Respuesta :
Answer:
α = 1.114 × 10⁻³ (°C)⁻¹
Explanation:
Given that:
Length of rod (L) = 1.5 m,
Diameter (d) = 0.55 cm,
Area (A) = [tex]\pi r^2[/tex]
Radius (r) = d / 2 = 0.275 cm,
Voltage across the rod (V) = 15.0 V.
At initial temperature (T₀) = 20°C, the current (I₀) = 18.8 A while at a temperature (T) = 92⁰C, the current (I) = 17.4 A
a) The resistance of the rod (R) is given as:
[tex]R=\frac{Voltage(V)}{I_0} \\R=\frac{15}{18.8}=0.798\Omega[/tex]
Therefore the resistivity and for the material of the rod at 20 °C (ρ) is:
[tex]\rho=\frac{RA}{L}=\frac{0.798*\pi *0.275^2}{1.5}=0.126\Omega m[/tex]
b) The temperature coefficient of resistivity at 20°C for the material of the rod (α) can be gotten from the equation:
[tex]R_T=R_0[1-\alpha (T-T_0)]\\but,R_T=\frac{V}{I}=\frac{15}{17.4}=0.862\\[/tex]
Rearranging to make α the subject of formula:
[tex]\frac{R_T}{R_0} =1+\alpha (T-T_0)\\\alpha (T-T_0)=\frac{R_T}{R_0}-1\\\alpha =\frac{\frac{R_T}{R_0}-1}{(T-T_0)} \\Substituting:\\\alpha =\frac{\frac{0.862}{0.798}-1 }{92-20} \\\alpha =\frac{0.0802}{72} =1.114*10^-3(^0C)^{-1[/tex]
The answer to the questions are:
1. The resistivity for the material of the rod at 20 °C (ρ) is
[tex]1.26378\times 10^{-5} \Omega m[/tex] .
2. The temperature coefficient of resistivity at 20 °C for the material of the
rod(α) is [tex]1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}[/tex].
Given to us:
Voltage across the rod, V = 15.0 V
Length of rod, L = 1.5 m = 150 cm,
Diameter, d = 0.55 cm,
[tex]Radius, r= \frac{d}{2}=\frac{0.550}{2}=0.275\ cm[/tex]
[tex]\begin{aligned}Area, A&= \pi r^2\\&=\pi\times (0.275)^2\\&=0.075625\pi\ cm^2\\\end{aligned}[/tex]
Initial temperature, [tex]T_o=20.0\ ^oC[/tex]
current at [tex]T_o[/tex], [tex]I_o= 18.8\ A[/tex]
Final temperature, [tex]T_1=92.0\ ^oC[/tex]
current at [tex]T_1[/tex], [tex]I_1=17.4\ A[/tex]
1.) To find out the resistivity of the rod(ρ),
Resistant of the rod(R),
[tex]\begin{aligned}\\R_o&=\frac{Voltage}{Current(I_o)}\\&=\frac{15.0}{18.8} \\&=0.7979\ \Omega \\\end{aligned}[/tex]
Resistivity of the rod at [tex]20^o\ C[/tex](ρ),
[tex]\begin{aligned}\\\rho&=\frac{RA}{L}\\&=\frac{0.7979\times 0.075625\pi}{150}\\&=0.00126378\ \Omega cm\\&=1.26378\times 10^{-3} \Omega cm\\&=1.26378\times 10^{-5} \Omega m\\\end{aligned}[/tex]
Hence, the resistivity for the material of the rod at 20 °C (ρ) is [tex]1.26378\times 10^{-5} \Omega m[/tex] .
2.) To find out the temperature coefficient of resistivity at 20°C for the material of the rod (α) can be gotten from the equation,
[tex]{R_1} ={R_o}[ 1+\alpha(T_1-T_o)][/tex]
we need resistant of the rod([tex]R_1[/tex]),
[tex]\begin{aligned}\\R_1&=\frac{Voltage}{Current(I_o)}\\&=\frac{15.0}{17.4} \\&=0.862\ \Omega \\\end{aligned}[/tex]
Now, solving to get the value of α
[tex]\begin{aligned}{R_1} &={R_o}[ 1+\alpha(T_1-T_o)]\\0.862&=0.7979[1+\alpha(92-20)]\\\frac{0.862}{0.7979}&= [1+\alpha(72)]\\1.0804&=[1+\alpha(72)]\\0.0804&=\alpha(72)\\\alpha&=0.0011169\ ^oC^{-1}\\\alpha&=1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}[/tex]
Hence, the temperature coefficient of resistivity at 20 °C for the material of the rod(α) is [tex]1.1169\times10^{-3}\ ^oC^{-1}\end{aligned}[/tex].
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