A circuit can be used to measure the resistance of a platinum resistance thermometer (PRT). AB is a uniform resistance wire of length 1.0mandCisa sliding contact on this wire. A standard resistor R is included in the circuit. The position of C is adjusted until the voltmeter V reads zero.
(i) By applying Kirchhoff’s laws to loops ADCA and BCDB, deduce an expression for the resistance of the PRT in terms of l1,land the value of the standard resistor.
(ii) The 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? Show that you have checked that the value for the resistivity and its unit are sensible
I'm not sure how tutors here can be of much help without the diagram for this one.
Though I suppose you're supposed to equate the potential drop across the circuit with that of the PRT, since the Voltmeter reads zero.
(i) Applying Kirchhoff's laws to the loops ADCA and BCDB, we can deduce an expression for the resistance of the PRT.
Loop ADCA:
Starting from point A and moving clockwise:
- We have a voltage drop across length 1 of the wire, given by V_1 = l_1 * (current through PRT)
- We have a voltage drop across length l of the wire, given by V = l * (current through PRT)
- We have a voltage drop across the standard resistor R, given by V_R = R * (current through standard resistor)
By Kirchhoff's voltage law, the sum of these voltage drops must be zero:
V_1 + V + V_R = 0
Given that V = 0 because the voltmeter reads zero, we can rewrite the equation as:
V_1 + V_R = 0
l_1 * (current through PRT) + R * (current through standard resistor) = 0
l_1 * (I) + R * (I) = 0 (assuming the current through both PRT and standard resistor is the same)
l_1 * I + R * I = 0
I * (l_1 + R) = 0
From this, we can deduce the expression for the resistance of the PRT:
Resistance of PRT = R_PRT = -(l_1 / R)
(ii) To find the resistivity of platinum, we need to use the known values and the expression for the resistance of the PRT from part (i).
Given:
Length of PRT wire, l = 9.0 m
Diameter of PRT wire, d = 8.0 × 10^(-2) mm
Using the diameter, we can find the radius, r:
radius, r = (d/2) = (8.0 × 10^(-2) mm) / 2 = 4.0 × 10^(-2) mm = 4.0 × 10^(-5) m
To find the length l1 when the voltmeter reads zero, we use the given condition: V = 0 when l1 = 0.44 m.
Using the expression for the resistance of the PRT from part (i):
R_PRT = -(l1 / R) = -((0.44 m) / 224 Ω) = -1.964 × 10^(-3)
Since resistance cannot be negative, we take the absolute value:
R_PRT = 1.964 × 10^(-3)
Now, we can use the formula for the resistance of a wire in terms of its resistivity, length, and cross-sectional area:
Resistance of a wire = (resistivity * length) / cross-sectional area
The resistance of the PRT wire is given by:
R_PRT = (resistivity * l) / (pi * r^2)
Rearranging the equation, we can solve for resistivity:
resistivity = (R_PRT * pi * r^2) / l
Substituting the known values:
resistivity = (1.964 × 10^(-3) * pi * (4.0 × 10^(-5))^2) / 9.0
Calculating the resistivity, we find:
resistivity ≈ 9.26 × 10^(-8) Ω·m
The value for the resistivity and its unit, Ω·m (ohm-meter), are sensible for a material like platinum.