Question #187490

1. Compute the current flowing through a copper wire of cross-sectional area 1.25×10−6 𝑚2 assuming that the drift velocity is 2×10−4 𝑚/𝑠 and the charge carrier density is 1.0×1029/𝑚3.

2. (a) Calculate the resistance per unit length of a 22-gauge Nichrome wire of radius 0.321 mm.

(b) If a potential difference of 10.0 V is maintained across a 1.00 m length of the Nichrome wire, what is the current in the wire?

3. A metal wire of diameter 2 𝑚𝑚 and length 300 𝑚 has a resistance of 1.642 Ω at 20 ℃ and 2.415 Ω at 150 ℃. Find the values of 𝛼0, 𝑅0 and 𝜌20.


1
Expert's answer
2021-05-03T10:31:38-0400

(1.) Here,

Cross- sectional area A=1.25×106m2A=1.25\times 10^{-6} m^2

Drift velocity vd=2×104 m/sv_d=2\times 10^{-4}\ m/s

Carrier Charge Density n=1.0\times 10^{29} \ \ \m3m^3

Using the formula,

Current flowing through a copper wire

I=neAvdI=neAv_d

=(1.0×1029)×(1.6×1019)×(1.25×106)×(2×104)= (1.0\times 10^{29})\times(1.6\times10^{-19})\times(1.25\times10^{-6})\times(2\times10^{-4})

=4 Amperes= 4\ Amperes

Hence, Current flowing through a copper wire is 4 A



(2.) (a) The cross-sectional area of the wire is

A=πr2=π(0.321×103m)2=3.24×107 m2A=\pi r^2=\pi(0.321\times 10^{-3}m)^2=3.24\times 10^{-7} \ m^2

and the resistivity of Nichrome is 1.5×106  Ωm1.5\times 10^{-6}\ \ \Omega\cdot m

So, the resistivity per unit length

Rl=ρA=1.5×106  Ωm3.24×107  m2=4.6  Ω/m\dfrac{R}{l}=\dfrac{\rho}{A}=\dfrac{1.5\times 10^{-6}\ \ \Omega\cdot m}{3.24\times 10^{-7}\ \ m^2}=4.6\ \ \Omega/m


(b) Current in wire I=ΔVR=10V4.6Ω=2.2AI=\dfrac{\Delta V}{R}=\dfrac{10V}{4.6\Omega}=2.2 A



(3.) R(150°C)=2.415Ω=R(0°C)(1+α(1500))   ......(i)R(150\degree C)=2.415 \Omega=R(0\degree C)(1+\alpha(150-0))\ \ \ ......(i)

R(20°C)=1.642 Ω=R(0°C)(1+α(200))    .....(ii)R(20\degree C)=1.642\ \Omega=R(0\degree C)(1+\alpha(20-0))\ \ \ \ .....(ii)

Substitute first equation in second and solve for α\alpha

On Solving, we get

α=3.9×103 °C1\alpha=3.9\times 10^{-3}\ \degree C^{-1}

and R(0°C)=1.5236  ΩR(0\degree C)=1.5236\ \ \Omega


With α known, use any of the above equations to get

R(0°C)=1.5236ΩR(0\degree C)=1.5236Ω . Then use R(0oC)=ρ(0°C)L/AR(0oC)=ρ(0\degree C)L/A

1.5236Ω=ρ(0°C)300m0.25π(2×103m)2ρ(0°C)=1.596×108 Ωm1.5236Ω=\dfrac{\rho(0\degree C)\cdot 300m}{0.25\pi(2\times 10^{-3} m)^2}\Rightarrow \rho(0\degree C)=1.596\times 10^{-8}\ \Omega m


and ρ(20°C)=ρ(0°C)(1+α20)=1.72×108 Ωm\rho(20\degree C)=\rho(0\degree C)(1+\alpha\cdot 20)=1.72\times 10^{-8}\ \Omega m


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Comments

reubenMushibwe
27.04.22, 11:34

Thank you very much for helping.

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