EG1216 · Physics 2 — Electricity & Magnetism
Theme 3 · Current & DC circuits

Resistance, resistivity & Ohm's law

Voltage pushes the current; resistance holds it back. Ohm's law ties the two together — the most-used equation in circuits.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §27.2–27.4.

Before you start

What you need first

  • Topic 16 — current \(I\) (amperes).
  • Topic 9 — voltage \(V\) as the "push."
  • Cross-section area of a wire.

What you'll be able to do

  • Use Ohm's law \(V = IR\) (and \(I=V/R\), \(R=V/I\)).
  • Find a wire's resistance from \(R = \rho L/A\).
  • Tell ohmic from non-ohmic behaviour.
  • Account for resistance rising with temperature.

The hold-back

Resistance — what slows the current

Resistance \(R\) is how strongly a material fights the flow of current, measured in ohms (Ω).

Water-pipe picture: a narrow pipe is hard to push water through — high resistance; a wide pipe is easy — low resistance. The battery's voltage is the push; resistance is what holds the current back.

Ohm's law

$$ V = IR $$
where:
SymbolMeaningSI unit
Vvoltage across the resistor (the push)V
Icurrent through it (the flow)A
Rresistance (the hold-back)Ω
Rearrange as needed: \(I = \dfrac{V}{R}\) and \(R = \dfrac{V}{I}\).
📐 Worked example 1

Using Ohm's law three ways

(a) \(I = 2.0\) A through \(R = 5.0\ \Omega\) — find \(V\). (b) \(V = 12\) V across \(R = 4.0\ \Omega\) — find \(I\). (c) \(V = 9.0\) V drives \(I = 0.50\) A — find \(R\).

1\(V = IR = (2.0)(5.0) = 10\ \text{V}\).
2\(I = V/R = 12/4.0 = 3.0\ \text{A}\).
3\(R = V/I = 9.0/0.50 = 18\ \Omega\).
$$ V=10\ \text{V}, \quad I=3.0\ \text{A}, \quad R=18\ \Omega $$

Resistance from material and shape

A wire's resistance grows with its length and shrinks with its thickness; the material sets the rest through its resistivity \(\rho\):

$$ R = \frac{\rho L}{A} $$
A length L
Long and thin → more resistance; short and fat → less.
where:
SymbolMeaningSI unit
ρresistivity of the material (here \(\rho\) means resistivity, not charge density)Ω·m
Llength of the wirem
Across-section aream²
Material\(\rho\) (Ω·m, about)
Copper\(1.7\times10^{-8}\)great conductor (wires)
Aluminium\(2.8\times10^{-8}\)good conductor
Iron\(1.0\times10^{-7}\)okay conductor
Glass\(\sim 10^{12}\)insulator (huge \(\rho\))
📐 Worked example 2

Resistance of a copper wire

A copper wire (\(\rho = 1.7\times10^{-8}\) Ω·m) is \(L = 2.0\) m long with cross-section \(A = 1.0\times10^{-6}\) m². Find its resistance.

1Use \(R = \rho L/A\):
$$ R = \frac{(1.7\times10^{-8})(2.0)}{1.0\times10^{-6}} $$
2Work it out:
$$ R \approx 0.034\ \Omega $$
A short copper wire barely resists — which is exactly why we make wires from copper.

Ohmic vs non-ohmic

A material is ohmic if its resistance stays constant — then \(I\) vs \(V\) is a straight line through the origin (slope \(1/R\)). Many devices are non-ohmic: a light-bulb filament heats up as the current rises, so its resistance climbs and the line bends.

V I ohmic non-ohmic
Ohmic: a straight \(I\)–\(V\) line. Non-ohmic (e.g. a bulb): it bends as \(R\) changes.

Resistance rises with temperature

In a metal, hotter atoms vibrate more and get in the electrons' way, so resistance increases with temperature. Over a normal range it grows almost linearly:

$$ R = R_0\big[\,1 + \alpha (T - T_0)\,\big] $$
where:
SymbolMeaningSI unit
R₀resistance at the reference temperature \(T_0\)Ω
αtemperature coefficient of resistance1/°C
T − T₀temperature rise above the reference°C
📐 Worked example 3

Heating a copper coil

A copper coil has \(R_0 = 10.0\ \Omega\) at \(T_0 = 20\,^\circ\text{C}\). Copper's \(\alpha = 3.9\times10^{-3}\,/^\circ\text{C}\). Find its resistance at \(80\,^\circ\text{C}\).

1Temperature rise \(T - T_0 = 60\,^\circ\text{C}\); apply the formula:
$$ R = 10.0\big[1 + (3.9\times10^{-3})(60)\big] = 10.0(1.234) $$
2Work it out:
$$ R \approx 12.3\ \Omega $$

✏️ Try it yourself

A nichrome heating wire (\(\rho = 1.1\times10^{-6}\) Ω·m) is \(L = 5.0\) m long with cross-section \(A = 2.0\times10^{-7}\) m².

(a) Find its resistance.
(b) Connected to a \(12\) V supply, what current flows?

(a) \(R = \dfrac{\rho L}{A} = \dfrac{(1.1\times10^{-6})(5.0)}{2.0\times10^{-7}} \approx 27.5\ \Omega.\) (b) \(I = \dfrac{V}{R} = \dfrac{12}{27.5} \approx 0.44\ \text{A}.\)

Common mistakes

MistakeFix
Mixing up \(R\) and \(\rho\).\(R\) (ohms) is for a particular wire; \(\rho\) (Ω·m) is the material property. \(R=\rho L/A\).
Putting \(A\) on top in \(R=\rho L/A\).Thicker wire (bigger \(A\)) → less resistance, so \(A\) is on the bottom.
Assuming every device is ohmic.Bulbs, diodes and more are non-ohmic — \(R\) is not constant.
Forgetting metals' \(R\) rises when hot.Use \(R = R_0[1+\alpha(T-T_0)]\) for a temperature change.

Recap — the whole topic on one screen

IdeaWhat you own now
Ohm's law\(V = IR\) (and \(I=V/R\), \(R=V/I\)).
From shape\(R = \rho L/A\); long-thin → big \(R\).
Ohmic?Straight \(I\)–\(V\) = ohmic; bent = non-ohmic.
Temperature\(R = R_0[1+\alpha(T-T_0)]\); metals rise when hot.

Next topic

Electrical power

A current through a resistance delivers energy — heating the element, lighting the bulb. Next we find the rate: \(P = IV\), and its handy cousins \(I^2R\) and \(V^2/R\).

→ Topic 18 · Electrical power