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(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in length in which a current of 0.25 A passes in an axial direction. A voltage of 24 V is measured across two probes that are separated by 45 mm (1.75 in.).(b)Compute the resistance over the entire 57 mm (2.25 in.) of the specimen.

Respuesta :

Answer:

a) [tex]\sigma = 12.2[/tex] (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

[tex]\sigma = \frac{I l}{V \pi r^2}[/tex]

[tex]\sigma  = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}[/tex]

[tex]\sigma = 12.2 [/tex](Ω-m)^{-1}[/tex]

b)

[tex]Resistance = \frac{l}{\sigma A}[/tex]

                  [tex]= \frac{l}{ \sigma \pi r^2}[/tex]

[tex]= \frac{57  \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}[/tex]

Resistance = 121.4 Ω

jayilych4real

The electrical conductivity of the cylindrical silicon specimen is 12.2(Ω-m)^{-1}

The resistance over the entire 57 mm (2.25 in.) of the specimen is 121.4 Ω

Calculations and Parameters:

  • The diameter is 7.0 mm
  • Length 57 mm
  • Current I = 0.25 A
  • Voltage v = 24 v
  • The distance between the probes is 45 mm

To find the electrical conductivity, we would use the formula

(0.25 x 45 x 10^-3)/24[tex]\pi[/tex][(7 x 10^-3)/2]²

=12.2(Ω-m)^{-1}

To find the resistance, we would use the formula

[tex]1/\alpha \pi r^{2}[/tex]

=> 57 x 10^-3/12.2 x [tex]\pi[/tex][(7 x 10^-3)]²

Therefore, the resistance is 121.4 Ω

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