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Q.
The resistance across $R$ and $Q$ in the figure.
Current Electricity
Solution:
Two resistances of each side of triangle are connected in parallel.
Therefore, the effective resistance of each arm of the triangle would be
$=\frac{r \times r}{r +r}=\frac{r}{2}$.
The two arms $AB$ and $A C$ are in series and they together are in parallel with third one.
$\therefore R'=(r / 2)+(r / 2)=r$
Total resistance
$\frac{1}{R}=\frac{1}{r}+\frac{2}{r}=\frac{3}{r} $
$R=\frac{r}{3}$
