In the following star circuit diagram (figure), the equivalent resistance between the points $A$ and $H$ will be ..............$r$
Diffcult
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(b) Resistance of $CD$ $arm$ = $2r \cos 72^o = 0.62r$
Resistance of $CBFC$ branch
$\frac{1}{R} = \frac{1}{{2r}} + \frac{1}{{0.62r}} = \frac{1}{r}\left( {\frac{{2.62}}{{2 \times 0.62}}} \right)$
$\frac{1}{R} = \frac{{2.62}}{{1.24r}}$
Resistance of $CBFC$ branch
$\frac{1}{R} = \frac{1}{{2r}} + \frac{1}{{0.62r}} = \frac{1}{r}\left( {\frac{{2.62}}{{2 \times 0.62}}} \right)$
$\frac{1}{R} = \frac{{2.62}}{{1.24r}}$
$R = \frac{{1.24r}}{{2.62}}$
Equivalent $R' = 2R + r = 2 \times \frac{{1.24r}}{{2.62}} + r$
$ = r\,\left( {\frac{{2.48}}{{2.62}} + 1} \right) = 1.946r$
Because the star circuit is symmetrical about the line $AH$
Equivalent resistance between $A$ and $H$
$\frac{1}{{{R_{eq}}}} = \frac{1}{{R'}} + \frac{1}{{R'}}$ $==>$ ${R_{eq}} = \frac{{R'}}{2} = \frac{{1.946}}{2}r = 0.973r$
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