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Q.
A wire is bent in the form of a triangle now the equivalent resistance $R$ between its one end and (he midpoint of the side is
Chhattisgarh PMTChhattisgarh PMT 2006
Solution:
Resistances $R,\, R$ and $\frac{R}{2}$ are in series,
the resultant resistance, $R'=R+R+\frac{R}{2}=\frac{5 R}{2}$
Resistances $R'$ and $\frac{R}{2}$ are in parallel.
Their resultant resistance
$\frac{1}{R}=\frac{2}{5 R}+\frac{2}{R}=\frac{2+10}{5 R}$
$R=\frac{5 R}{12}$