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Resistivity and superconductivity

Resistance belongs to a particular wire; reshape it and R changes with the geometry. Divide the shape out and what remains is resistivity, the material's own property. Temperature changes that property, and in a few materials it falls to zero below a critical temperature.

Builds on Current, charge and the direction problem and Stress, strain and the Young modulus.

IN THIS TOPIC

  • Move between a wire's resistance and its material's resistivity with ρ = RA/L, in Ω m.
  • Say what warming does to a metal's resistance, and why an ntc thermistor does the opposite.
  • Give thermistor applications, reading a resistance-temperature graph as the calibration.
  • Model an LDR's fall in resistance as light freeing extra conduction electrons, and contrast it with the metal.
  • Describe superconductivity below a critical temperature, with its two examinable applications.

COMMON MISCONCEPTION

A material has a resistance.

From the object to the material

Resistance is a property of one particular object. Make the wire longer and R rises in proportion, because the carriers undergo more collisions with the lattice on the way through; make it fatter and R falls in proportion, because a larger cross-sectional area provides more parallel conducting paths. Divide the geometry out, the same move that turned a spring constant into the Young modulus, and the material's own number is left behind.

ρ=RAL\rho = \frac{RA}{L}ON THE AQA DATA SHEET
Resistance depends on the wire's shape: longer means more, fatter means less; resistivity divides the shape outLAdouble L: double Rdouble A: half R
FIG. 1A wire's resistance scales with its shape: proportional to length, inversely proportional to cross-sectional area. Resistivity is what remains when both are divided away.

Resistivity ρ belongs to the material alone, at a stated temperature, and its unit is the ohm metre, Ω m, not ohms per metre. The values span an astonishing range. Copper sits near 1.7 × 10−8 Ω m, good insulators reach 1016 Ω m, and those two lie some twenty-four orders of magnitude apart.

WORKED EXAMPLE

Scaling a wire without a calculator

A wire has resistance R. A second wire of the same material is twice as long and twice the diameter. Find its resistance in terms of R.

Work through R = ρL/A one factor at a time. Doubling the length doubles the resistance.

Doubling the diameter quadruples the area, since A depends on d2, and that divides the resistance by four.

Together, R × 2/4 = R/2. No numbers were needed. Proportional reasoning like this comes up as often as the plug-in version.

Metals and temperature

Warm a metal and its resistivity rises. The lattice ions vibrate with greater amplitude, the drifting carriers collide with them more often, and the same pd drives less current. Over everyday ranges the rise is roughly linear. It is the same mechanism that curved the filament lamp's characteristic.

Required practical 5 pins the number down. Measure the wire's diameter with a micrometer at several places and average, take V and I readings for a series of lengths, then plot R against L. The gradient is ρ/A, so ρ comes out as gradient × A. Nearly all the uncertainty arrives through the diameter, because A depends on d2 and so doubles whatever percentage error d carries.

Thermistors

A thermistor is a semiconductor component whose resistance falls steeply as it warms; only this negative temperature coefficient (ntc) type is considered. Heating a semiconductor frees extra charge carriers, and the flood of new carriers outweighs the extra lattice collisions, so the net resistance drops.

Resistance against temperature: a metal rises gently, an ntc thermistor falls steeplytemperatureRmetalntc thermistor
FIG. 2Resistance against temperature: the metal climbs gently, the ntc thermistor falls steeply as heat frees additional charge carriers.

That steep, reliable fall is what makes the component useful. A thermistor works as a resistance thermometer. Put one where the temperature matters, a car engine, an incubator, a smart thermostat, and read the temperature off the resistance, with the component's own resistance-temperature graph as the calibration.

GUIDED PRACTICE

A thermistor warms up

A thermistor connected to a steady 6.0 V supply has resistance 1200 Ω when cool and 400 Ω when warm. Find the current in each case, and state the direction of the change.

Show the working

Cool: I = 6.0/1200 = 5.0 mA. Warm: I = 6.0/400 = 15 mA.

Warming the thermistor tripled the current. Falling resistance with rising temperature is the ntc signature, the opposite of the metal wire, because heat here liberates charge carriers instead of merely scattering them.

Light-dependent resistors

Change the stimulus and the same argument runs a second time. A light dependent resistor (LDR) is a wafer of semiconductor, and in the dark almost every one of its electrons is locked into a bond. The number density of conduction electrons, the n in I = nAvq, is minute, so the wafer's resistance is enormous, a hundred kilohms and upwards.

Light undoes that, one electron at a time. 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 to carry a current. A larger n means more current for the same pd, which is to say a lower resistance: the same LDR that reads 100 kΩ in darkness reads about 1 kΩ in daylight. Cover it again and the freed electrons settle back into bonds within milliseconds, and the resistance climbs.

Inside an LDR: light frees extra conduction electrons, so the number density rises and the resistance fallsphotons inin the darkin daylightfew free electronsmany free electronsR ≈ 100 kΩR ≈ 1.0 kΩelectron counts are schematic, not to scale
FIG. 3The model behind the LDR. In the dark only a handful of electrons are free to carry current and the resistance is high; photons arriving free many more, and the resistance falls.

Notice what never entered that account. The lattice did not soften and nothing was heated. In a metal the carriers are all free already, so warming can only shake the ions harder and scatter them more, and the resistance rises. In both semiconductor components it is the supply of carriers that changes instead: heat frees them in the thermistor, light frees them in the LDR, and either way n climbs while R falls.

GUIDED PRACTICE

Three components, two mechanisms

An LDR and an ntc thermistor both show a falling resistance, one against illumination and one against temperature. Explain, in terms of conduction electrons, why both fall, and why a copper wire's resistance does the opposite as it warms.

Show the working

Both components are semiconductors in which most electrons start bound. Absorbed light in the LDR, and thermal energy in the thermistor, free extra conduction electrons, so n rises, more current flows for the same pd and the resistance falls.

Copper has its full complement of free electrons at any temperature, so n barely changes. Warming it only makes the lattice ions vibrate with larger amplitude, the electrons collide with them more often, and the resistance rises. One story is about how many carriers there are, the other about how far they get.

Superconductivity

Cool certain materials below a critical temperature and their resistivity does something no ordinary conductor manages. It drops to exactly zero. Not small, not negligible, zero. A current started in a superconducting loop circulates without measurable loss for years, which is how MRI magnets run. The zero has conditions beyond temperature that A-level does not examine: push the magnetic field or the current density past the material's own critical values and the superconductivity is lost too.

A superconductor's resistivity drops to exactly zero at its critical temperaturetemperatureρcritical temperaturezero, exactlyordinary metalsuperconductor, normal above Tc
FIG. 4An ordinary metal's resistivity falls smoothly as it cools; a superconductor's drops discontinuously to zero at the critical temperature and stays there.

The critical temperature depends on the material, from a few kelvin for simple metals to above 130 K for certain ceramic compounds. Two applications carry the marks. Superconducting coils make the very strong magnetic fields inside MRI scanners and maglev systems, since currents up to the material's limit can flow without heating anything. Superconducting cables would allow power transmission with no resistive loss, because I2R vanishes when R does.

INDEPENDENT PRACTICE

The power lost in a cable

A copper cable of resistance 0.020 Ω carries 100 A to a building. Find the power lost in the cable, and state what the loss becomes if the cable is replaced by a superconductor below its critical temperature.

Show the working

P = I2R = 1002 × 0.020 = 200 W, warming the street instead of the building.

Below the critical temperature the resistance is exactly zero rather than very small, so the loss is 0 W. The limitation is the energy needed for refrigeration to stay that cold, which is why superconducting transmission is not in general use.

ASSESSMENT FOCUS

  • The unit of resistivity is the ohm metre, Ω m. Writing Ω m−1 counts as a definition error and it is tested directly.
  • Keep the nouns attached. R belongs to the wire, ρ belongs to the material, and “the resistivity of the wire is 5 Ω” loses marks twice over.
  • Doubling a wire's diameter quarters its resistance, because A depends on d2. Radius mistaken for diameter, and a forgotten square, are two traps sitting in one line.
  • The thermistor mechanism is about carrier numbers. Heating frees more charge carriers. “The ions vibrate less” is the metal story told backwards, not the carrier-number mechanism the question asks for.
  • The LDR mechanism is the same. Light frees extra conduction electrons, n rises and the resistance falls; “the light warms it up” is not the mechanism and gains no credit. Edexcel asks for that model by name, and on the other boards it is the reason behind the direction you are expected to state.
  • For superconductivity, say the resistivity is zero at and below the critical temperature and that the critical temperature depends on the material. Zero, not very small.
  • In the resistivity practical, take the micrometer reading to the nearest 0.01 mm at several points and average. Asked which measurement dominates the uncertainty, answer the diameter, and say why.

CHECK YOURSELF

A wire of length 2.5 m and diameter 0.40 mm has a resistance of 1.2 Ω. Find the resistivity of its material.

Show a hint

Area from the diameter first, in metres, then rearrange the resistivity equation.

Show the answer

Area first, from the diameter. A = πd2/4 = π × (0.40 × 10−3)2 / 4 = 1.26 × 10−7 m2.

ρ=RAL\rho = \frac{RA}{L} = 1.2 × 1.26 × 10−7 / 2.5 = 6.0 × 10−8 Ω m, a value typical of a pure metal.

R belongs to the wire.

ρ belongs to the material.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

18 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

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  • Move between a wire's resistance and its material's resistivity with ρ = RA/L, in Ω m.
  • Say what warming does to a metal's resistance, and why an ntc thermistor does the opposite.
  • Give thermistor applications, reading a resistance-temperature graph as the calibration.
  • Model an LDR's fall in resistance as light freeing extra conduction electrons, and contrast it with the metal.
  • Describe superconductivity below a critical temperature, with its two examinable applications.

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