Themes › Theme B The particulate nature of matter
B.5Current and circuits
Every phone, torch and car depends on charge flowing round a circuit. This topic explains what current and potential difference really are, why materials resist the flow of charge, how to work out currents and voltages in series and parallel circuits, and how cells supply the energy. It follows the two lesson decks: B.5a Electric circuits and B.5b Cells. There is no additional HL content in B.5.
Knowledge and science
Nature of science
Models. We picture a current as a flow of tiny charged particles, and resistance as those particles bumping into vibrating ions. The same particle model explains thermal conduction in B.1: metals conduct both heat and electricity well because of their free electrons.
Conventions. Benjamin Franklin chose to call one kind of charge "positive" long before electrons were discovered. Conventional current, from + to −, is still used, even though in metals the electrons actually move the other way. A convention doesn't have to match nature, as long as everyone uses it consistently.
Measurement and patterns. In the 1820s Georg Ohm measured currents through wires of different lengths and found that, at constant temperature, current is proportional to potential difference. His results were at first dismissed, partly because good meters hardly existed.
Global impact. Better batteries make electric cars and phones possible, and cheaper solar cells are changing how the world generates electricity.
ToK: questions to think about
- Does it matter that conventional current is "backwards"? Physics still works perfectly with Franklin's choice. What does that tell us about the role of conventions in knowledge?
- Can we know what we can't observe? Nobody has watched an electron drift through a wire. How do we come to trust models of things we infer only from meter readings?
- When is a "law" not a law? Ohm's law only holds for some components, and only at constant temperature. Should it be called a law at all?
- Who decides which energy source is best? Comparing batteries, solar cells and fossil fuels involves cost, safety and environmental damage as well as physics. Which of these can science measure?
1. Charge, conductors and insulators
Electric charge comes in two kinds, positive and negative. Like charges repel; unlike charges attract. Charge is measured in coulombs (C).
- The smallest free charge is the elementary charge, $e = 1.60 \times 10^{-19}$ C. An electron has charge $-e$ and a proton $+e$. Every charge is a whole number of $e$: charge is quantised.
- Charge is conserved: it can't be created or destroyed, only moved. Rubbing a balloon on wool moves electrons from the wool to the balloon, leaving the wool positive by exactly as much as the balloon is negative.
Whether a material conducts depends on its charge carriers, charged particles that are free to move (are mobile):
- Conductors, such as metals, have huge numbers of free electrons that can drift through the material. In salt solutions and molten salts the charge carriers are ions.
- Insulators, such as plastic, rubber and glass, have almost no mobile charge carriers: their electrons are tightly bound to their atoms.
- Semiconductors, such as silicon, have far fewer free charge carriers than metals, and the number rises with temperature or light. They are used in thermistors, LDRs, LEDs and solar cells.
2. Electric current
Direct current (dc) is a flow of charge carriers in one direction. The current $I$ is the rate of flow of charge past a point:
$I$ in amperes (A), $\Delta q$ in coulombs, $\Delta t$ in seconds. $1\ \text{A} = 1\ \text{C s}^{-1}$.
By convention, current is drawn from the positive terminal round to the negative terminal: conventional current is the direction positive charge would flow. In a metal wire, the electrons actually drift the opposite way. Both descriptions give the same answers.
The electrons in a wire drift surprisingly slowly, typically less than a millimetre per second. A lamp still lights almost instantly because the wire is full of electrons already: when the circuit is completed, they all start moving at once, like water in a pipe that is already full.
Worked example: charging a phone
A phone charges with a current of 2.0 A for 30 minutes. How much charge flows, and how many electrons is that?
$\Delta q = I\Delta t = 2.0 \times 30 \times 60 = 3600$ C.
Number of electrons $= \dfrac{3600}{1.60 \times 10^{-19}} = 2.3 \times 10^{22}$.
3. Potential difference
As charge flows round a circuit it carries energy from the cell to the components. The potential difference (p.d.) $V$ between two points is the work done per unit charge on moving a positive charge between those points:
$V$ in volts (V), $W$ in joules, $q$ in coulombs. $1\ \text{V} = 1\ \text{J C}^{-1}$. A lamp with 6 V across it transfers 6 J of energy for every coulomb that passes through it.
"Voltage" is the everyday word for p.d. Potential difference is always measured between two points: it makes no sense to talk about the p.d. "at" a single point or "through" a component.
Worked example: energy from a battery
The 3600 C in the phone example passes through a 5.0 V charger. How much energy is transferred?
$W = qV = 3600 \times 5.0 = 1.8 \times 10^4$ J.
4. Circuit diagrams and meters
Circuit diagrams use standard symbols, which are listed in the data booklet:
- An ammeter measures current, so it goes in series: the current must flow through it. An ideal ammeter has zero resistance, so it doesn't reduce the current it measures.
- A voltmeter measures the p.d. between two points, so it goes in parallel, across the component. An ideal voltmeter has infinite resistance, so no current is diverted through it.
Unless a question says otherwise, treat meters as ideal. If a non-ideal meter is used, its resistance is given and stays constant: include it in the circuit like any other resistor.
5. Resistance and resistivity
The resistance $R$ of a component is the ratio of the p.d. across it to the current through it:
$R$ in ohms (Ω). $1\ \Omega = 1\ \text{V A}^{-1}$.
Where resistance comes from: as free electrons drift through a metal, they keep colliding with the positive ions of the metal lattice, which vibrate about fixed positions. Each collision transfers some of the electrons' kinetic energy to the lattice, so the ions vibrate more: the metal gets hotter. This is the heating effect of a current. In a hotter metal the ions vibrate more, collisions happen more often, and the resistance increases.
The resistance of a wire depends on its material, its length $L$ and its cross-sectional area $A$:
$\rho$ is the resistivity of the material, in Ω m. It is a property of the material alone (at a given temperature). A longer wire has more resistance; a thicker wire has less.
| Material | Resistivity at 20 °C / Ω m |
|---|---|
| copper | $1.7 \times 10^{-8}$ |
| aluminium | $2.8 \times 10^{-8}$ |
| tungsten | $5.6 \times 10^{-8}$ |
| nichrome | $1.1 \times 10^{-6}$ |
| silicon (pure) | about $10^3$ |
| glass | about $10^{12}$ |
Worked example: a heater element
A heater element is made of nichrome wire of diameter 0.40 mm. It needs a resistance of 24 Ω. What length of wire is needed?
$A = \pi r^2 = \pi \times (0.20 \times 10^{-3})^2 = 1.26 \times 10^{-7}\ \text{m}^2$. (Use the radius, in metres.)
$L = \dfrac{RA}{\rho} = \dfrac{24 \times 1.26 \times 10^{-7}}{1.1 \times 10^{-6}} = 2.7$ m.
A copper wire of the same size would need to be over 170 m long, which is why heaters use high-resistivity alloys.
6. Ohm's law: ohmic and non-ohmic conductors
$R = \dfrac{V}{I}$ is the definition of resistance and works for any component at any instant. Ohm's law is the extra claim that $R$ stays the same as $V$ changes. A metal wire at constant temperature is ohmic.
Why a filament lamp is non-ohmic: a bigger current heats the tungsten filament (to over 2000 °C). The lattice ions vibrate more, the electrons collide more often, and the resistance increases. Equal increases in $V$ therefore give smaller and smaller increases in $I$. Reversing the p.d. gives the same shape upside down.
On an I–V graph, the resistance at any point is $\dfrac{V}{I}$ for that point, not the inverse of the gradient of the curve.
7. Electrical power
Power is the rate of energy transfer. Combining $V = \dfrac{W}{q}$ and $I = \dfrac{\Delta q}{\Delta t}$ gives:
$P$ in watts. Use whichever form contains the quantities you know. In a resistor, all this power is dissipated as thermal energy.
Worked example: a kettle
A kettle is rated 2.3 kW at 230 V. Find the current, the resistance of its element, and the energy used in 4.0 minutes, in joules and in kilowatt-hours.
$I = \dfrac{P}{V} = \dfrac{2300}{230} = 10$ A, $R = \dfrac{V}{I} = 23\ \Omega$.
Energy $= Pt = 2300 \times 240 = 5.5 \times 10^5$ J. In kWh: $2.3\ \text{kW} \times \dfrac{4.0}{60}\ \text{h} = 0.15$ kWh. Electricity companies charge per kWh ($1\ \text{kWh} = 3.60 \times 10^6$ J).
8. Series and parallel circuits
Series (one path)
$$I = I_1 = I_2 = \ldots \qquad V = V_1 + V_2 + \ldots$$ $$R_s = R_1 + R_2 + \ldots$$Parallel (separate branches)
$$I = I_1 + I_2 + \ldots \qquad V = V_1 = V_2 = \ldots$$ $$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots$$These rules come from two conservation laws:
- Conservation of charge: charge can't pile up or vanish, so the current into a junction equals the current out of it.
- Conservation of energy: round any complete loop, the energy given to each coulomb by the cell equals the energy it transfers to the components, so the p.d.s add up to the emf.
Useful checks: adding a resistor in series always increases the total resistance; adding a resistor in parallel always decreases it (there's an extra path for the current). The total resistance of a parallel combination is always smaller than its smallest resistor. Two equal resistors in parallel give half the resistance of one.
Worked example: a combination circuit
A 12 V battery (no internal resistance) is connected to a 4.0 Ω resistor in series with a parallel pair of 6.0 Ω and 12 Ω resistors. Find the current from the battery, and the current in each branch.
Parallel pair: $\dfrac{1}{R_p} = \dfrac{1}{6.0} + \dfrac{1}{12} = \dfrac{3}{12}$, so $R_p = 4.0\ \Omega$.
Total: $R = 4.0 + 4.0 = 8.0\ \Omega$, so $I = \dfrac{12}{8.0} = 1.5$ A from the battery.
p.d. across the 4.0 Ω resistor: $1.5 \times 4.0 = 6.0$ V, which leaves $12 - 6.0 = 6.0$ V across the parallel pair.
Branch currents: $\dfrac{6.0}{6.0} = 1.0$ A and $\dfrac{6.0}{12} = 0.50$ A. Check: $1.0 + 0.50 = 1.5$ A ✓.
9. Variable resistors and potential dividers
Some resistors are designed to change their resistance:
- A thermistor (the IB uses the common "NTC" type) is made of a semiconductor. As its temperature rises, more charge carriers are freed, so its resistance falls.
- A light-dependent resistor (LDR) is also a semiconductor. As the light on it gets brighter, its resistance falls.
- A potentiometer has a sliding contact that can be moved along a resistor, giving any fraction of the total p.d.
A potential divider is two (or more) resistors in series across a supply. The supply p.d. is shared between them in proportion to their resistances, so the output p.d. across one of them is a fraction of the input:
where $V_{\text{out}}$ is the p.d. across $R_2$. This follows from the series rules: the current is the same in both resistors, so $V \propto R$. It isn't in the data booklet, so be ready to derive it.
Worked example: a fire alarm sensor
In the circuit above, $V_{\text{in}} = 9.0$ V and $R_2 = 2.0$ kΩ. The thermistor's resistance is 18 kΩ at room temperature and 1.0 kΩ in a fire. Find $V_{\text{out}}$ in each case. The alarm switches on when $V_{\text{out}}$ exceeds 5.0 V. Does it work?
Room temperature: $V_{\text{out}} = 9.0 \times \dfrac{2.0}{18 + 2.0} = 0.90$ V.
Fire: $V_{\text{out}} = 9.0 \times \dfrac{2.0}{1.0 + 2.0} = 6.0$ V.
Yes: in a fire the thermistor's resistance falls, it takes a smaller share of the p.d., so $V_{\text{out}}$ rises above 5.0 V. If the thermistor and the fixed resistor were swapped, $V_{\text{out}}$ would fall when it got hot, which would suit a circuit that switches on a heater in the cold.
10. Cells and emf
A cell is a source of electrical energy. Inside it, energy from another form does work on the charges, moving positive charge from the negative terminal to the positive terminal (to a higher potential). The charges then transfer that energy to the external circuit.
- In a chemical cell, chemical reactions between two different electrodes and an electrolyte convert chemical energy into electrical energy. A primary cell (such as an alkaline battery) is used once and thrown away; a secondary cell (such as a lithium-ion battery) can be recharged by pushing current through it backwards, reversing the reactions.
- A battery is strictly several cells connected in series. Their emfs add: three 1.5 V cells make a 4.5 V battery.
The difference between emf and p.d. is the direction of the energy transfer: emf is energy converted into electrical energy (in a source); p.d. is electrical energy converted into other forms (in a component).
11. Internal resistance
Real cells aren't perfect. The charges also have to pass through the materials inside the cell, which have some resistance: the internal resistance $r$. Some of the emf is used up driving current through $r$, so less is available for the external circuit. This energy warms the cell, which is why batteries get hot when they deliver large currents.
By conservation of energy, the emf equals the sum of the p.d.s round the circuit:
The p.d. across the external resistance is the terminal p.d., $V = IR = \varepsilon - Ir$. The part $Ir$ is sometimes called the "lost volts". When no current flows, $V = \varepsilon$: a high-resistance voltmeter connected straight across a cell reads (almost exactly) its emf.
Worked example: finding the internal resistance
A 9.0 V battery is connected to a 12 Ω resistor. The terminal p.d. is then 8.4 V. Find the current, the internal resistance, and the power wasted inside the battery.
$I = \dfrac{V}{R} = \dfrac{8.4}{12} = 0.70$ A.
Lost volts: $9.0 - 8.4 = 0.6$ V, so $r = \dfrac{0.6}{0.70} = 0.86\ \Omega$.
Power wasted in the battery: $I^2r = 0.70^2 \times 0.86 = 0.42$ W.
Measuring emf and internal resistance
Vary the external resistance $R$ and record pairs of values of $I$ and the terminal p.d. $V$. Rearranging gives $V = -rI + \varepsilon$, so a graph of $V$ against $I$ is a straight line:
12. Solar cells
A solar cell (photovoltaic cell) converts light energy directly into electrical energy. It is made of layers of a semiconductor, usually silicon with small amounts of other elements added. Absorbed photons free electrons, which are pushed through the external circuit. Many cells are connected together to make a solar panel.
- Its output depends on the intensity of the light reaching it, so it changes with the weather, the time of day, the season and the angle of the panel. It produces nothing at night.
- Typical panels convert about 20% of the incident light energy into electrical energy.
Worked example: a solar panel
A solar panel of area 1.6 $\text{m}^2$ has an efficiency of 20%. Sunlight of intensity 800 $\text{W m}^{-2}$ falls on it at right angles. Find its electrical power output.
Input power $= 800 \times 1.6 = 1280$ W. Output $= 0.20 \times 1280 = 260$ W.
13. Comparing sources of electrical energy
- Primary chemical cells. For: portable, cheap, give a steady p.d. and work anywhere at any time. Against: store little energy, can't be recharged, and contain chemicals that harm the environment if thrown away.
- Secondary (rechargeable) cells. For: portable, reusable hundreds of times, and can store energy from renewable sources for later. Against: expensive, slow to recharge, lose capacity over time, and mining lithium and cobalt has environmental and human costs.
- Solar cells. For: renewable, no fuel or emissions while running, little maintenance, useful in remote places. Against: nothing at night and less on cloudy days, need large areas, and their manufacture uses energy and materials.
- Mains electricity from power stations. For: a large, continuous supply that is cheap per kWh. Against: not portable, and often generated from fossil fuels, which add to the enhanced greenhouse effect (B.2).
14. Common mistakes
- Saying current is "used up" in a component. Current is the same before and after a component in series. It is energy that is transferred.
- Connecting meters the wrong way. Ammeters in series, voltmeters in parallel.
- Forgetting to invert when using $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$. The answer is $R_p$, not $\frac{1}{R_p}$.
- Using the diameter instead of the radius in $A = \pi r^2$, or forgetting to convert mm to m.
- Finding resistance from the gradient of a curved I–V graph. Use $R = \frac{V}{I}$ at the point.
- Saying Ohm's law is $V = IR$. $V = IR$ defines resistance. Ohm's law says $R$ is constant (at constant temperature).
- Confusing emf and terminal p.d. They are equal only when no current flows. With current, $V = \varepsilon - Ir$.
- Getting thermistors and LDRs backwards. Hotter thermistor or brighter LDR means lower resistance.
- Mixing up conventional current and electron flow. Conventional current goes from + to − outside the cell.
15. Check your understanding
Three 6.0 Ω resistors are connected in parallel. What is their total resistance? What if they are in series?
Parallel: $\frac{1}{R} = 3 \times \frac{1}{6.0} = \frac{1}{2.0}$, so $R = 2.0\ \Omega$. Series: $18\ \Omega$.
A wire is stretched to twice its length, and its volume stays the same. By what factor does its resistance change?
The length doubles and, because the volume is fixed, the area halves. $R = \frac{\rho L}{A}$ is multiplied by $2 \times 2 = 4$.
Why does a filament lamp have a lower resistance just after it is switched on?
The filament is still cold, so its lattice ions vibrate less and the electrons collide less often. As it heats up, its resistance rises. (That's why lamps most often fail at the moment they're switched on: the current is briefly large.)
An LDR and a fixed resistor form a potential divider, with $V_{\text{out}}$ taken across the fixed resistor. What happens to $V_{\text{out}}$ when it gets darker?
The LDR's resistance increases, so it takes a bigger share of the supply p.d. The p.d. across the fixed resistor, $V_{\text{out}}$, decreases.
A cell of emf 1.5 V and internal resistance 0.50 Ω is short-circuited with a thick copper wire of negligible resistance. What current flows, and why is this dangerous?
$I = \frac{\varepsilon}{R + r} = \frac{1.5}{0 + 0.50} = 3.0$ A. All the power ($I^2r = 4.5$ W) is dissipated inside the cell, which heats up quickly and can leak, or even catch fire for larger batteries.
Why does the terminal p.d. of a battery fall when it supplies a larger current?
$V = \varepsilon - Ir$. A larger current means more "lost volts" across the internal resistance, so less p.d. is left for the external circuit.