A PRT consists of 9.0 m of wire of diameter 8.0×10−2 mm. The voltmeter reads 0 V when l1 =0 .44m. If the standard resistor, R, has a resistance of 224Ω, what is the resistivity of platinum?
PRT? Is that a bridge circuit?
apologies it stands for Platinum Resistance Thermometer.
Given:
L = 9.0 m.
Dia. = 8.0*10^-2 mm = 8.0*10^-5 m.
R = 224 Ohms.
R = p*L/A.
p = A*R/L. = Resistivity.
A = pi*r^2 = 3.14 * (4*10^-5)^2 = 50.3*10^-10m^2.
p = (50.3*10^-10 * 224)/9 = 1.3*10^-7 Ohm-meters.
To find the resistivity of platinum, we can use the formula:
ρ = (RA) / (L)
where:
ρ = resistivity of the material (in ohm-meters)
R = resistance of the material (in ohms)
A = cross-sectional area of the wire (in square meters)
L = length of the wire (in meters)
First, let's find the cross-sectional area of the wire.
Given:
Diameter of wire = 8.0 × 10^(-2) mm = 8.0 × 10^(-5) m
Radius of wire = diameter / 2 = 8.0 × 10^(-5) m / 2 = 4.0 × 10^(-5) m
Area of wire = π * (radius)^2 = π * (4.0 × 10^(-5))^2 = 1.6 × 10^(-9) m^2
Next, we need to find the resistance of the wire using the given values for length (l1) and resistance (R).
Given:
Length of wire (l1) = 0.44 m
Resistance of resistor (R) = 224 Ω
Resistance of wire (RW) = R * (l1 / L)
Substituting the values, we get:
224 Ω = RW * (0.44 m / 9.0 m)
Solving for RW:
RW = (224 Ω) / (0.44 m / 9.0 m) = 408 Ω
Now, we can substitute the values of RW, A, and L into the formula for resistivity (ρ) to calculate the resistivity of platinum.
ρ = (RA) / L
ρ = (408 Ω) * (1.6 × 10^(-9) m^2) / (9.0 m)
Calculating this expression, we find:
ρ ≈ 7.20 × 10^(-8) Ω-m
Therefore, the resistivity of platinum is approximately 7.20 × 10^(-8) ohm-meters.