What's the total resistance in this circuit?

[lbwet]

Homework Statement

https://imgur.com/3H3pCkD

R is the resistance of each resistor, determine total resistance in the circuit. (see the image in the link above)

Homework Equations

R=1/R₁+1/R₂+1/R₃+... (in parralel circuit)
R=R₁+R₂+R₃... (in series circuit)

The Attempt at a Solution

I'm assuming that flow of electricity starts from point A. R₁ and R₂ should be considered parallel because charge can go in R₁ or R², thus resistance in those two resistors: 1/Rₜₒₜₐₗ=1/R₁+1/R₂ ==> Rₜₒₜₐₗ=R/2.
R₃ and R₄ also should be considered parallel for the same reason, thus in those two resistors Rₜₒₜₐₗ=R/2. Now, solving for resistance in the whole circuit I get: Rₜₒₜₐₗ=R/2+R/2=R, but that answer does not seem to be correct. Any help will be appreciated!

 

[phinds]

R¹ and R² should be considered parallel

You really need to go back and get solid on exactly what series and parallel ARE. Without that knowledge, such problems will always be a mess for you but with it they become trivial.

 

[FactChecker]

I'm assuming that flow of electricity starts from point A. R¹ and R² should be considered parallel because charge can go in R¹ or R², thus resistance in those two resistors: 1/Rₜₒₜₐₗ=1/R¹+1/R² ==> Rₜₒₜₐₗ=R/2.

This is wrong. R² does not stand alone. If you consider what is in parallel with the R¹ resister, then it is the entire part of the network that includes R², R³, and R⁴. You need to approach this with a strategy where you can collapse simple parallel resisters into one and simple series resisters into one. Each step should be simple (all in parallel or all in series) and not a combination of parallel and series.

PS. Most people prefer that you use subscripts for the resisters because superscripts can be confused with powers.

 

[lbwet]

You really need to go back and get solid on exactly what series and parallel ARE. Without that knowledge, such problems will always be a mess for you but with it they become trivial.

Thanks for the tip. So "a" is branching location (the image again, PS: i couldn't upload local image, so i have to use off-site link), because charge can choose which branch will it flow, and also in each branch here should be one resistor otherwise they would be considered series, so that makes R₁ parallel to R₂, R₃ and R₄? Now another branching location aka node is "b" because charge can 'choose' to flow through resistor R₃ or R₄, thus they also should be parallel circuits. I think R₂ should be in series circuit with R₃ and R₄, because charge that will flow through R₂ will also flow in R₃ or R₄.

https://imgur.com/nQVYmK1

 

[Chandra Prayaga]

I am afraid your reasoning for what makes resistors in series and what makes them parallel is not correct. As phinds above says, you really should look at your textbook and figure that out first.

 

[lbwet]

This is wrong. R² does not stand alone. If you consider what is in parallel with the R¹ resister, then it is the entire part of the network that includes R², R³, and R⁴. You need to approach this with a strategy where you can collapse simple parallel resisters into one and simple series resisters into one. Each step should be simple (all in parallel or all in series) and not a combination of parallel and series.

PS. Most people prefer that you use subscripts for the resisters because superscripts can be confused with powers.

I edited subscripts. How can I collapse parallel/series resistors into one when single resistor is parallel and also series, depending resistor I consider it with.

 

[lbwet]

I think R₁, R₃ and R₄ are parallel, then we calculate total resistance of these resistors which is R/3, and now we are left with two 'main' resistors in series, we add them up and get R*4/3

 

[Chandra Prayaga]

Why do you think those three are parallel?

 

[jbriggs444]

I edited subscripts. How can I collapse parallel/series resistors into one when single resistor is parallel and also series, depending resistor I consider it with.

Here is an algorithm, off the top of my head.

First pass: Look for a wiring path that goes directly from one end of a resistor to its other end without passing through any other device. If there is such a path then that resistor is shorted. Remove it from the diagram. Keep looking at the remaining resistors.

Second pass: Look for devices that are connected on one end only. If there is such a device, it's an open circuit, doing nothing. Remove it from the diagram. Restart the second pass on the simplified diagram.

Third pass: Look for pieces of the network that are completely isolated from the piece you care about. No wires or devices going between the one part and the other. If found, remove the detached piece from the diagram and keep looking for other isolated pieces.

Fourth pass: Focus on a single pair of resistors at a time.

Look for a wiring path between one resistor and the other. Is there a wiring path that leads from one to the other that does not cross any other devices? If no such path exists then they are neither series nor parallel. STOP. Try another pair of resistors.

If such a path is found, then the pair might be either series or parallel. Keep considering. We know that the resistors are connected to each other at one end. Look at the opposite ends.

If a path exists connecting the opposite ends to one another then that's a parallel pair of resistors. Compute the equivalent resistance of the parallel pair and replace the pair with the equivalent single resistor. STOP. Restart the search for pairs of resistors in the new simplified diagram.

If no such path exists connecting the opposite ends then the pair still might be series. Look for other devices attached to the wire between the two ends that are connected to one another. If no other devices are connected on this path then the resistors are in series. Compute the equivalent resistance of the series pair and replace it with the equivalent single resistor. STOP. Restart the search for pairs of resistors in the new simplified diagram.

If there are other devices hanging off of the path between the connected ends then the pair is neither series nor parallel. STOP. Try another pair of resistors.

If you run out of resistor pairs, then the network is either down to one resistor (you win!) or will need to be simplified with more sophisticated methods.

 

[lbwet]

Why do you think those three are parallel?

Here is an algorithm, off the top of my head.

We take R₃ and R₄, which are parallel, compute resistance which is R/2, and replace with one resistor. Now this resistor is series with R₂, compute again and we get 3R/2. This resistor will be parallel with R₁, thus Rₜₒₜₐₗ=(1/3R/2+1/R)⁻¹=(2/3R+1/R)⁻¹=(5/3R)⁻¹=R*3/5.

 

[FactChecker]

We take R₃ and R₄, which are parallel, compute resistance which is R/2, and replace with one resistor. Now this resistor is series with R₂, compute again and we get 3R/2. This resistor will be parallel with R₁, thus Rₜₒₜₐₗ=(1/3R/2+1/R)⁻¹=(2/3R+1/R)⁻¹=(5/3R)⁻¹=R*3/5.

I haven't checked your calculations, but that is the right approach. Instead of looking at the top-level whole diagram, look for any little simple pieces and simplify them (R3 & R4 parallel => 1/2R first). Make pass after pass (R2 & 1/2R in series next), each time doing what you can to simplify little pieces. Eventually the whole diagram is one resistance.