Help with Simple Resistor problem from physics resistor chapter

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The discussion revolves around solving a resistor network problem involving series and parallel configurations. The initial calculations incorrectly combine resistances, leading to confusion about the equivalent resistance. The correct approach involves systematically reducing the network by treating R1 and R2 in the middle segment as a single resistor in series, then calculating the parallel combination with the left segment's R1. The final effective resistance is found by adding the resulting parallel resistance to the remaining series resistance. This method clarifies the path to arrive at the correct equivalent resistance of 32Ω.
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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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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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