Question

A 1.50 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 15.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 C) the ammeter reads 18.8 A, while at 92.0 C it reads 17.4 A. You can ignore any thermal expansion of the rod.
1.Find the resistivity and for the material of the rod at 20 C. (rho= ? Ω*m)2.Find the temperature coefficient of resistivity at 20 C for the material of the rod.α= ? (C)^-1)

Answer 1

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) = $\pi r^2$

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:

$R=\frac{Voltage(V)}{I_0} \\R=\frac{15}{18.8}=0.798\Omega$

Therefore the resistivity and for the material of the rod at 20 °C (ρ) is:

$\rho=\frac{RA}{L}=\frac{0.798*\pi *0.275^2}{1.5}=0.126\Omega m$  

b) The temperature coefficient of resistivity at 20°C for the material of the rod (α) can be gotten from the equation:

$R_T=R_0[1-\alpha (T-T_0)]\\but,R_T=\frac{V}{I}=\frac{15}{17.4}=0.862$

Rearranging to make α the subject of formula:

$\frac{R_T}{R_0} =1+\alpha (T-T_0)\\\alpha (T-T_0)=\frac{R_T}{R_0}-1\\\alpha =\frac{\frac{R_T}{R_0}-1}{(T-T_0)} \\Substituting:\\\alpha =\frac{\frac{0.862}{0.798}-1 }{92-20} \\\alpha =\frac{0.0802}{72} =1.114*10^-3(^0C)^{-1$