Hello! I need help simplifying this electrical diagram. Thank you!

 

Have you covered star-delta transforms yet?
Or alternatively, start by looking which resistors are in parallel with others and simplify that way.

 

Hello! I need help simplifying this electrical diagram. Thank you!

I have included a cleaned up drawing of your circuit.

You should be able to collapse the network down to a single resistor working from right to left. You just need to combine them either in parallel or in series and that will accomplish your goal.

Here is the drawing:

 

Combine two and only two resistors at a time.
Combine two resistors in series by adding their resistance.
Combine two resistors in parallel by adding their conductance.



R1 and R2 are in parallel.
R3 and R4 are in series.

 

Hello again,

One thing i forgot to mention.
I dont see any resistor values given in the first post. For example, 10 Ohms, 100 Ohms 1.2k, etc. I only see the resistor labels R1, R2, R3, R4, etc.
If there are no values given this becomes a symbolic calculation which will involve every resistor in the network, and because they are connected together in a somewhat complex way, the resulting symbolic expression will be very complicated with a lot of characters even when factored and simplified. This means you probably have to use a math program that can handle symbolic algebra.

An alternate way however is to just keep lumping resistor combinations as you go, and renaming those lumped combinations with a new 'R' number.
For an example unrelated directly to the network in the first post, say we have R1 and R2 in series and then we have to put that set in parallel to R3. This would come out to (factored on way):
RT=((R2+R1)*R3)/(R3+R2+R1)

and now say we have another resistor R4 in series with that. The total would come out to:
RT=((R3+R2+R1)*R4+(R2+R1)*R3)/(R3+R2+R1)

when factored one way. This means every time we involve another resistor the final expression becomes more and more complex.
If instead we lumped the first calculation with R1, R2, and R3 we could define:
Ra=((R2+R1)*R3)/(R3+R2+R1)

and now when we add R4 we get:
RT=Ra+R4

a much simpler expression.
You do have to show all your definitions though or else nobody would know what Ra is and so the above would not really be a solution.

For the problem in the first post you would end up with several definitions:
Ra=...
Rb=...
Rc=...
.
.
Rn=...
then
RT=...

 

Hello again,

One thing i forgot to mention.
I dont see any resistor values given in the first post. For example, 10 Ohms, 100 Ohms 1.2k, etc. I only see the resistor labels R1, R2, R3, R4, etc.
If there are no values given this becomes a symbolic calculation which will involve every resistor in the network, and because they are connected together in a somewhat complex way, the resulting symbolic expression will be very complicated with a lot of characters even when factored and simplified. This means you probably have to use a math program that can handle symbolic algebra.

An alternate way however is to just keep lumping resistor combinations as you go, and renaming those lumped combinations with a new 'R' number.
For an example unrelated directly to the network in the first post, say we have R1 and R2 in series and then we have to put that set in parallel to R3. This would come out to (factored on way):
RT=((R2+R1)*R3)/(R3+R2+R1)

and now say we have another resistor R4 in series with that. The total would come out to:
RT=((R3+R2+R1)*R4+(R2+R1)*R3)/(R3+R2+R1)

when factored one way. This means every time we involve another resistor the final expression becomes more and more complex.
If instead we lumped the first calculation with R1, R2, and R3 we could define:
Ra=((R2+R1)*R3)/(R3+R2+R1)

and now when we add R4 we get:
RT=Ra+R4

a much simpler expression.
You do have to show all your definitions though or else nobody would know what Ra is and so the above would not really be a solution.

For the problem in the first post you would end up with several definitions:
Ra=...
Rb=...
Rc=...
.
.
Rn=...
then
RT=...

You just gave the answer away. How could it have been anything but Rab.

 

For example, you can

 

One wonders where the thread starter is.

 

One wonders where the thread starter is.

Homework was assigned on Friday - he/she will be back the evening before it's due.

unless it was due on Friday and he had to drop e class because he completely failed te test.

 

You just gave the answer away. How could it have been anything but Rab.

You can not state one value to solve this symbolically. That's because if you do state one value, you MUST include all the definitions that trickle back to every single resistor in the circuit. In other words, you also have to show the derivation of each lumped value.
For example, in a series circuit of R1, R2, and R3, we cant just say:
RT=Rab

we have to show somewhere where that Rab came from based on the actual resistors. This means we would have to show:
Rab=R1+R2+R3
RT=Rab

or if we did it in small steps:
Raa=R1+R2
Rab=Raa+R3
RT=Rab

WE can not just show RT=Rab because we cant know where Rab came from by looking at that alone the defining lumped resistances must be shown with it.

 

For example, you can
View attachment 262605

Hi,

I must have missed this post earlier.

Even though we still have to do some algebraic work on that expression yet, that's a neat way to show how this network problem can be solved.

We can also save on parentheses making it a little simpler to read out by assigning a higher precedence to the symbol for the parallel operator than for the symbol for addition:
RT=((((R3+R4)||R5+R9)||R6+R10)||R7+R2)||R8+R1+R11

If we renumber the resistors we could write it out as:
RT=((((R1+R2)||R3+R4)||R5+R6)||R7+R8)||R9+R10+R11

Alternately, if we use a different numbering we can write it out as:
RT=((((R1+R3)||R2+R4)||R5+R7)||R6+R8)||R9+R10+R11
where we work with an enumeration plus the second next enumeration in parallel with the next available enumeration with the second next added to that, etc. The end is different because we have two in series not just one. This is not quite as neat as the 1 through N enumeration though.

The form comes out quite large when all that is evaluated. The simplest is probably a graphic not a string of text, where we have multiple division levels.

We could also collapse the network from left to right starting with a voltage source between A and B, starting with:
V2=V1/(R1+R11)*(R1+R11)||R8
where V1 is the voltage source and V2 is the next voltage down the line.
We'd have to define the parallel symbol to have higher precedence than the multiplication and division operations also for this one, and of course multiplication is higher than addition. So for the above we would start out with (R1+R11)||R8.

It is interesting also that the parallel operator comes up now and then in these discussions even though i dont think it is a formal math thing.

 

R_AB = V_AB / I1 ≈ 123.8185 V / 1 A ≈ 123.8185 Ω
View attachment 262870

Hi,

Where you getting the resistor values from?

 

Hi,
Where you getting the resistor values from?

In post #1:
R_n+1 = R_n + 5 Ω
R4 = 37 Ω

So answer is 123.45 Ω

 

Gosh, it is there. I didn't realize we had to do photo editing to find the secret resistor value decoder.

 

In post #1:
R_n+1 = R_n + 5 Ω
R4 = 37 Ω
View attachment 262877View attachment 262895
So answer is 123.45 Ω

You mean this:



How did you see that text in the drawing? It shows up on my monitor as a big black box.
Your image does too.
I had to pull the text in my image above out of the shadows
The image in the first post is probably the worst image i have ever seen on this forum to date.
Can you see the text in my image above or is it too light on your monitor?

 

How did you see that text in the drawing? It shows up on my monitor as a big black box.
Your image does too.
I had to pull the text in my image above out of the shadows
The image in the first post is probably the worst image i have ever seen on this forum to date.
Can you see the text in my image above or is it too light on your monitor?

I always tune monitors to see full gamma of hues.
I see TS's image and your light variant of image very well.
My monitor has brightness 30% and contrast 50%.

 

I always tune monitors to see full gamma of hues.
I see TS's image and your light variant of image very well.
My monitor has brightness 30% and contrast 50%.

Oh ok thanks, i even had trouble seeing it when i turned my brightness up.
Well thanks for posting that i would have never known

 

Hello, forgot to do my homework, teacher gave me a little bit of time to do it, i don't quite get it. Can you help me?

 

In order to get help with a problem you need to show us what part of it you understand. How would you approach this problem?

 

In your previous thread @MrSalts showed you way to solve your problem:

For example, you can

Definitely, you can solwing your last problem by the same way:
R = R1 + (R2 || (R3 || R4))