Need help with resistor network from hell - Page 1
Forum section: electronics / beginners

author: modalenjr
author_rank: Newbie
timestamp: November 19, 2015, 11:26:40 pm
content: Does anyone have a math formula to solve this one? (see attachment)

author: fivefish
author_rank: Frequent Contributor
timestamp: November 19, 2015, 11:31:07 pm
content: 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

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 19, 2015, 11:31:43 pm
content: Quote from: modalenjr on November 19, 2015, 11:26:40 pm23.3... kohmSpot the resistors in parallel and use the parallel formula:1/Req = 1/R1 + 1/R2 + ... + 1/RnThen spot the resistors in series and use the addition formula:Req = R1 + R2 + ... + RnTim

author: modalenjr
author_rank: Newbie
timestamp: November 19, 2015, 11:43:01 pm
content: Sorry i draw it wrong

author: Andy Watson
author_rank: Super Contributor
timestamp: November 19, 2015, 11:50:55 pm
content: Easy! Remove R2. Work-out the resulting network.Now put R2 back ... if you want - it makes no difference.

author: modalenjr
author_rank: Newbie
timestamp: November 19, 2015, 11:56:34 pm
content: Quote from: Andy Watson on November 19, 2015, 11:50:55 pmDon't mind 10K, Every resistor can have a different value.

author: wraper
author_rank: Unknown
timestamp: November 20, 2015, 12:02:18 am
content: For easier understanding the same thing drawn differently:

author: helius
author_rank: Super Contributor
timestamp: November 20, 2015, 12:03:51 am
content: 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__/

author: Andy Watson
author_rank: Super Contributor
timestamp: November 20, 2015, 12:16:16 am
content: Quote from: modalenjr on November 19, 2015, 11:56:34 pmYou 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.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 20, 2015, 02:00:10 am
content: For reducing circuits of this shape, you need nodal analysis (the sledge hammer), or if you prefer a graphical approach, applydelta-wye transformationsas mentioned byQuote from: Andy Watson on November 20, 2015, 12:16:16 amApplication 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

author: Syntax_Error
author_rank: Regular Contributor
timestamp: November 20, 2015, 03:03:24 am
content: Quote from: blueskull on November 20, 2015, 02:07:14 amIn 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.

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: November 20, 2015, 12:28:14 pm
content: You can use y-delta tramsform. Use this to turn r1,r2 and r5 intoa y.

author: Electric flower
author_rank: Regular Contributor
timestamp: November 20, 2015, 09:02:36 pm
content: 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 transformationSide 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