Question

please answer this?!..(chapter: electricity)

Answer 1

resistance = 26 ohm
length = 1m
area = π × ( 1.5 × 10^(-4) )^2 m^2
$Resistance =\frac{ \rho L}{A}\\ \\ \Rightarrow 26 = \frac{ \rho \times 1}{ \pi \times {(1.5 \times 10^{-4})}^2} \\ \\ \Rightarrow 26 = \frac{ \rho}{ \pi\times 2.25 \times 10^{-8}} \\ \\ \Rightarrow \rho = 26 \times \pi \times 2.25 \times {10}^{ - 8} \\ \\ \Rightarrow \rho = 26 \times \frac{22}{7} \times 2.25 \times {10}^{ - 8} \\ \\ \Rightarrow \rho = 26 \times \frac{22}{7} \times \frac{9}{4} \times {10}^{ - 8} \\ \\ \Rightarrow \rho = \frac{13 \times 11 \times 9 \times {10}^{ - 8} }{7} \\ \\ \Rightarrow \rho = \frac{1287 \times {10}^{ - 8} }{7} \\ \\ \Rightarrow \rho = 183.86 \times {10}^{ - 8} \\ \\ \Rightarrow \rho= 1.84 \times {10}^{ - 6} \: \Omega m$

Check the resistivity value with Table 12.2.

Make sure the temperature is 20℃. If it is not 20℃, then calculate resistivity at 20℃ using the temperature coefficient.

Answer 2

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resistance = 26 ohm

length = 1m

area = π × ( 1.5 × 10^(-4) )^2 m^2

$Resistance =\frac{ \rho L}{A}\\ \\ \Rightarrow 26 = \frac{ \rho \times 1}{ \pi \times {(1.5 \times 10^{-4})}^2} \\ \\ \Rightarrow 26 = \frac{ \rho}{ \pi\times 2.25 \times 10^{-8}} \\ \\ \Rightarrow \rho = 26 \times \pi \times 2.25 \times {10}^{ - 8} \\ \\ \Rightarrow \rho = 26 \times \frac{22}{7} \times 2.25 \times {10}^{ - 8} \\ \\ \Rightarrow \rho = 26 \times \frac{22}{7} \times \frac{9}{4} \times {10}^{ - 8} \\ \\ \Rightarrow \rho = \frac{13 \times 11 \times 9 \times {10}^{ - 8} }{7} \\ \\ \Rightarrow \rho = \frac{1287 \times {10}^{ - 8} }{7} \\ \\ \Rightarrow \rho = 183.86 \times {10}^{ - 8} \\ \\ \Rightarrow \rho= 1.84 \times {10}^{ - 6} \: \Omega m$

Check the resistivity value with Table 12.2.

Make sure the temperature is 20℃. If it is not 20℃, then calculate resistivity at 20℃ using the temperature coefficient.

HOPE SO IT WILL HELP.....