Answer:
ρ(20°C) = 9.7 * 10^-6 Ω
m
α = 0.0013 /°C
Explanation:
<u>Step 1:</u> Data given
length of the rod = 1.70 meters
diameter = 0.55 cm
Potential difference = 13.0 V
Temperature = 20°C →18.7 A
At 92°C → 17.1 A
<u>Step 2:</u> Calculate the resistivity
ρ = (RA)/L
⇒ with R = the resistance of the rod
⇒ R = V/I0 = 13/18.7 = 0.695 Ω
⇒ with L = length in meters = 1.70 m
⇒ with A = cross sectional area in m^2 = π * r²
⇒ r = d/2 = 0.550/2 = 0.275 cm = 0.00275m
ρ(20°C) = (R*π * r² )/L
ρ(20°C) = (0.695 * π 0.00275²)/1.7
ρ(20°C) = 9.7 * 10^-6 Ω
m
ρ(92°C) = (R*π * r² )/L
⇒ R = 13/17.1 = 0.76
ρ(92°C) =(0.76 * π 0.00275²)/1.7
ρ(92°C) = 1.06 *10^-5 Ω
m
<em></em>
<em>2.Find the temperature coefficient of resistivity at 20 ^\circ {\rm C} for the material of the rod.(α= ? (C)^-</em>
RT = R0*[1 + α(T-T0)]
⇒ with RT = the resistance of the rod at tmperature T = 92.0°C
⇒ with R0= the resistance of the rod at temperature T = 20°C
⇒ with α = the temperature coefficient of resisivity
⇒ with T = 92°C
⇒ with t0 = 20°C
RT = V/I = 13.0/ 17.1 =0.76 Ω
α =((RT/R0)-1)/(T-T0)
α = ((0.76/0.695)-1)/(92-20)
α = 0.0013 /°C
