Question

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

Answer 1

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`