Question
When a wire of uniform cross-section $$a,$$ length $$l$$ and resistance $$R$$ is bent into a complete circle, resistance between two of diametrically opposite points will be
A.
$$\frac{R}{4}$$
B.
$$\frac{R}{8}$$
C.
$$4R$$
D.
$$\frac{R}{2}$$
Answer :
$$\frac{R}{4}$$
Solution :
When wire is bent to form a complete circle, then Total circumference, $$2\pi r = $$ Total length, $$R$$
$$ \Rightarrow r = \frac{R}{{2\pi }}$$
Resistance of each semicircle $$ = \pi r$$
$$ = \frac{{\pi R}}{{2\pi }} = \frac{R}{2}$$
Thus, net resistance in parallel combination of two semicircular resistances
$$R' = \frac{{\frac{R}{2} \times \frac{R}{2}}}{{\frac{R}{2} + \frac{R}{2}}} = \frac{{\frac{{{R^2}}}{4}}}{R} = \frac{R}{4}$$
When wire is bent to form a complete circle, then Total circumference, $$2\pi r = $$ Total length, $$R$$
$$ \Rightarrow r = \frac{R}{{2\pi }}$$
Resistance of each semicircle $$ = \pi r$$
$$ = \frac{{\pi R}}{{2\pi }} = \frac{R}{2}$$
Thus, net resistance in parallel combination of two semicircular resistances
$$R' = \frac{{\frac{R}{2} \times \frac{R}{2}}}{{\frac{R}{2} + \frac{R}{2}}} = \frac{{\frac{{{R^2}}}{4}}}{R} = \frac{R}{4}$$
