Notes · PhysicsUK · A-Levels

Electricity

This chapter develops the physics of electric circuits from the microscopic flow of charge to whole-circuit behaviour. It defines current, potential difference and resistance, explains the I-V characteristics of standard components, and treats resistivity, series and parallel networks, power, electromotive force with internal resistance, and the potential divider - the sensor circuit at the heart of electronic control.

5 sections~14 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 2

T·0555 / 13

Exam profile

AO1 · Define current, charge, potential difference, resistance, resistivity and emf, and describe the I-V characteristics of standard componentsAO2 · Calculate combined resistance, power, resistivity, terminal pd and potential-divider outputs and apply I = nAvqAO3 · Analyse the emf/internal-resistance and resistivity required practicals and interpret I-V graphs

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basic level

AS-Level requires current and charge, resistance and resistivity, the standard I-V characteristics, series/parallel circuits, power, emf and internal resistance, and the potential divider.

higher level

The full A-Level applies these quantitatively in multi-loop circuits and sensor systems and links resistivity and internal resistance to the required practicals with their graphical analysis.

Depth

Text

Contents · 5 sections

Electricity

Current, charge and potential difference#

FoundationAQA 7408 3.5.1.1DfE GCE Physics - current and charge

Electric current is the rate of flow of electric charge,

I = \dfrac{\Delta Q}{\Delta t}

, measured in amperes (one ampere is one coulomb per second). Charge is quantised in whole numbers of the elementary charge

e = 1.60\times10^{-19}\,\text{C}

, so any charge is

Q = ne

for an integer

n

. Conventional current is defined as the direction positive charge would flow - opposite to the actual drift of electrons in a metal.

At the microscopic level the current in a conductor depends on how many charge carriers there are, how fast they drift and how much charge each carries:

I = nAvq

, where

n

is the number of carriers per unit volume,

A

the cross-sectional area,

v

the mean drift velocity and

q

the charge on each carrier. Because

n

is enormous in a metal, the drift velocity is surprisingly small - a fraction of a millimetre per second - even for a substantial current.

Potential difference (voltage) between two points is the energy transferred per unit charge as charge moves between them:

V = \dfrac{W}{Q}

, measured in volts (one volt is one joule per coulomb). A component with

3\,\text{V}

across it transfers

3\,\text{J}

of energy for every coulomb that passes. The electronvolt, the energy gained by an electron moving through one volt, is

1\,\text{eV} = 1.60\times10^{-19}\,\text{J}

Because

n

differs hugely between materials, the same current gives very different drift velocities in a metal, a semiconductor and an electrolyte. This is why semiconductors, with far fewer carriers, show markedly different electrical behaviour - the basis of the diode and transistor met later.

I = \dfrac{\Delta Q}{\Delta t}

Current

The rate of flow of charge.

I = nAvq

Microscopic current

Carrier density, area, drift velocity and charge per carrier.

V = \dfrac{W}{Q}

Potential difference

Energy transferred per unit charge.

Worked example

Drift velocity in a copper wire

A copper wire of cross-sectional area $2.0\times10^{-6}\,\text{m}^2$ carries $5.0\,\text{A}$ . The carrier density is $8.5\times10^{28}\,\text{m}^{-3}$ and $e = 1.60\times10^{-19}\,\text{C}$ . Find the mean drift velocity.

Rearrange the microscopic equation
From $I = nAvq$ , $v = \dfrac{I}{nAq}$ .
Substitute
$v = \dfrac{5.0}{(8.5\times10^{28})(2.0\times10^{-6})(1.60\times10^{-19})}$ .
$v = \dfrac{5.0}{2.72\times10^{4}} = 1.8\times10^{-4}\,\text{m}\,\text{s}^{-1}$

Result: The electrons drift at only $1.8\times10^{-4}\,\text{m}\,\text{s}^{-1}$ , about $0.18\,\text{mm}\,\text{s}^{-1}$ .

Typical mistakes

Confusing conventional current (direction of positive charge flow) with the direction of electron drift, which is opposite.
Forgetting to use the cross-sectional area in $\text{m}^2$ (not $\text{mm}^2$ ) in $I = nAvq$ .

Active revision

A copper wire of cross-sectional area $1.0\times10^{-6}\,\text{m}^2$ carries a current of $3.0\,\text{A}$ . If $n = 8.5\times10^{28}\,\text{m}^{-3}$ , find the mean drift velocity of the electrons.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Physics 7408 specification (AQA)

Resistance, I-V characteristics and resistivity#

StandardAQA 7408 3.5.1.2AQA 7408 3.5.1.3DfE GCE Physics - resistance and resistivity

I-V characteristics of an ohmic conductor and a filament lamp

Fig. 1The ohmic conductor gives a straight line (constant resistance). The filament lamp curves towards the voltage axis as its resistance rises with temperature.

The resistance of a component is the ratio of the potential difference across it to the current through it,

R = V/I

, measured in ohms. Ohm's law states that for a metallic conductor at constant temperature the current is directly proportional to the potential difference, so

R

is constant and the I-V graph is a straight line through the origin. Ohm's law is a special case, not a universal rule.

Different components have characteristic I-V graphs. An ohmic conductor gives a straight line. A filament lamp curves: as the current heats the filament, its resistance rises, so the graph bends towards the voltage axis (the current increases less than proportionally). A semiconductor diode conducts only one way - almost no current in reverse, and negligible forward current until the threshold of about

0.6\,\text{V}

, after which it rises steeply.

Resistance depends on the material and the dimensions:

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

, where

\rho

is the resistivity (in ohm metres),

L

the length and

A

the cross-sectional area. A long, thin wire of a high-resistivity material has a large resistance. Resistivity is a property of the material at a given temperature; rearranged,

\rho = RA/L

is measured in the resistivity required practical.

Temperature changes resistance. In a metal, heating increases the vibration of the lattice ions, which impede the electrons more, so resistance rises with temperature. In a thermistor (a semiconductor), heating frees many more charge carriers, so resistance falls sharply with temperature - the basis of temperature sensors. In a superconductor, below a critical temperature the resistance falls abruptly to zero, allowing loss-free currents used in powerful electromagnets.

R = \dfrac{V}{I}

Resistance

Potential difference divided by current.

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

Resistivity

Resistance from resistivity, length and cross-sectional area.

I-V characteristic of a semiconductor diode

Fig. 2The diode conducts negligibly in reverse and forward, until the threshold near

0.6\,\text{V}

, beyond which the current rises steeply.

Worked example

Resistivity of a wire

A metal wire of length $1.5\,\text{m}$ and diameter $0.40\,\text{mm}$ has a resistance of $8.0\,\Omega$ . Find its resistivity.

Cross-sectional area
Radius $= 0.20\,\text{mm} = 2.0\times10^{-4}\,\text{m}$ , so $A = \pi r^2 = \pi(2.0\times10^{-4})^2 = 1.257\times10^{-7}\,\text{m}^2$ .
Rearrange for resistivity
From $R = \rho L/A$ , $\rho = RA/L$ .
$\rho = \dfrac{8.0 \times 1.257\times10^{-7}}{1.5} = 6.7\times10^{-7}\,\Omega\,\text{m}$

Result: The resistivity is $6.7\times10^{-7}\,\Omega\,\text{m}$ , typical of a resistance-wire alloy.

Typical mistakes

Treating a filament lamp as ohmic - its resistance rises with temperature, so the I-V graph is curved.
Using diameter instead of radius when finding $A = \pi r^2$ for a wire in a resistivity calculation.

Active revision

A wire of length $1.2\,\text{m}$ and diameter $0.30\,\text{mm}$ has resistance $9.6\,\Omega$ . Calculate the resistivity of the metal.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Series and parallel circuits, energy and power#

StandardAQA 7408 3.5.1.4DfE GCE Physics - circuits and power

In a series circuit the same current passes through every component, the potential differences add to the supply voltage, and the resistances add:

R_{\text{series}} = R_1 + R_2 + \dots

. In a parallel circuit each branch has the full supply voltage across it, the branch currents add to the total, and the reciprocals of the resistances add:

\dfrac{1}{R_{\text{parallel}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots

, so the combined resistance is always less than the smallest branch.

These rules follow from two conservation laws (Kirchhoff's laws). Conservation of charge means the current into a junction equals the current out of it. Conservation of energy means that around any complete loop the sum of the potential differences equals the sum of the emfs. Applying these two statements systematically solves any network.

Electrical power is the rate at which a component transfers energy,

P = VI

. Using

V = IR

this can be written in the equivalent forms

P = I^2 R = V^2/R

; which to choose depends on which quantities are known. The energy transferred in a time

t

W = VIt = Pt

, in joules (the kilowatt-hour is the domestic unit,

1\,\text{kWh} = 3.6\times10^{6}\,\text{J}

The three power forms explain everyday circuits. A device labelled by its power and voltage draws a definite current; a heating element of low resistance dissipates a large

V^2/R

; and a long supply cable, carrying the current

I

, wastes

I^2 R

as heat, which is why power is transmitted at high voltage and low current to reduce these losses.

R_{\text{series}} = R_1 + R_2, \qquad \dfrac{1}{R_{\text{parallel}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}

Combining resistors

Series resistances add; parallel reciprocals add.

P = VI = I^2 R = \dfrac{V^2}{R}

Electrical power

Three equivalent forms of the power dissipated.

Worked example

A mixed circuit

A $4.0\,\Omega$ and a $12\,\Omega$ resistor are in parallel; this pair is in series with a $2.0\,\Omega$ resistor across a $9.0\,\text{V}$ battery of negligible internal resistance. Find the total current and the power in the $2.0\,\Omega$ resistor.

Parallel combination
$\dfrac{1}{R_p} = \dfrac{1}{4.0} + \dfrac{1}{12} = \dfrac{3 + 1}{12} = \dfrac{4}{12}$ , so $R_p = 3.0\,\Omega$ .
Total resistance
$R = R_p + 2.0 = 3.0 + 2.0 = 5.0\,\Omega$ .
Total current
$I = V/R = 9.0/5.0 = 1.8\,\text{A}$ .
Power in the 2 ohm resistor
The full current passes through it: $P = I^2 R = (1.8)^2 \times 2.0$ .
$P = 3.24 \times 2.0 = 6.5\,\text{W}$

Result: The total current is $1.8\,\text{A}$ and the $2.0\,\Omega$ resistor dissipates $6.5\,\text{W}$ .

Typical mistakes

Adding parallel resistances directly instead of adding their reciprocals.
Assuming components in parallel share the current equally when their resistances differ - the smaller resistance takes the larger current.

Active revision

A $6.0\,\Omega$ and a $3.0\,\Omega$ resistor are connected in parallel, and this combination is in series with a $2.0\,\Omega$ resistor across a $12\,\text{V}$ supply. Find the total current and the power dissipated in the $2.0\,\Omega$ resistor.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

EMF and internal resistance#

AdvancedAQA 7408 3.5.1.5DfE GCE Physics - emf and internal resistance

Cell with internal resistance driving a load

Fig. 3A cell of emf and internal resistance

r

drives a current round the circuit through an external resistor

R

; the terminal pd is what appears across the load.

The electromotive force (emf)

\varepsilon

of a source is the energy it gives to each coulomb of charge, in volts. A real cell has internal resistance

r

, so some of that energy is dissipated inside the cell itself. When the cell drives a current

I

through an external resistance

R

, the emf divides between the external and internal resistances:

\varepsilon = I(R + r) = IR + Ir

The terminal potential difference is the voltage actually available at the cell's terminals,

V = \varepsilon - Ir

. The term

Ir

is the 'lost volts' dropped across the internal resistance. As the current drawn increases, the lost volts increase and the terminal pd falls; on open circuit (no current) the terminal pd equals the emf. This is why a car's headlights dim when the starter motor - a heavy load - is engaged.

Plotting terminal pd

V

against current

I

gives a straight line: the intercept on the voltage axis (at

I = 0

) is the emf, and the gradient is

-r

. This is exactly the emf/internal-resistance required practical, in which a variable load resistor is used to vary the current while

V

and

I

are measured. The maximum current, on short circuit, is

\varepsilon/r

Internal resistance sets a limit on the power a source can deliver. The power transferred to the load is

P = I^2 R

; analysis shows it is greatest when the load resistance equals the internal resistance, though this is only

50\%

efficient because equal power is then wasted inside the source. Batteries for high-current use are therefore designed with very low internal resistance.

\varepsilon = I(R + r)

EMF equation

The emf drives the current through the external and internal resistances.

V = \varepsilon - Ir

Terminal potential difference

The emf minus the 'lost volts' across the internal resistance.

Terminal pd against current

Fig. 4A straight line of terminal pd against current: the y-intercept is the emf and the gradient is minus the internal resistance.

Worked example

Finding emf and internal resistance

A cell drives $0.40\,\text{A}$ through a $14\,\Omega$ resistor and $0.25\,\text{A}$ through a $23\,\Omega$ resistor. Find the emf and internal resistance.

Write the emf equation twice
$\varepsilon = 0.40(14 + r) = 5.6 + 0.40r$ and $\varepsilon = 0.25(23 + r) = 5.75 + 0.25r$ .
Eliminate emf
Set the two equal: $5.6 + 0.40r = 5.75 + 0.25r$ , so $0.15r = 0.15$ and $r = 1.0\,\Omega$ .
Find the emf
Substitute back: $\varepsilon = 5.6 + 0.40(1.0)$ .
$\varepsilon = 5.6 + 0.40 = 6.0\,\text{V}$

Result: The cell has an emf of $6.0\,\text{V}$ and an internal resistance of $1.0\,\Omega$ .

Typical mistakes

Forgetting the internal resistance and equating the terminal pd to the emf when a current flows.
Reading the gradient of the $V$ - $I$ graph as $+r$ - it is $-r$ , since terminal pd falls as current rises.

Active revision

A cell drives $0.50\,\text{A}$ through a $10\,\Omega$ resistor and $0.30\,\text{A}$ through a $19\,\Omega$ resistor. Determine the emf and internal resistance of the cell.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA) · GCE AS and A level subject content for the sciences (Department for Education)

Potential dividers and sensor circuits#

AdvancedAQA 7408 3.5.1.6DfE GCE Physics - potential dividers

A thermistor potential-divider sensor

Fig. 5A thermistor and a fixed resistor form a divider across the supply; the output tapped between them varies with temperature and can trigger a control circuit.

A potential divider is two (or more) resistors in series across a supply, used to produce a chosen fraction of the supply voltage. The current is common to both, so the voltage across each is in proportion to its resistance. The output taken across the lower resistor

R_2

V_{\text{out}} = \dfrac{R_2}{R_1 + R_2}\,V_{\text{in}}

. A variable resistor (potentiometer) makes this output continuously adjustable, as in a volume control.

The divider becomes a sensor circuit when one resistor is a transducer whose resistance changes with a physical quantity. Replacing

R_1

with a thermistor (resistance falls as temperature rises) makes

V_{\text{out}}

change with temperature; replacing it with a light-dependent resistor (resistance falls as light increases) makes

V_{\text{out}}

change with illumination. Choosing which resistor is the sensor decides whether the output rises or falls with the measured quantity.

The output voltage can then switch or control other electronics - triggering a heater, a fan or an alarm when the temperature or light level crosses a set value. Because the sensor and fixed resistor form a ratio, the design decision is the value of the fixed resistor: it sets the operating point (the temperature or light level at which the output reaches the switching voltage).

A subtlety is loading: connecting a device across the output draws current and lowers

V_{\text{out}}

below the unloaded value, because the load is effectively in parallel with

R_2

. The effect is small if the load's resistance is much larger than

R_2

, which is why sensor outputs feed high-resistance inputs such as an operational amplifier or a voltmeter.

V_{\text{out}} = \dfrac{R_2}{R_1 + R_2}\,V_{\text{in}}

Potential divider

The output across $R_2$ is the supply voltage scaled by the resistance ratio.

Worked example

Output of a light sensor

A $12\,\text{V}$ supply drives a potential divider in which an LDR ( $R_1$ ) is in series with a fixed $1.0\,\text{k}\Omega$ resistor ( $R_2$ ), the output taken across $R_2$ . In darkness the LDR has resistance $11\,\text{k}\Omega$ ; in bright light $200\,\Omega$ . Find the output in both cases.

In darkness
$V_{\text{out}} = \dfrac{R_2}{R_1 + R_2}V_{\text{in}} = \dfrac{1.0}{11 + 1.0}\times12 = 1.0\,\text{V}$ .
In bright light
$V_{\text{out}} = \dfrac{1.0}{0.20 + 1.0}\times12 = \dfrac{1.0}{1.2}\times12$ .
$V_{\text{out}} = 10\,\text{V}$

Result: The output rises from $1.0\,\text{V}$ in darkness to $10\,\text{V}$ in bright light - a signal a control circuit can use.

Typical mistakes

Getting the output ratio upside down - the output across $R_2$ uses $R_2$ in the numerator.
Ignoring the loading effect of a device connected across the output.

Active revision

In a potential divider across a $9.0\,\text{V}$ supply, a thermistor of resistance $2.0\,\text{k}\Omega$ is in series with a fixed $4.0\,\text{k}\Omega$ resistor, with the output taken across the fixed resistor. Find the output voltage, and state how it changes as the thermistor warms.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Physics 7408 specification

Electricity

5 sections~14 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 2

Electricity

Key points

Drift velocity in a copper wire

Key points

Resistivity of a wire

Key points

A mixed circuit

Key points

Finding emf and internal resistance

Key points

Output of a light sensor

Sources

Electricity

Key points

Drift velocity in a copper wire

Key points

Resistivity of a wire

Key points

A mixed circuit

Key points

Finding emf and internal resistance

Key points

Output of a light sensor

Sources