Calculate the resistance in the circuit

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Basel H
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Homework Statement


All resistors are ## R = 10 \Omega##
Resistance.PNG

Homework Equations


Calculate the total Resistance

The Attempt at a Solution


all resistors are parallel, so
$$ \dfrac{1}{R_{AB}} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} $$
## R_{AB} = \dfrac{R}{6}##
 

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Why do you think they are all in parallel? They arent, for example R2 and R3 are in series.

Hint: The circuit has some symmetry. If you applied a voltage to AB what would be the voltage across R6?
 
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Basel H said:

Homework Statement


All resistors are ## R = 10 \Omega##
View attachment 217281

Homework Equations


Calculate the total Resistance

The Attempt at a Solution


all resistors are parallel, so
$$ \dfrac{1}{R_{AB}} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} $$
## R_{AB} = \dfrac{R}{6}##
Look up 'balanced Wheatstone bridge'.
 
Basel H said:
all resistors are parallel ...
As has already been pointed out, this is incorrect. I strongly suggest that you review, several times, the definitions of serial and parallel. Until you have those down cold, you will continue to have problems with circuit analysis.
 
Since no current will flow through R6 you may delete it [R6].
 
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As @Babadag stated, you can ignore R6. But suppose it was not balanced (or you didn't recognize that). You could do a delta-Y transformation with R2 R4 & R6, to simplify things, if you have learned that.