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

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Answer 1 · 2018-04-19T12:50:56

The equivalent resistance is $=7/12R$

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

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