Help with Simple Resistor problem from physics resistor chapter

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SUMMARY

The discussion focuses on solving a resistor network problem involving series and parallel resistors. The user initially miscalculated the equivalent resistance, arriving at 17.14Ω instead of the correct 32Ω. The correct approach involves systematic reduction of the resistor network, where R1 and R2 are combined to form a new resistor R3, which is then analyzed in parallel with another R1. The final effective resistance is determined by adding R4 (the parallel combination of R3 and R1) to the horizontal R1 in series.

PREREQUISITES
  • Understanding of series and parallel resistor combinations
  • Familiarity with Ohm's Law
  • Basic knowledge of electrical circuits
  • Ability to perform algebraic calculations with resistances
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  • Study the systematic reduction techniques for resistor networks
  • Learn about equivalent resistance calculations in complex circuits
  • Explore advanced circuit analysis methods such as Thevenin's and Norton's Theorems
  • Practice problems involving series and parallel resistors to reinforce concepts
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Students studying physics or electrical engineering, educators teaching circuit analysis, and anyone looking to improve their understanding of resistor networks.

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Homework Statement




http://imageshack.us/a/img69/3384/rseistors.jpg


Homework Equations



Resistors in series and parallel formulas.


The attempt at a solution

Very confused with part (b)!

So in the first square or w/e you call it (left side), R1 in series with R1 so that's 20Ω + 20Ω = 40Ω. Then the second square (middle), R2 in series with R1 so 10Ω + 20Ω = 30Ω. then 40 and 30 are in parallel 40-1 + 30-1 = 7/120-1 = 17.14Ω.
But my solution Manuel has 32Ω as the resistor equivalent. I know I am making a mistake somewhere but i don't know what my mistake is.

help please!
 
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then 40 and 30 are in parallel...

Nope.

The 30Ω is only in parallel with one of the R1's not both.

It should be (30//20)+20
 
nchin said:

Homework Statement

http://imageshack.us/a/img69/3384/rseistors.jpg

Homework Equations



Resistors in series and parallel formulas. The attempt at a solution

Very confused with part (b)!

So in the first square or w/e you call it (left side), R1 in series with R1 so that's 20Ω + 20Ω = 40Ω. Then the second square (middle), R2 in series with R1 so 10Ω + 20Ω = 30Ω. then 40 and 30 are in parallel 40-1 + 30-1 = 7/120-1 = 17.14Ω.
But my solution Manuel has 32Ω as the resistor equivalent. I know I am making a mistake somewhere but i don't know what my mistake is.

help please!

When you get this kind of problem, you need to do systematic reduction of the resistor network. The third (rightmost) segment can be ignored, so just consider the left and middle.

Basically, when switch S2 is closed, that R1 and R2 in the middle segment are in series. The effective resistance of those two taken together (R1 + R2) can be represented by a new resistor called R3. This resistor R3 is in parallel with the vertical R1 in the left segment. So calculate the effective resistance of R3 and R1 in parallel, calling it R4 (symbolically represented as R3 || R1 = R4). Finally, R4 is in series with that horizontal R1 in the left segment - just add those up (R4 + R1) to get the final effective resistance.
 
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