Answer to Question #187490 in Electric Circuits for Muzala Seleji

Question

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.

Answer 1

(1.) Here,

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

Drift velocity $v_d=2\times 10^{-4}\ m/s$

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

Using the formula,

Current flowing through a copper wire

$I=neAv_d$

$= (1.0\times 10^{29})\times(1.6\times10^{-19})\times(1.25\times10^{-6})\times(2\times10^{-4})$

$= 4\ Amperes$

Hence, Current flowing through a copper wire is 4 A

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

$A=\pi r^2=\pi(0.321\times 10^{-3}m)^2=3.24\times 10^{-7} \ m^2$

and the resistivity of Nichrome is $1.5\times 10^{-6}\ \ \Omega\cdot m$

So, the resistivity per unit length

$\dfrac{R}{l}=\dfrac{\rho}{A}=\dfrac{1.5\times 10^{-6}\ \ \Omega\cdot m}{3.24\times 10^{-7}\ \ m^2}=4.6\ \ \Omega/m$

(b) Current in wire $I=\dfrac{\Delta V}{R}=\dfrac{10V}{4.6\Omega}=2.2 A$

(3.) $R(150\degree C)=2.415 \Omega=R(0\degree C)(1+\alpha(150-0))\ \ \ ......(i)$

$R(20\degree C)=1.642\ \Omega=R(0\degree C)(1+\alpha(20-0))\ \ \ \ .....(ii)$

Substitute first equation in second and solve for $\alpha$

On Solving, we get

$\alpha=3.9\times 10^{-3}\ \degree C^{-1}$

and $R(0\degree C)=1.5236\ \ \Omega$

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

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

$1.5236Ω=\dfrac{\rho(0\degree C)\cdot 300m}{0.25\pi(2\times 10^{-3} m)^2}\Rightarrow \rho(0\degree C)=1.596\times 10^{-8}\ \Omega m$

and $\rho(20\degree C)=\rho(0\degree C)(1+\alpha\cdot 20)=1.72\times 10^{-8}\ \Omega m$