Question 411: Current Electricity

Question

A rectangular parallelopiped is measured as $$1 \mathrm{~cm} 	imes 1 \mathrm{~cm} 	imes 100 \mathrm{~cm}$$. If its specific resistance is $$3 	imes 10^{-7} ~\Omega \mathrm{m}$$, then the resistance between its two opposite rectangular faces will be ___________  $$	imes 10^{-7} ~\Omega$$.

Accepted Answer

The resistance of a material can be calculated using the formula:
$ R = ho \frac{L}{A} $
where
$R$ is the resistance,
$ho$ (rho) is the resistivity or specific resistance of the material,
$L$ is the length (or distance over which the resistance is being measured), and
$A$ is the cross-sectional area through which the current flows.
In the context of a rectangular parallelepiped, the "length" and "cross-sectional area" can vary depending on which faces of the shape are considered.
In this particular calculation, the resistance is being measured between the two smaller faces of the parallelepiped, which are squares of side length 1 cm:
The cross-sectional area $A$ should be:
$A = 100 \, 	ext{cm} 	imes 1 \, 	ext{cm} = 100 \, 	ext{cm}^2$
Converting this to meters gives:
$A = 100 \, 	ext{cm}^2 = 1 \, 	ext{m} 	imes 0.01 \, 	ext{m} = 0.01 \, 	ext{m}^2$
So, if we use these values for the length $L$ and cross-sectional area $A$ in the resistance formula, we get:
$ R = ho \frac{L}{A} = 3 	imes 10^{-7} \Omega \, m 	imes \frac{0.01 \, m}{0.01 \, 	ext{m}^2} = 3 	imes 10^{-7} \Omega $
Therefore, the resistance between the two smaller faces of the rectangular parallelepiped is $3 	imes 10^{-7} \Omega$.

Current Electricity

A rectangular parallelopiped is measured as $1 \mathrm{~cm} \times 1 \mathrm{~cm} \times 100 \mathrm{~cm}. If its specific resistance is 3 \times 10^{-7} ~\Omega \mathrm{m}, then the resistance between its two opposite rectangular faces will be ___________ \times 10^{-7} ~\Omega$.

Choose the correct answer: