Find the net resistance in the second diagram (right in the centre of the image) with explanation?

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1 Answer
Apr 19, 2018

The equivalent resistance is #=7/12R#

Explanation:

There are #2# equilateral triangles.

The #6# resistors are equivalent.

Convert from #Delta# to #Y# configuration the #2# triangles

For each equilateral triangle

#R_a=(R*R)/(R+R+R)=R^2/(3R)=R/3#

#R_b=(R*R)/(R+R+R)=R^2/(3R)=R/3#

#R_c=(R*R)/(R+R+R)=R^2/(3R)=R/3#

Then,

You'll get #2# more triangles with sides #2/3R#, and #R/3# and #R/3#

Convert from #Delta# to #Y# configuration the #1# triangle

#R_a'=(R/3*2R/3)/(R/3+R/3+2/3R)=(2R^2/9)/(4/3R)=1/6R#

#R_b'=(R/3*2R/3)/(R/3+R/3+2/3R)=(2R^2/9)/(4/3R)=1/6R#

#R_c'=(R/3*R/3)/(R/3+R/3+2/3R)=(R^2/9)/(4/3R)=1/12R#

Then,

There is one group in parallel

#1/r=1/((1/6+1/3)R)+1/((1/6+1/3)R)=2/(R)#

#r=1/2R#

And finally

#R_(eq)=1/12R+1/2R=7/12R#