Is R = V/I the same as Ohm’s law?

No. R = V/I defines resistance and always applies. Ohm’s law is the stronger claim that R stays constant as V changes, which only holds for ohmic conductors at constant temperature.

Why does resistance depend on length and area?

A longer path means more opportunities for carriers to be scattered; a larger cross-section provides more parallel routes. Hence R L and R 1/A.

Why does a diode conduct in only one direction?

Its p-n junction allows carriers to cross easily in forward bias once the threshold voltage is exceeded, but in reverse bias the depletion region widens and blocks the flow.

What is special about a superconductor?

Below its critical temperature the resistivity is exactly zero, so current flows with no energy dissipation. That allows very strong electromagnets without the heat that would otherwise be produced.

Back to course

33, 38–41, 47–48ElectricityCore

Resistance and Resistivity

$\textbf{Resistance}$ is a property of a particular component — this wire, that lamp. $\textbf{Resistivity}$ is a property of the material it is made from, independent of its shape. Keeping the two apart is what lets you predict how a component will behave before building it.

35 min Topic 3 · Electric Circuits

What you'll be able to do

Define resistance and state the conditions under which Ohm’s law applies
Use $R = \rho L / A$ and distinguish resistivity from resistance
Sketch and explain $I$ – $V$ characteristics for a resistor, lamp and diode
Explain how temperature affects the resistance of metals and thermistors

Resistance and Ohm’s law

Resistance is defined as $R = \dfrac{V}{I}$ , measured in ohms. This definition always holds — it is how resistance is defined, and it applies to any component at any instant.

$\textbf{Ohm’s law}$ is a separate and stronger claim: that for some components the current is $\textit{proportional}$ to the potential difference, so $R$ stays constant. That holds only for $\textbf{ohmic}$ conductors at constant temperature.

The distinction matters. You can always compute $V/I$ for a lamp at a given point, but the answer changes with the operating point, so the lamp does not obey Ohm’s law.

R = \dfrac{V}{I}

The definition — always valid.

V \propto I \text{ at constant } T

Ohm’s law — only for ohmic conductors.

Resistivity: the material property

Resistance depends on the object’s dimensions as well as its material. A longer wire has more resistance because carriers travel further against opposition; a thicker wire has less, because there are more parallel paths.

Those dependences combine into $R = \dfrac{\rho L}{A}$ , where $\rho$ is the $\textbf{resistivity}$ in ohm-metres — an intrinsic property of the material, independent of shape.

Resistivity spans an extraordinary range: about $1.7\times10^{-8}\ \Omega\,\mathrm{m}$ for copper and around $10^{16}\ \Omega\,\mathrm{m}$ for a good insulator — twenty-four orders of magnitude, which is why the same p.d. produces wildly different currents.

R = \dfrac{\rho L}{A}

Resistance from material and geometry.

\rho = \dfrac{RA}{L}

Rearranged; units

\Omega\,\mathrm{m}

Example 1A copper wire is 2.0 m long with cross-sectional area

1.5\times10^{-6}\ \mathrm{m^{2}}

. Copper has

\rho = 1.7\times10^{-8}\ \Omega\,\mathrm{m}

. Find its resistance.

R = \dfrac{\rho L}{A} = \dfrac{1.7\times10^{-8} \times 2.0}{1.5\times10^{-6}}

= \dfrac{3.4\times10^{-8}}{1.5\times10^{-6}}

Answer

R = 0.023\ \Omega

(2 s.f.)

Example 2A wire is stretched to twice its length, keeping its volume constant. By what factor does its resistance change?

1Constant volume means doubling

L

halves

A

R = \dfrac{\rho L}{A}

, so doubling the numerator and halving the denominator.

3The factor is

2 \times 2

AnswerResistance increases by a factor of 4.

Tip — The stretched-wire question is a favourite. Remember that changing the length also changes the area if the volume is fixed — both effects act in the same direction.

I–V characteristics

A $\textbf{metallic conductor at constant temperature}$ gives a straight line through the origin. Constant gradient means constant resistance, so it is ohmic.

A $\textbf{filament lamp}$ gives a curve that flattens as current increases. Higher current heats the filament, the metal ions vibrate more, electrons collide with them more often, and the resistance rises. The gradient falls because resistance is increasing.

A $\textbf{semiconductor diode}$ conducts in one direction only. Below the threshold voltage — about 0.6 V for silicon — almost no current flows; above it the current rises steeply. In reverse bias the resistance is extremely high and the current is negligible.

\text{gradient of } I\text{-}V = \dfrac{1}{R}

Steeper means lower resistance.

Example 1Explain why the

I

–

V

graph of a filament lamp curves towards the

V

-axis at higher voltages.

1Larger current transfers more energy to the filament, raising its temperature.

2Metal ions vibrate with greater amplitude.

3Electrons collide with them more frequently, impeding the flow.

4So resistance increases and the graph flattens.

AnswerRising temperature increases lattice vibration, which increases resistance, so the gradient decreases.

Temperature, thermistors and superconductors

For a $\textbf{metal}$ , resistance increases with temperature. The number of carriers is essentially fixed, but hotter ions vibrate more and scatter electrons more effectively, reducing the drift velocity for a given field.

For a $\textbf{negative temperature coefficient thermistor}$ — a semiconductor — resistance $\textit{decreases}$ sharply with temperature. Heating releases many more charge carriers, and that increase in $n$ overwhelms the extra scattering. Referring back to $I = nAvq$ makes the contrast clear: metals have fixed $n$ and falling $v$ ; semiconductors have rapidly rising $n$ .

Below a $\textbf{critical temperature}$ , some materials become $\textbf{superconductors}$ with exactly zero resistivity. Current then flows with no energy dissipation, which is what makes powerful electromagnets in MRI scanners and particle accelerators practical.

\text{metal: } T \uparrow \Rightarrow R \uparrow

More lattice vibration.

\text{NTC thermistor: } T \uparrow \Rightarrow R \downarrow

More charge carriers released.

Example 1Explain why a metal and an NTC thermistor respond oppositely to a temperature rise.

1In a metal,

n

is essentially constant, so the increased scattering reduces drift velocity and raises resistance.

2In a semiconductor, heating frees many additional carriers, increasing

n

substantially.

3That increase in

n

dominates the increased scattering, so resistance falls.

AnswerMetals have fixed carrier density and increased scattering; semiconductors gain carriers faster than scattering increases.

Tip — Answer temperature questions through $I = nAvq$ . Saying which of $n$ and $v$ changes, and which dominates, is what earns the explanation marks.

Equation recap

R = \tfrac{V}{I}

Definition of resistance.

R = \tfrac{\rho L}{A}

From material and geometry.

\text{gradient of } I\text{-}V = \tfrac1R

For

I

plotted vertically.

V \propto I

Ohmic, at constant temperature.

Common mistakes to avoid

Saying every component obeys Ohm’s law.

Only ohmic conductors at constant temperature. Lamps and diodes do not.

Using resistance and resistivity interchangeably.

Resistance depends on dimensions; resistivity is a material property in

\Omega\,\mathrm{m}

Forgetting that stretching a wire also reduces its area.

At constant volume, doubling the length halves the area, so resistance quadruples.

Claiming a thermistor behaves like a metal.

An NTC thermistor’s resistance falls with temperature because carrier density rises.

Key takeaways

$R = V/I$ defines resistance; Ohm’s law adds that $R$ is constant for ohmic conductors.
Resistivity is a material property; $R = \rho L/A$ brings in the geometry.
A metallic conductor gives a straight $I$ – $V$ line; a lamp curves as it heats; a diode conducts one way.
Metal resistance rises with temperature; NTC thermistor resistance falls.
Explain temperature effects through $n$ and $v$ in $I = nAvq$ .

Test yourself

Answer a few questions, then review anything you missed.

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