Question 130: Current Electricity

Question

A wire of length 25 m and cross-sectional area $5 \mathrm{~mm}^2$ having resistivity of $2 	imes 10^{-6} \Omega \mathrm{~m}$ is bent into a complete circle. The resistance between diametrically opposite points will be

Accepted Answer

Length of the wire, $L = 25m\,m$
CrossÃ¢ÂÂsectional area, $A = 5m\,mm^2 = 5	imes10^{-6}\,m^2$
Resistivity, $ho = 2	imes10^{-6}\,\Omega\cdot m$
Resistance of the entire wire:
$$R_{m total}=\frac{ho\,L}{A}    =\frac{2	imes10^{-6}	imes25}{5	imes10^{-6}}    =\frac{50	imes10^{-6}}{5	imes10^{-6}}    =10\ \Omega$$
When bent into a circle, points diametrically opposite divide the ring into two equal halves.
Each half-circle has length $L/2 = 12.5m\,m$, so its resistance is half of $R_{m total}$:
$$R_{m half}=\frac{R_{m total}}{2}    =\frac{10}{2}    =5\ \Omega$$
These two $5\,\Omega$ halves are in parallel between the opposite points.
Equivalent resistance $R$ is given by:
$$\frac{1}{R}=\frac{1}{R_{m half}}+\frac{1}{R_{m half}}    =\frac{1}{5}+\frac{1}{5}    =\frac{2}{5}$$
$$R=\frac{5}{2}=2.5\ \Omega$$
Answer: 2.5 ÃÂ© (Option B)

Answer

$100 \Omega$

Answer

$2.5 \Omega$

Answer

$12.5 \Omega$

Answer

$50 \Omega$

Current Electricity

A wire of length 25 m and cross-sectional area 5 \mathrm{~mm}^2 having resistivity of 2 \times 10^{-6} \Omega \mathrm{~m} is bent into a complete circle. The resistance between diametrically opposite points will be

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