First slide

combination of resistance

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

Easy

A

$\frac{R}{4}$
B

$\frac{R}{8}$
C

4R
D

$\frac{R}{2}$

Solution

When the wire is bent to form a complete circle, then
$2 πr = R \Rightarrow r = \frac{R}{2 π}$
Resistance of each semicircle
$= πr = \frac{πR}{2 π} = \frac{R}{2}$
Thus, net resistance in parallel combination of two semicircular resistances
$R^{'} = \frac{\frac{R}{2} \times \frac{R}{2}}{\frac{R}{2} + \frac{R}{2}} = \frac{\frac{R^{2}}{4}}{R} = \frac{R}{4}$

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