Trying to grasp a wiring concept

[Stuartambient]

Learning series - parallel circuits can be a bit confusing.  I know schematics can look different when you actually lay things out.  Hopefully the picture is clear enough.  Basically there are 2 pairs of resistors in parallel but in between them they are back to serial.   Putting it the way I did seems fine but not sure if I have the current going back to serial between them..

[Zero999]

The circuit on the right is a short circuit which will blow the fuse/breaker/battery/PSU.

[Stuartambient]

Ok thanks! Back to the drawing board.

[Mechatrommer]

the current will find the least resistance path.

[suicidaleggroll]

I'm not sure what you're trying to accomplish?  All you did is add a wire to short circuit each set of parallel resistors.

[Kilrah]

The left part of the picture seems to say everything that's needed... no idea what you mean with the 2nd one. Is that supposed to be your understanding of when the current goes? If yes, then no... just follow the paths on the left schematic, there's nothing more to it.

[Stuartambient]

I'm not sure what you're trying to accomplish?  All you did is add a wire to short circuit each set of parallel resistors.

The circuit is a simple demonstration of a series / parallel circuit.  In between the parallel resistors the energy is flowing back into a serial.
There should be positive going across one side of the resistors and negative on the other side.  What is confusing me is how to make it serial again in between the pair.  Maybe I'm still being confusing.

[Stuartambient]

The left part of the picture seems to say everything that's needed... no idea what you mean with the 2nd one. Is that supposed to be your understanding of when the current goes? If yes, then no... just follow the paths on the left schematic, there's nothing more to it.

The left is the circuit ,on the right my attempt at the wiring. The left to me isn't showing me much but a single connection between the two. If it was just 4 resistors in parallel I get it.  However it is the middle of them that has me stumped.

[suicidaleggroll]

The left one already shows you how it's wired.  The "inner" leg of all four resistors connects to the same place, there's nothing more to it than that.

There are three nodes in the system.  The battery + and one leg of two of the resistors connect to node A, one leg of all four resistors connects to node B, and the battery - and one leg of the last two resistors connects to node C.

[MarkF]

Think of it the same way you would analyze the circuit.

[tautech]

The left part of the picture seems to say everything that's needed... no idea what you mean with the 2nd one. Is that supposed to be your understanding of when the current goes? If yes, then no... just follow the paths on the left schematic, there's nothing more to it.

The left is the circuit ,on the right my attempt at the wiring. The left to me isn't showing me much but a single connection between the two. If it was just 4 resistors in parallel I get it.  However it is the middle of them that has me stumped.

Break it into the various components for simplicity.
Essentially it's just 2 resistors in series, the middle point can be thought of as the divided voltage.

Solve the 2 sets of paralleled resistors equations first.
http://www.electronics-tutorials.ws/resistor/res_4.html

[Mechatrommer]

and what you were trying to do is: (rightmost)
btw, how old are you?

[Stuartambient]

btw, how old are you?

Thanks but what do you ask my age?

[timb]

btw, how old are you?

Thanks but what do you ask my age?

Most likely so we'd know how in-depth the explanation should be. Throwing a bunch of equations at a 13 year old who hasn't take Trig yet might be frustrating, whereas a 20 year old in College might understand it better. So, if we know you're young (or older but not strong in the math department) we can explain the fundamentals first.

You should look into a program like iCircuit to help you visualize things. There's also a good JavaScript based circuit simulator that will run right in your browser.

Here's your initial schematic, simulated:



Now, we can simplify it by replacing the parallel pairs of resistors with their equivalent:



But wait! We can drill down even further by replacing the series pair with a single resistor:



Hopefully that helps you visualize it a bit better.

[Stuartambient]

Thank you for all the detail.  I had tried to attach what I worked out earlier but for some reason it wasn't posting. 
Your diagrams are way better and combined with my table (which hopefully now you'll see) might be a good combo for analysis.  I'm 60 and learning but not completely foreign to electronics.  Done some soldering, worked with a few boards and micro-controllers, just not at any deeper level.  Haven't fried anything yet or seen the magic smoke.  I'm going to try harder.

Math is rusty with no trig but some recent algebra review.  I'm making progress.  Funny thing is all the earlier replies still left me baffled, until I placed 4 resistors and 2 pieces of wire in front of myself and then... it all made sense.   

I added the pic of the circuit on the breadboard. Yellow to + Green to -

[rstofer]

Math is rusty with no trig but some recent algebra review.  I'm making progress.  Funny thing is all the earlier replies still left me baffled, until I placed 4 resistors and 2 pieces of wire in front of myself and then... it all made sense.   

I also learn better when I can see the results.

[Kirr]

Here is one possibly helpful tool for understanding resistor circuits : Resistor Network Solver. It computes the equivalent resistance between any two nodes of a resistor network.

For the OP's original network, it would be written as: "a 100 b a 250 b b 350 c b 200 c". Then specify nodes "a" and "c" as terminals. Also check the "Explain each step" checkbox - it will then explain how it's solved.

EDIT: Or just click this link to explore the original network.

[Stuartambient]

Here is one possibly helpful tool for understanding resistor circuits : Resistor Network Solver. It computes the equivalent resistance between any two nodes of a resistor network.

For the OP's original network, it would be written as: "a 100 b a 250 b b 350 c b 200 c". Then specify nodes "a" and "c" as terminals. Also check the "Explain each step" checkbox - it will then explain how it's solved.

EDIT: Or just click this link to explore the original network.

Hmmm, it showed me combined R3//R4 but not R1//R2 for some reason.  I'll have to play around with it some. The book I'm learning from uses a table approach to (so far) to analysis and also  SPICE to confirm the tallied findings. I haven't really tried SPICE yet but created an account here https://easyeda.com/ which is a SPICE derivative or so it seems.  I can't import Netlists though but can get one.  Once I figure out how to use it, it should be helpful.  I'll also look into Icircuit. 

[Kirr]

Hmmm, it showed me combined R3//R4 but not R1//R2 for some reason.  I'll have to play around with it some.

That's curious. Because to me it clearly shows that the network has 3 nodes and 4 resistors, that it is solved in 3 steps, and that the final resistance is 15300/77. Among the 3 steps used for solving, the first one combines your R1 and R2, the second one combines R3 and R4, and the last step combines the two resulting resistors, connected in series, into one. Do you see anything different there?

[Stuartambient]

I'll have to play around with it some.

That's curious. Because to me it clearly shows that the network has 3 nodes and 4 resistors, that it is solved in 3 steps, and that the final resistance is 15300/77. Among the 3 steps used for solving, the first one combines your R1 and R2, the second one combines R3 and R4, and the last step combines the two resulting resistors, connected in series, into one. Do you see anything different there?

Got it now.  Not sure what was going on earlier, sorry!  Curious why the solver does not compute a final value?

[Kirr]

Got it now.  Not sure what was going on earlier, sorry!

Great to hear, and no problem!

Curious why the solver does not compute a final value?

Simply because I haven't coded it yet. I'll probably add it soon.