 videos and the first video from chapter 4.
• #11 Reply
  Posted by TheUnnamedNewbie on 20 Nov, 2015 12:28
• You can use y-delta tramsform. Use this to turn r1,r2 and r5 intoa y.
• #12 Reply
  Posted by Electric flower on 20 Nov, 2015 21:02
• On first picture you see how to transform delta configuration of resistors into star (or Y)
  
  
  NOTE: If you had ditched R2 you would still get same result of 20Kohm beacuse all resistors are of the same value, but if they were different you would have to use delta to Y transformation
  
  Side question, if picture is worth thousand words, how much is worth picture of words?
  
  Edit: On the second picture where is correctet exponent on 100 should be 100*10^6, not 100*^5

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