Question

The resistance across $R$ and $Q$ in the figure.

• A. r/3
• B. r/2
• C. 2r
• D. 6r

Answer 1

Answer: (A)

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 $AC$ are in series and they together are in parallel with third one.
$\therefore R^{\prime }=(r/2)+(r/2)=r$
Total resistance
$\frac{1}{R}=\frac{1}{r}+\frac{2}{r}=\frac{3}{r}$
$R=\frac{r}{3}$