# How can I simplify it? equivalent resistance

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#### riko

Jan 21, 2021
5

Moderators note : shown image full size

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Moderator
Nov 8, 2019
3,156

#### Martaine2005

May 12, 2015
4,783
Start from the far right. Calculate resistors in series and replace those with a single resistor value. Then working left, calculate the resistors in parallel and replace them with a single value.

Martin

#### riko

Jan 21, 2021
5
R57 = R5 + R7
1/R576 = 1/R57 + 1/R6
R5764 = R576 + R4
R12 = R1 + R2
1/R123 = 1/R12 + 1/R3
R1234567 = R5764 + R123

Like that?

#### bertus

Moderator
Nov 8, 2019
3,156
Hello,

That is not all correct.
Have a look at the schematic when we draw it a little different:

Can you now tell wich resistor is in series and wich in parallel?

Bertus

#### riko

Jan 21, 2021
5
(R3,R4,R6,R57) = x = parallel, then series = x + R2 and x + R2 are parallel if think good

Moderator
Nov 8, 2019
3,156
Hello,

Bertus

#### riko

Jan 21, 2021
5
Oh i forgot to write it. (x + R2) = y, y and R1 are parallel

#### bertus

Moderator
Nov 8, 2019
3,156
Hello,

That is correct when R57 is R5+R7.

Bertus

#### riko

Jan 21, 2021
5
Yes!
Thank you very much for your help

#### TommyN

Jan 21, 2021
1
View attachment 50625

Moderators note : shown image full size

Riko, there appear to be three nodes which are disguised by the schematic layout HINT

Another way of looking at the network of resistors may quickly simplify how the resistances in the series and parallel configuration they are shown in. Hope that helps TommyN

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#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,476
That is what @bertus did in post #5.

Jan 21, 2023
1

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,476
@equ : This thread will soon be 2 years old without any further contribution -> locked

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