Worked examples · Sheet 2

Resistance of wires and temperature dependence

Theory used: Resistance of a wire Ohm's law in local form Conductors and insulators Temperature dependence of resistivity Resistors in series

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

To build a $R = 50 Ω$ resistor a nichrome wire of diameter $d = 0.40 mm$ is used, with resistivity $ρ = 1.1 \cdot 1 0^{- 6} Ω \cdot m$ . Compute the length of wire needed.

R = ρ \frac{ℓ}{A} ⟹ ℓ = \frac{R A}{ρ} = \frac{π d ^{2} R}{4 ρ}

A ℓ = π \cdot (0.20 \cdot 1 0^{- 3})^{2} m^{2} \approx 1.257 \cdot 1 0^{- 7} m^{2} = \frac{50 \cdot 1.257 \cdot 1 0 ^{- 7}}{1.1 \cdot 1 0 ^{- 6}} m \approx 5.7 m

The wire is cylindrical and homogeneous, therefore the formula for the resistance of a wire applies, solved for the length. The cross-sectional area is computed with the radius $0.20 mm = 0.20 \cdot 1 0^{- 3} m$ . The units are consistent: $\frac{Ω \cdot m ^{2}}{Ω \cdot m} = m$ . With a copper wire of the same diameter, whose resistivity is about $65$ times smaller, more than $350 m$ of wire would be needed; for this reason high-resistivity alloys are used in heating elements.

Exercise 2

A copper wire has resistance $R = 2.0 Ω$ . The wire is drawn until its length triples, with no change in volume or resistivity. Compute the new resistance.

{ℓ^{'} = 3 ℓ ℓ^{'} A^{'} = ℓ A ⟹ A^{'} = \frac{A}{3}

R^{'} = ρ \frac{ℓ ^{'}}{A ^{'}} = ρ \frac{3 ℓ}{\frac{A}{3}} = 9 ρ \frac{ℓ}{A} = 9 R = 18 Ω

The volume of the wire is the product of the length and the cross-sectional area, and it stays constant; since the length triples, the cross-sectional area becomes one third. In the formula for the resistance of a wire the length appears in the numerator and the area in the denominator, therefore both changes increase the resistance by a factor $3$ , and the resistance becomes $9$ times the initial one. In general, stretching the wire by a factor $k$ at constant volume increases the resistance by a factor $k^{2}$ .

Exercise 3

A platinum resistance thermometer has resistance $R_{0} = 100.0 Ω$ at $T_{0} = 0^{\circ} C$ and temperature coefficient $α = 3.85 \cdot 1 0^{- 3} K^{- 1}$ . Immersed in a liquid, the thermometer has resistance $R = 138.5 Ω$ . Compute the temperature of the liquid and the resistance the thermometer would have at $37.0^{\circ} C$ .

R = R_{0} [1 + α (T - T_{0})] ⟹ T = T_{0} + \frac{1}{α} (\frac{R}{R _{0}} - 1)

T R (37.0^{\circ} C) = 0^{\circ} C + \frac{1.385 - 1}{3.85 \cdot 1 0 ^{- 3}}^{\circ} C = 100^{\circ} C = 100.0 \cdot (1 + 3.85 \cdot 1 0^{- 3} \cdot 37.0) Ω \approx 114.2 Ω

The empirical law of the temperature dependence of resistivity applies, which for a wire of constant dimensions holds for the resistance as well; the expansion of the wire is neglected. Solving for $T$ gives the temperature of the liquid, $100^{\circ} C$ . A temperature difference in kelvins coincides with the same difference in degrees Celsius, therefore $α$ can be used with temperatures in degrees Celsius. In the second calculation the law is used in the direct direction; the resistance increases by about $0.385 Ω$ per degree, and a change of $0.1 Ω$ corresponds to about a quarter of a degree.

Exercise 4

We want to make a resistor of $10.0 Ω$ at $20^{\circ} C$ whose resistance does not depend on temperature, by connecting in series a carbon resistor, with temperature coefficient $α_{C} = - 5.0 \cdot 1 0^{- 4} K^{- 1}$ , and a copper resistor, with $α_{Cu} = 3.9 \cdot 1 0^{- 3} K^{- 1}$ . Compute the resistances of the two resistors at $20^{\circ} C$ and check the result at $70^{\circ} C$ .

R (T) = R_{C} [1 + α_{C} Δ T] + R_{Cu} [1 + α_{Cu} Δ T] = (R_{C} + R_{Cu}) + (α_{C} R_{C} + α_{Cu} R_{Cu}) Δ T

{R_{C} + R_{Cu} = R_{tot} α_{C} R_{C} + α_{Cu} R_{Cu} = 0 ⟹ R_{C} = \frac{α _{Cu}}{α _{Cu} - α _{C}} R_{tot}, R_{Cu} = \frac{- α _{C}}{α _{Cu} - α _{C}} R_{tot}

R_{C} R (70^{\circ} C) = \frac{3.9 \cdot 1 0 ^{- 3}}{4.4 \cdot 1 0 ^{- 3}} \cdot 10.0 Ω \approx 8.86 Ω, R_{Cu} = \frac{0.50 \cdot 1 0 ^{- 3}}{4.4 \cdot 1 0 ^{- 3}} \cdot 10.0 Ω \approx 1.14 Ω = [8.864 \cdot (1 - 0.025) + 1.136 \cdot (1 + 0.195)] Ω = (8.642 + 1.358) Ω = 10.0 Ω

The resistors are in series, therefore by the rule for resistors in series the total resistance is the sum of the two resistances, each of which depends on temperature according to the linear law, with $Δ T = T - 20^{\circ} C$ . Collecting terms, the total resistance is a first-degree polynomial in $Δ T$ . For it not to depend on temperature the coefficient of $Δ T$ must vanish, and the constant term must equal $10.0 Ω$ : the two conditions must hold together and form the system.

The system has a solution because the two temperature coefficients have opposite signs: the increase in the resistance of the copper compensates the decrease in that of the carbon. At $70^{\circ} C$ , that is, with $Δ T = 50 K$ , the carbon loses $0.222 Ω$ and the copper gains the same amount, therefore the total resistance remains $10.0 Ω$ . The compensation is exact as long as the linear law holds for both materials.