Finding the equivalent resistance of a circuit!!

MrChips

Joined Oct 2, 2009
30,712
How would you write it?
R1 is in series with the rest

Req = R1 + [???]

R8 is in parallel with resistors on the left and resistors on the right. Do one at a time.

R11 = R2//R3
R10 = R8//( R11 + R4)

R9 = R7 + R10

See if you can write out the equation for Req using R5, R6, R9
Req = R1 + [ ???]
 

Thread Starter

AntonioDuarte2001

Joined Nov 1, 2020
47
R1 is in series with the rest

Req = R1 + [???]

R8 is in parallel with resistors on the left and resistors on the right. Do one at a time.

R11 = R2//R3
R10 = R8//( R11 + R4)

R9 = R7 + R10

See if you can write out the equation for Req using R5, R6, R9
Req = R1 + [ ???]
20201102_165245.jpg
Like this? I am not sure, i find this very difficult...
 

MrChips

Joined Oct 2, 2009
30,712
No.
R7 is not in parallel with R5 and R6.

Rather that trying to write it all out in one equation, do one step at a time and redraw the circuit using the combined equivalent resistor.

For example,
R11 = R2//R3

Redraw the circuit using R11 instead of R2 and R3.
 

WBahn

Joined Mar 31, 2012
29,978
Hi. thank you both for your help so far. I apologize for not responding faster. simplifying the circuit makes it easier. I think the equivalent resistance would look like this: (it looks a bit confuse)
View attachment 221251
So you are saying that R1 is in series with (R5||R6) and that this is then in series with R7.

Remember what it means for two resistors to be in series: Any and all current that flows through one of them must flow through the other.

So imagine replacing R5 and R6 with their parallel equivalent (let's call it R56). Is it possible for an electron that leaves the right side of R1 to travel a path through the circuit that does not involve going through R56? Sure! It could go through R1 and then over to R8 and down. So R1 is not in series with R56.

Try simplifying the diagram in stages. First, identify those resistors that are clearly in series or clearly in parallel and combine them and redraw the circuit using the combined components.

In this case, R5 and R6 are clearly in parallel, so replace them with R56. Also, R2 and R3 are clearly in parallel, so replace them with R23.

Now you have a circuit that has 6 resistors instead of the original 8. What can you do with R23 and R4 (to get R234)? Then redraw the circuit. Now what can you do with R234 and R8 (to get R2348)?

Keep doing this process until you have a single resistor (namely R12345678, using this naming convention).
 

Thread Starter

AntonioDuarte2001

Joined Nov 1, 2020
47
So you are saying that R1 is in series with (R5||R6) and that this is then in series with R7.

Remember what it means for two resistors to be in series: Any and all current that flows through one of them must flow through the other.

So imagine replacing R5 and R6 with their parallel equivalent (let's call it R56). Is it possible for an electron that leaves the right side of R1 to travel a path through the circuit that does not involve going through R56? Sure! It could go through R1 and then over to R8 and down. So R1 is not in series with R56.

Try simplifying the diagram in stages. First, identify those resistors that are clearly in series or clearly in parallel and combine them and redraw the circuit using the combined components.

In this case, R5 and R6 are clearly in parallel, so replace them with R56. Also, R2 and R3 are clearly in parallel, so replace them with R23.

Now you have a circuit that has 6 resistors instead of the original 8. What can you do with R23 and R4 (to get R234)? Then redraw the circuit. Now what can you do with R234 and R8 (to get R2348)?

Keep doing this process until you have a single resistor (namely R12345678, using this naming convention).
I think i understood now. Please see if its correct.


20201102_173751.jpg
 

ericgibbs

Joined Jan 29, 2010
18,766
hi A,

Check this
1604339448774.png

Update:
Is the last symbol a '1' or a poorly written Ohms symbol.?:rolleyes: Ω

So what is final equivalent numeric value.?
 
Last edited:

ericgibbs

Joined Jan 29, 2010
18,766
hi,
I agree the final is 11 Ohms, you have an error in your equation construction.

So that we can re-examine the circuit reduction, check this image against your incorrect equation
E
It maybe frustrating to be wrong, but keep trying.:)

E
 

Attachments

Thread Starter

AntonioDuarte2001

Joined Nov 1, 2020
47
hi,
I agree the final is 11 Ohms, you have an error in your equation construction.

So that we can re-examine the circuit reduction, check this image against your incorrect equation
E
It maybe frustrating to be wrong, but keep trying.:)

E
Yes, i forgot the "R7". Thank you!
In this circuit i think that R2//R3, R4//R5 and R7, R8, R9 and R10 are in serie.
What do you think?
20201102_213749.jpg
 

ericgibbs

Joined Jan 29, 2010
18,766
hi AD,
Have you now completed the 11Ω problem, if so, please post your final equation.

Will look over the new circuit Fig 8,e.
Post what you think the equation should be.

E

BTW:
The latest circuit you have posted is drawn correctly.
 

Attachments

Last edited:

Thread Starter

AntonioDuarte2001

Joined Nov 1, 2020
47
hi AD,
Have you now completed the 11Ω problem, if so, please post your final equation.

Will look over the new circuit Fig 8,e.
Post what you think the equation should be.

E

BTW:
The latest circuit you have posted is drawn correctly.
Hi.
Sorry for taking so long to answer.
The 11 ohm exercise, i understood it by taking little steps, the whole equation looks a bit confuse but i think this is right:
20201103_150013.jpg
The equation of the exercise 8e i think is like this:
20201103_150025.jpg
 
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