Electric Current and Ohm's Law

Lessons

Senior 1 · Physics

Electric Current and Ohm's Law

5 lessons

Lesson 4 of 5

Resistance, Resistivity and Ohm's Law

25 min

Why does the heating element glow but the cable stay cool?

The same current flows through the cable of an electric kettle and through the coiled element inside it, yet the element becomes red hot while the cable stays cool. A long, thin extension cable makes a phone charge more slowly than a short, thick one. Something in each conductor opposes the current, and that opposition depends on what the conductor is made of and on its shape. This lesson measures it.

Experiment: how does the current depend on the p.d.?

Figure 4.1: Changing the rheostat changes the p.d. across R; the ammeter and voltmeter show the current through R and the p.d. across it.

Set up the circuit with a resistor (or a length of constantan wire) as R.
Close the switch and adjust the rheostat until the voltmeter reads $0.5\,\mathrm{V}$ . Record the ammeter reading.
Repeat for p.d.s up to $3.0\,\mathrm{V}$ . Open the switch between readings so that R does not warm up.

p.d. $V$ (V)	0.5	1.0	1.5	2.0	2.5	3.0
current $I$ (A)	0.10	0.21	0.30	0.41	0.50	0.60
$\frac{V}{I}$ ( $\Omega$ )	5.0	4.8	5.0	4.9	5.0	5.0

Doubling the p.d. doubles the current, and $\frac{V}{I}$ stays the same (about $5.0\,\Omega$ ) within the error of reading the meters.

Figure 4.2: V against I. The resistor's points lie on a straight line through the origin, so V is proportional to I; its gradient is the resistance. A filament lamp's line curves because its resistance rises as it heats up.

Definition

The resistance $R$ of a conductor is the ratio of the p.d. across it to the current through it:

R = \frac{V}{I}

The SI unit is the ohm ( $\Omega$ ): a conductor has a resistance of $1\,\Omega$ if a p.d. of $1\,\mathrm{V}$ drives a current of $1\,\mathrm{A}$ through it ( $1\,\Omega = 1\,\mathrm{V\,A^{-1}}$ ).

Definition

Ohm's law: the current through a metallic conductor is directly proportional to the potential difference across it, provided its temperature and other physical conditions stay the same.

V = I R

A conductor that obeys Ohm's law (a straight-line V–I graph through the origin) is ohmic, for example a resistor or a metal wire at constant temperature. A filament lamp is non-ohmic: as it gets hotter its resistance increases.

Worked example 1: using V = IR three ways

(a) A lamp of resistance $24\,\Omega$ carries $0.25\,\mathrm{A}$ . Find the p.d. across it: $V = IR = 0.25 \times 24 = 6.0\,\mathrm{V}$ .

(b) A $12\,\mathrm{V}$ supply drives $0.40\,\mathrm{A}$ through a heater. Find its resistance: $R = \frac{V}{I} = \frac{12}{0.40} = 30\,\Omega$ .

(c) A $46\,\Omega$ heating element is connected to the $230\,\mathrm{V}$ mains. Find the current: $I = \frac{V}{R} = \frac{230}{46} = 5.0\,\mathrm{A}$ .

Worked example 2: resistance from the graph

Use the gradient triangle on Figure 4.2 to find the resistance of R.

R = \text{gradient} = \frac{\Delta V}{\Delta I} = \frac{3.0\,\mathrm{V} - 1.0\,\mathrm{V}}{0.60\,\mathrm{A} - 0.20\,\mathrm{A}} = \frac{2.0\,\mathrm{V}}{0.40\,\mathrm{A}} = 5.0\,\Omega

Using the whole line averages out the small errors in single readings.

What decides the resistance of a wire?

Figure 4.3: A wire of length L and cross-sectional area A.

Tests with wires of different sizes show that the resistance of a wire:

is proportional to its length $L$ : twice as long, twice the resistance (the charges must pass through twice as much material);
is inversely proportional to its cross-sectional area $A$ : twice the area, half the resistance (there is twice as much room for the charges);
depends on the material, through a number called its resistivity $\rho$ ;
increases with temperature for metals.

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

Definition

The resistivity $\rho$ (the Greek letter rho) of a material is the resistance of a piece of it $1\,\mathrm{m}$ long with a cross-sectional area of $1\,\mathrm{m^2}$ . From $\rho = \frac{RA}{L}$ , its unit is the ohm metre ( $\Omega\,\mathrm{m}$ ). A good conductor has a very small resistivity; an insulator has a huge one.

Material	Resistivity at 20 °C ( $\Omega\,\mathrm{m}$ )	Typical use
copper	$1.72 \times 10^{-8}$	house wiring, cables
aluminium	$2.65 \times 10^{-8}$	overhead power lines (light and cheap)
tungsten	$5.6 \times 10^{-8}$	lamp filaments (melts only above $3400\,^\circ\mathrm{C}$ )
iron	$9.71 \times 10^{-8}$	nails, steel structures
nichrome	$1.00 \times 10^{-6}$	heating elements in kettles, irons and cookers
glass	$10^{9}$ to $10^{14}$	insulators on power poles
hard rubber	$10^{13}$ to $10^{16}$	insulating covers and gloves

Worked example 3: the resistance of a copper cable

A copper cable is $100\,\mathrm{m}$ long and has a cross-sectional area of $1.5\,\mathrm{mm^2}$ . Find its resistance.

Change the area into square metres: $1\,\mathrm{mm^2} = 10^{-6}\,\mathrm{m^2}$ , so $A = 1.5 \times 10^{-6}\,\mathrm{m^2}$ .

R = \rho\frac{L}{A} = 1.72 \times 10^{-8} \times \frac{100}{1.5 \times 10^{-6}} = 1.1\,\Omega

Worked example 4: designing a heating element

A heater needs an element of resistance $46\,\Omega$ , made from nichrome wire of cross-sectional area $0.20\,\mathrm{mm^2}$ . What length of wire is needed?

L = \frac{RA}{\rho} = \frac{46 \times (0.20 \times 10^{-6})}{1.00 \times 10^{-6}} = 9.2\,\mathrm{m}

The wire is coiled so that 9.2 m fits inside the heater.

Resistors in series and in parallel

Figure 4.4: The same two resistors combined in two ways.

In series, the charges must pass through both resistors, like one longer wire, so the resistances add:

R = R_{1} + R_{2}

In parallel, the charges have two paths, like one wider wire, so the combined resistance is LESS than either resistor alone:

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}

Worked example 5: series and parallel on a 12 V supply

Resistors of $4\,\Omega$ and $6\,\Omega$ are connected to a $12\,\mathrm{V}$ supply, first in series, then in parallel. Find the current from the supply each time.

Series: $R = 4 + 6 = 10\,\Omega$ , so $I = \frac{12}{10} = 1.2\,\mathrm{A}$ . The p.d.s are $1.2 \times 4 = 4.8\,\mathrm{V}$ and $1.2 \times 6 = 7.2\,\mathrm{V}$ , which add up to $12\,\mathrm{V}$ .

Parallel:

\frac{1}{R} = \frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \quad\Rightarrow\quad R = \frac{12}{5} = 2.4\,\Omega

$I = \frac{12}{2.4} = 5.0\,\mathrm{A}$ . Check with the branches: $\frac{12}{4} = 3.0\,\mathrm{A}$ and $\frac{12}{6} = 2.0\,\mathrm{A}$ , total $5.0\,\mathrm{A}$ .

Warning

Common mistakes. (1) Forgetting to change $\mathrm{mm^2}$ into $\mathrm{m^2}$ : multiply by $10^{-6}$ , not $10^{-3}$ . (2) Stopping at $\frac{1}{R}$ in a parallel calculation: turn the fraction upside down at the end. A parallel combination is always SMALLER than the smallest resistor. (3) Applying Ohm's law to a lamp as if its resistance were fixed: it changes as the filament heats up.

Tip

Cover-up triangle: write V on top and I and R side by side underneath. Cover the one you want: $V = I R$ , $I = \frac{V}{R}$ , $R = \frac{V}{I}$ .

Choosing the right conductor

Cables are made of thick copper because its low resistivity and large area give a tiny resistance, so they waste little energy and stay cool. Heating elements are made of nichrome, whose resistivity is about 60 times that of copper and which does not rust when red hot. Overhead power lines are aluminium: a little more resistive than copper, but much lighter and cheaper for the long spans between pylons. Explore with PhET: Ohm's Law (version 1.4.28) and PhET: Resistance in a Wire (version 1.6.25): change V, R, the length and the area, and watch the current and the resistance.

Voltage, current and resistance, V = IR and I–V graphs (Cognito). Watch the three rearrangements of V = IR and the shapes of the graphs for a resistor and a filament lamp.

✦

Practice

State Ohm's law, and the condition under which it holds.
A $9.0\,\mathrm{V}$ battery drives a current of $0.30\,\mathrm{A}$ through a resistor. Find its resistance.
What current flows through a $5.0\,\Omega$ resistor connected to a $1.5\,\mathrm{V}$ cell?
A current of $2.0\,\mathrm{A}$ flows through a $15\,\Omega$ heater. What is the p.d. across it?
Find the resistance of a copper wire $20\,\mathrm{m}$ long with a cross-sectional area of $0.50\,\mathrm{mm^2}$ ( $\rho = 1.72 \times 10^{-8}\,\Omega\,\mathrm{m}$ ).
A wire has a resistance of $8.0\,\Omega$ . What is the resistance of a wire of the same material that is (a) twice as long; (b) the same length but with twice the cross-sectional area?
Find the combined resistance of (a) $3\,\Omega$ and $6\,\Omega$ in parallel; (b) $10\,\Omega$ , $10\,\Omega$ and $5\,\Omega$ in parallel.
A student's V–I graph for a torch lamp is a curve, and she concludes that her meters are faulty. Evaluate her conclusion.

Show answersHide answers

The current through a metallic conductor is directly proportional to the p.d. across it, provided the temperature (and other physical conditions) stay constant.
$R = \frac{9.0}{0.30} = 30\,\Omega$
$I = \frac{1.5}{5.0} = 0.30\,\mathrm{A}$
$V = 2.0 \times 15 = 30\,\mathrm{V}$
$R = 1.72 \times 10^{-8} \times \frac{20}{0.50 \times 10^{-6}} = 0.69\,\Omega$
(a) $16\,\Omega$ (resistance is proportional to length); (b) $4.0\,\Omega$ (resistance is inversely proportional to area).
(a) $\frac{1}{R} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6}$ , so $R = 2\,\Omega$ ; (b) $\frac{1}{R} = \frac{1}{10} + \frac{1}{10} + \frac{1}{5} = \frac{4}{10}$ , so $R = 2.5\,\Omega$ .
Her conclusion is probably wrong. A lamp filament gets hotter as the current increases, and the resistance of a metal rises with temperature, so the lamp does not obey Ohm's law: a curved graph is the expected result.

Key points

Resistance $R = \frac{V}{I}$ , in ohms ( $\Omega$ ); Ohm's law: $V = I R$ for a metal at constant temperature.
The resistance of a wire is $R = \rho\frac{L}{A}$ : longer wire, more resistance; thicker wire, less.
Series: $R = R_{1} + R_{2}$ . Parallel: $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ , smaller than either resistor.

Sources and credits

20.2: Ohm's Law: Resistance and Simple Circuits, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
20.3: Resistance and Resistivity, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
21.1: Resistors in Series and Parallel, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Electronic-Axial-Lead-Resistors-Array.jpg by Evan-Amos, Wikimedia Commons, in the public domain.
Ohm's Law (version 1.4.28). Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC BY 4.0 (https://phet.colorado.edu)
Resistance in a Wire (version 1.6.25). Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC BY 4.0 (https://phet.colorado.edu)
Video: GCSE Physics - Voltage, Current & Resistance | V = IR Equation | IV Graphs, Cognito, embedded from YouTube.

The experiment write-up and results, the diagrams and graph, the worked examples and the practice exercises are written by Ganzaa.

1Why does the heating element glow but the cable stay cool?

2Experiment: how does the current depend on the p.d.?

3What decides the resistance of a wire?

4Resistors in series and in parallel

5Choosing the right conductor

Why does the heating element glow but the cable stay cool?

Experiment: how does the current depend on the p.d.?

What decides the resistance of a wire?

Resistors in series and in parallel

Choosing the right conductor