# Help with Simple Resistor problem! from physics resistor chapter!

1. Apr 4, 2013

### nchin

1. The problem statement, all variables and given/known data

http://imageshack.us/a/img69/3384/rseistors.jpg [Broken]

2. Relevant 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 thats 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 im making a mistake somewhere but i dont know what my mistake is.

Last edited by a moderator: May 6, 2017
2. Apr 4, 2013

### CWatters

Nope.

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

It should be (30//20)+20

3. Apr 4, 2013

### Curious3141

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.

Last edited by a moderator: May 6, 2017