Help with Simple Resistor problem from physics resistor chapter

In summary: When switch S2 is open, the two resistors in the left and middle segments are in parallel. The effective resistance of those two taken together (R1 + R2) can be represented by a new resistor called R5. This resistor R5 is in series with the vertical R1 in the right segment. So calculate the effective resistance of R5 and R1 in parallel, calling it R6 (symbolically represented as R5 || R1 = R6). Finally, R6 is in series with that horizontal R1 in the right segment - just add those up (R6 + R1) to get the final effective resistance.When switch S2 is closed, the two resistors in the right
  • #1
nchin
172
0

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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  • #2
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
 
  • #3
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.
 
Last edited by a moderator:

1. What is the purpose of resistors in a circuit?

Resistors are used to control the flow of electric current in a circuit. They resist the flow of electrons, which can help regulate the amount of current and prevent damage to other components in the circuit.

2. How do I calculate the resistance of a resistor?

The resistance of a resistor can be calculated using Ohm's law: R = V/I, where R is resistance in ohms, V is voltage in volts, and I is current in amps. You can also use the color code on the resistor to determine its resistance value.

3. What is the difference between series and parallel resistors?

In a series circuit, resistors are connected end-to-end, and the total resistance is the sum of each individual resistance. In a parallel circuit, resistors are connected side-by-side, and the total resistance is calculated using the formula 1/Rtotal = 1/R1 + 1/R2 + ... + 1/Rn.

4. Can resistors be used in AC and DC circuits?

Yes, resistors can be used in both AC (alternating current) and DC (direct current) circuits. However, the resistance may vary in AC circuits due to the changing direction of current flow.

5. What is the power rating of a resistor and how is it determined?

The power rating of a resistor is the maximum amount of power it can dissipate without getting damaged. It is usually measured in watts and can be determined using the formula P = V^2/R, where P is power in watts, V is voltage in volts, and R is resistance in ohms.

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