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Bingel [31]
3 years ago
13

A lower coefficient of thermal conductivity along with a higher coefficient of thermal expansion and higher electrical resistanc

e are all characteristics of?
Engineering
1 answer:
HACTEHA [7]3 years ago
5 0

Answer:

stainless steel

Explanation:

Conductivity refers to the degree to which a specified material conducts electricity. It is the ratio of the current density in the material and the electric field.

The thermal conductivity of a material measures its ability to conduct heat.

In materials of low thermal conductivity, heat transfer occurs at a lower rate as compared to materials of high thermal conductivity.

Thermal expansion of the material refers to its tendency to change its shape, area, and volume as a result of change in temperature.

The electrical resistance of a material refers to the measure of its opposition to the flow of electric current.

<u>Stainless steel</u> has a lower coefficient of thermal conductivity along with a higher coefficient of thermal expansion and higher electrical resistance.

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If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less
liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19}  } =0.93ohm*cm

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })

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E = 11.68*8.85x10¹⁴ f/cm

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L = 10 - 0.335 = 9.665 um

The resistance is:

Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm

7 0
3 years ago
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