Questions on resistivity and temperature

Question 1

State the equation relating the resistance of a wire to its resistivity, and give the unit of resistivity.

Accepted Answer

R = ρL/A, where ρ is the resistivity, L the length and A the cross-sectional area (1). Resistivity has the unit ohm metre, Ω m (1).

Question 2

A copper wire (resistivity 1.7 × 10⁻⁸ Ω m) has length 2.0 m and cross-sectional area 1.0 × 10⁻⁶ m². Calculate its resistance.

Accepted Answer

R = ρL/A = (1.7 × 10⁻⁸ × 2.0)/(1.0 × 10⁻⁶) (1) R = 0.034 Ω (1)

Question 3

State what is meant by superconductivity.

Accepted Answer

Below a critical temperature the resistivity of a superconductor falls to zero (1), so it conducts electricity with no resistance and no energy dissipated (1).

Question 4

State how the resistance of an ntc thermistor changes as its temperature rises, and give one practical use of a thermistor.

Accepted Answer

The resistance falls (steeply) as the temperature rises (1). Used as a temperature sensor, for example in a thermostat, a car engine monitor or an incubator (1).

Question 5

A uniform wire of resistance 3.2 Ω is cut into two equal lengths. State the resistance of each half.

Accepted Answer

1.6 Ω, since R ∝ L and each piece has half the length (1).

Question 6

Sketch a graph of resistivity against temperature for a material that becomes superconducting, labelling the critical temperature T_c on the temperature axis.

Accepted Answer

Both axes labelled, with the resistivity falling as the temperature is reduced and dropping abruptly at a marked T_c (1). Below T_c the line lies along the temperature axis, since the resistivity there is exactly zero rather than merely small (1).

Question 7

A wire of length 1.5 m and diameter 0.40 mm has a resistance of 0.50 Ω. Calculate (a) its cross-sectional area and (b) the resistivity of its material.

Accepted Answer

(a) A = π(d/2)² = π × (0.20 × 10⁻³)² (1)
A = 1.26 × 10⁻⁷ m² (1)
(b) ρ = RA/L = (0.50 × 1.26 × 10⁻⁷)/1.5 (1)
ρ = 4.2 × 10⁻⁸ Ω m, to the two significant figures the data carry (1)

Question 8

A wire of resistivity 2.8 × 10⁻⁸ Ω m and cross-sectional area 2.0 × 10⁻⁶ m² is to have a resistance of 0.10 Ω. Calculate the length required.

Accepted Answer

L = RA/ρ (1) L = (0.10 × 2.0 × 10⁻⁶)/(2.8 × 10⁻⁸) (1) L = 7.14 m (1)

Question 9

A wire's length is doubled and its cross-sectional area is halved. State the factor by which its resistance changes, and explain why.

Accepted Answer

Resistance increases by a factor of 4 (1). From R = ρL/A, doubling L doubles R (1), and halving A doubles R again, giving 2 × 2 = 4 times the original resistance (1).

Question 10

Describe qualitatively how the resistance of a metal wire changes as its temperature increases, and why.

Accepted Answer

The resistance increases (1). Higher temperature makes the metal ions vibrate more, so the conduction electrons collide with them more often (1), impeding their flow and raising the resistance (1).

Question 11

A heater manufacturer needs a 1.2 m length of nichrome wire (resistivity 1.1 × 10⁻⁶ Ω m) to have a resistance of 15 Ω. Calculate the diameter of wire required.

Accepted Answer

A = ρL/R (1) = (1.1 × 10⁻⁶ × 1.2)/15 = 8.8 × 10⁻⁸ m² (1); d = √(4A/π) (1) = 0.33 mm (1).

Question 12

The mass of a 4.00 m length of copper wire is 2.85 g. The density of copper is 8930 kg m⁻³ and its resistivity is 1.7 × 10⁻⁸ Ω m. Calculate (a) the cross-sectional area of the wire and (b) its resistance.

Accepted Answer

(a) A = volume/L = m/(density × L), with the mass in kg (1); A = (2.85 × 10⁻³)/(8930 × 4.00) = 8.0 × 10⁻⁸ m² (1). (b) R = ρL/A (1) = (1.7 × 10⁻⁸ × 4.00)/(8.0 × 10⁻⁸) = 0.85 Ω (1).

Question 13

The main coil of an MRI scanner carries a current of 480 A and is normally superconducting. During a fault the coil warms above its critical temperature and develops a resistance of 0.25 Ω. Calculate the power dissipated in the coil during the fault, and state the power dissipated during normal operation.

Accepted Answer

P = I²R (1); P = 480² × 0.25 = 58 kW (1); in normal operation the resistance is zero, so the power dissipated is 0 W (1).

Question 14

The resistance of a light-dependent resistor falls from about 100 kΩ in darkness to about 1 kΩ in daylight. Explain this fall in terms of the charge carriers in the LDR, and explain why the resistance of a copper wire does not change in the same way when it is illuminated.

Accepted Answer

An LDR is a wafer of semiconductor in which almost every electron is locked into a bond, so in the dark the number density n of conduction electrons is minute and the resistance is very high (1). A photon absorbed in the wafer can transfer its energy to a bound electron and free it, so the brighter the illumination the more conduction electrons there are and the larger n becomes (1). From I = nAvq, a larger n means a larger current for the same pd, which is to say a lower resistance (1). Copper already has its full complement of free electrons at any illumination, so n cannot rise appreciably and the resistance of the wire is effectively unchanged; in a metal it is the lattice vibrations, not the supply of carriers, that decide the resistance (1).

Question 15

A light-dependent resistor is connected on its own across a 4.5 V supply. Its resistance is 150 kΩ in darkness and 1.2 kΩ in daylight. Calculate the current in the LDR in each case, and state what has happened to the number of conduction electrons in the wafer.

Accepted Answer

Darkness: I = V/R = 4.5/(150 × 10³) = 30 μA (1)
Daylight: I = 4.5/(1.2 × 10³) = 3.75 mA (1)
Photons absorbed in the wafer have freed extra conduction electrons, so the number of carriers available has risen and the same pd drives a much larger current (1)

Question 16

The light falling on an LDR is switched off. Describe what happens to its resistance over the following few milliseconds, and explain the change in terms of the conduction electrons in the wafer.

Accepted Answer

The resistance rises, back to its high dark value, within a few milliseconds (1). The electrons that the light had freed settle back into bonds, so the number density n of conduction electrons falls (1). From I = nAvq a smaller n carries a smaller current at the same pd, which is a larger resistance (1).

Question 17

A cable is to be kept superconducting by cooling it with liquid nitrogen, which boils at 77 K. Material P has a critical temperature of 92 K and material Q has a critical temperature of 18 K. Deduce which material the cable should be made from, and state the resistivity of that material at 77 K.

Accepted Answer

A material is superconducting only at or below its critical temperature (1). Liquid nitrogen holds the cable at 77 K, which is below 92 K but above 18 K, so P is superconducting at that temperature and Q is not; the cable should be made from P (1). At 77 K the resistivity of P is exactly zero, not merely a very small value (1).

Question 18

Two wires are made of the same material. Wire B has twice the length and twice the diameter of wire A. If wire A has a resistance of 8.0 Ω, calculate the resistance of wire B.

Accepted Answer

Doubling the diameter multiplies the area by 4 (1) R = ρL/A, so R_B = R_A × 2/4 (1) = 8.0 × 0.5 (1) R_B = 4.0 Ω (1)

Question 19

A heating element is made from wire of resistivity 1.1 × 10⁻⁶ Ω m, length 5.0 m and cross-sectional area 2.0 × 10⁻⁷ m². Calculate its resistance and its power when connected to a 230 V supply.

Accepted Answer

R = ρL/A = (1.1 × 10⁻⁶ × 5.0)/(2.0 × 10⁻⁷) (1) R = 27.5 Ω (1) P = V²/R = 230²/27.5 (1) P = 1924 W, about 1.9 kW to the two significant figures the data carry (1)

Question 20

Describe what happens to a superconductor at its critical temperature and give one practical use of superconductors.

Accepted Answer

As it is cooled to its critical temperature (1) the resistivity suddenly drops to zero (1). Uses include the powerful electromagnets in MRI scanners and particle accelerators, and lossless power transmission cables (1).

Question 21

A student needs a 2.0 m length of wire with a resistance between 4.5 Ω and 5.5 Ω. Four reels are available. Wire W: copper (ρ = 1.7 × 10⁻⁸ Ω m), diameter 0.25 mm. Wire X: constantan (ρ = 4.9 × 10⁻⁷ Ω m), diameter 0.50 mm. Wire Y: constantan, diameter 0.25 mm. Wire Z: nichrome (ρ = 1.1 × 10⁻⁶ Ω m), diameter 0.50 mm. Deduce which wire the student should use.

Accepted Answer

Use of A = πd²/4 with the diameters in metres (1); use of R = ρL/A (1); R_W = 0.69 Ω, R_X = 5.0 Ω, R_Y = 20 Ω, R_Z = 11 Ω (1); only X lies between 4.5 and 5.5 Ω, so the student should use wire X (1).

Question 22

A metal wire and an ntc thermistor happen to have the same resistance at 20 °C. Both are then warmed to 80 °C. Compare what happens to the resistance of each, explaining both changes in terms of the charge carriers and the lattice.

Accepted Answer

Metal: the resistance rises (1). The lattice ions vibrate with greater amplitude, so the conduction electrons collide with them more frequently, impeding the drift (1). The number of charge carriers in the metal is (almost) unchanged (1). Thermistor: the resistance falls steeply (1). Heating the semiconductor frees many more charge carriers (1). The increase in carrier numbers far outweighs the extra lattice scattering, so the net resistance drops (1).

Question 23

A student measures the resistance R of several different lengths L of the same wire and plots R against L, obtaining a straight line through the origin with gradient 2.4 Ω m⁻¹. The wire's diameter is 0.20 mm. Determine the resistivity of the wire's material.

Accepted Answer

A = πd²/4 = π × (0.20 × 10⁻³)²/4 = 3.1 × 10⁻⁸ m² (1); from R = ρL/A the gradient of R against L is ρ/A (1); ρ = gradient × A = 2.4 × 3.14 × 10⁻⁸ (1) = 7.5 × 10⁻⁸ Ω m (1).

Question 24

An LDR has a resistance of 96 kΩ in darkness and 1.6 kΩ in daylight. The pd across it is the same in both cases. Deduce the factor by which the number density of conduction electrons in the wafer increases when the light is turned on, and state the assumption you have made.

Accepted Answer

At the same pd, I = V/R, so the current rises by the factor by which the resistance falls (1)
From I = nAvq with A and q fixed, the current is proportional to n (1)
Factor = 96/1.6 = 60, so n is about 60 times larger in daylight (1)
This assumes the drift speed of the carriers is unchanged, so that the whole of the increase in current is due to the extra conduction electrons (1)

Question 25

An aluminium cable of resistivity 2.8 × 10⁻⁸ Ω m has a cross-sectional area of 4.0 × 10⁻⁴ m² and a total conductor length of 12 km. It carries a current of 250 A at a transmission voltage of 33 kV. Calculate the power dissipated in the cable and express it as a percentage of the power transmitted, and state what each of those becomes if the cable is replaced by a superconductor held below its critical temperature.

Accepted Answer

R = ρL/A = (2.8 × 10⁻⁸ × 12 000)/(4.0 × 10⁻⁴) = 0.84 Ω (1)
P = I²R = 250² × 0.84 = 52.5 kW (1)
Power transmitted = VI = 33 000 × 250 = 8.25 MW (1)
Percentage lost = 100 × (52.5 × 10³)/(8.25 × 10⁶) = 0.64% (1)
Below its critical temperature the superconductor's resistance is exactly zero, so I²R and the percentage lost are both zero; the cost is the energy needed to keep the cable that cold (1)

Question 26

A manufacturer supplies the calibration curve shown with an ntc thermistor. The thermistor is used as a resistance thermometer, and in use its resistance is measured as 2.0 kΩ. Determine the temperature of the thermistor. The same thermistor is offered for two jobs, one near 10 °C and one near 90 °C. Using values taken from the curve, determine at which end of the range the thermometer is the more sensitive, and justify the choice.

Accepted Answer

Reading the curve at R = 2.0 kΩ gives θ ≈ 52 °C (allow 50 °C to 54 °C) (1)
Take the resistance change over a stated temperature interval at each end of the curve, e.g. 12.0 kΩ to 6.0 kΩ over the first 20 °C and 0.75 kΩ to 0.4 kΩ over the last 20 °C (1)
About 0.30 kΩ °C⁻¹ near 10 °C against about 0.02 kΩ °C⁻¹ near 90 °C (allow 0.015 to 0.025) (1)
The thermometer is more sensitive near 10 °C, because the curve is steeper there, so the same temperature change produces a far larger change in resistance to measure (1)