IB DP Physics · Theme B · The particulate nature of matter

Current and circuits.

B.5 · Charge, current and potential difference, resistance, circuits, and cells with internal resistance
Standard level and higher level

SL + HL · 6 hours D.2 Electric and magnetic fields ↗

The journey

Four parts.

Part 01

Current and potential difference

Charge carriers, cells and circuit diagrams

Part 02

Resistance and power

Ohm’s law, resistivity and the heating effect

Part 03

Series and parallel

Combining resistors, and the potential divider

Part 04

Real cells and sensors

Internal resistance, thermistors, LDRs and potentiometers

B.5 CURRENT AND CIRCUITS · SL + HL

Part 01

Current and potential difference.

What flows round a circuit, and what makes it flow

B.5 CURRENT AND CIRCUITS · SL + HL

The story · Italy, 1800

Background story · not assessed

Volta builds the first battery.

Engraved portrait of Alessandro Volta, a man in a high-collared coat — **Alessandro Volta**, engraved in 1837

A tall stack of metal disks held upright between four thin rods on a wooden base, in a museum case — A **voltaic pile** in the Tempio Voltiano, Como

Luigi Galvani had made frogs’ legs twitch with two metals. He and Volta argued over why
Volta stacked disks of zinc and copper (or silver), separated by cloth soaked in brine
20 March 1800: he describes it in a letter to the Royal Society
Within months it was used to split water into hydrogen and oxygen

The first source of a continuous current. Every circuit starts here.

Portrait: Gaetano Bonatti, 1837, CC0. Photograph: GuidoB, CC BY-SA 3.0 (cropped).

B.5 CURRENT AND CIRCUITS · SL + HL

A flow of charge carriers

Electric current.

\(I = \dfrac{\Delta q}{\Delta t}\)

Unit: the ampere, \(1 \,\mathrm{A} = 1 \,\mathrm{C} \,\mathrm{s}^{-1}\)
In a metal the carriers are free electrons
Conventional current: from + to −, opposite to the electrons

INSIDE A METAL WIRE

B.5 CURRENT AND CIRCUITS · SL + HL

Fast, slow and instant

Three speeds are easy to confuse. The free electrons in a metal move at random at about \(10^{6}\) m s⁻¹, with or without a current. A current is a slow drift added to that motion: typically a fraction of a millimeter each second. Yet a lamp lights the moment the switch closes, because the electric field that pushes the electrons travels along the wire at nearly the speed of light, and every electron in the circuit starts to drift at once.

Why two directions

The direction of current was fixed, as from positive to negative, long before the electron was discovered. Electrons are negative, so they move the other way. Every rule in the course, including the left-hand rule in D.3, uses conventional current.

Other carriers

In salt solutions and in batteries the carriers are positive and negative ions, moving in opposite directions. Both contribute to the current in the same sense.

One electron carries \(1.60 \times 10^{-19}\) C, so one coulomb is \(6.25 \times 10^{18}\) electrons.

Mobility of charge carriers

Conductors and insulators.

Conductor

Many carriers, free to move

A metal: about one free electron for every atom, moving through the whole solid

Insulator

Carriers bound in place

Plastic, glass, dry air: the electrons are held to their atoms

In between

A few carriers, set free by energy

Semiconductors such as silicon: heat or light frees more carriers

What matters is how many charge carriers there are and how freely they can move.

B.5 CURRENT AND CIRCUITS · SL + HL

The size of the difference

Copper conducts about \(10^{20}\) times better than glass. Few physical properties vary so much from one material to another. The reason is not that the charges in glass are any different: every material is full of electrons. In glass they are all attached to particular atoms. In copper the outermost electron of each atom belongs to the whole piece of metal.

The third card matters later

A thermistor and a light-dependent resistor, in Part 4, are made of semiconductor. Warming one, or shining light on the other, frees more carriers, so its resistance falls. A metal behaves in the opposite way when heated, as Part 2 explains.

The same free electrons carry thermal energy, which is why good electrical conductors are also good thermal conductors (B.1).

Energy per unit charge

Potential difference.

\(V = \dfrac{W}{q}\)

The work done per unit charge in moving a positive charge between two points
Unit: the volt, \(1 \,\mathrm{V} = 1 \,\mathrm{J} \,\mathrm{C}^{-1}\)
Always between two points: “across” a component

A lamp with 6.0 V across it carries 0.25 A for 2.0 minutes \(\Delta q = I \Delta t = 0.25 \times 120 = \class{res}{30 \,\mathrm{C}}\) \(W = qV = 30 \times 6.0 = \class{res}{180 \,\mathrm{J}}\) Number of electrons \(N = \dfrac{\Delta q}{e} = \dfrac{30}{1.60 \times 10^{-19}} = \class{res}{1.9 \times 10^{20}}\)

B.5 CURRENT AND CIRCUITS · SL + HL

What the volt tells you

6.0 V across the lamp means that every coulomb passing through it transfers 6.0 J, as light and heating. Current tells you how many coulombs pass each second. Potential difference tells you how much energy each one delivers. A torch bulb and a car headlamp may carry similar currents; the headlamp is brighter because each coulomb gives up more energy.

Current is not used up

The same 30 C that entered the lamp left it: charge is conserved. What the charge lost on the way through was energy. A common picture is water turning a wheel: the same water flows on downstream, but it is lower than before.

Speaking of the potential difference “through” a component or the current “across” it. Current goes through; potential difference is across.

D.2 (HL) defines electric potential at a point. The potential difference is the difference between its values at two points.

Where the energy comes from

Cells as sources of emf.

A cell does work on charge: its emf is the energy it supplies to each coulomb, in volts

	Chemical cell	Solar (photovoltaic) cell
Energy from	A chemical reaction inside the cell	Light falling on a semiconductor
For	Portable; works anywhere, at any time; steady emf	No fuel and nothing emitted in use; lasts for decades
Against	Runs down; limited energy stored; materials must be mined and recycled	Output depends on the light: none at night; low emf per cell, so many are joined in series; little power from each square meter, so large areas are needed

B.5 CURRENT AND CIRCUITS · SL + HL

Emf and potential difference

Both are measured in volts and both are energy per unit charge, so they are easy to mix up. The emf of a source is the energy given to each coulomb by the source: chemical or light energy becoming electrical energy. A potential difference across a component is the energy each coulomb gives up there. Round a complete circuit, the energy supplied to a coulomb equals the energy it gives up.

“Electromotive force” is not a force

The name is historical. An emf is measured in volts, not newtons, and the abbreviation is now used as a word in its own right.

Primary and secondary cells

A primary cell is thrown away when its chemicals are used up. In a secondary cell the reaction can be reversed by driving a current backwards through it, which is recharging. Solar cells are often paired with rechargeable cells so that energy collected by day is available at night.

The guide asks for advantages and disadvantages of different sources of electrical energy. Have two of each ready for both kinds of cell.

Circuit diagrams

Circuit symbols.

AS IN THE DATA BOOKLETTHE LONGER LINE OF A CELL IS POSITIVE

Both mean resistor. Many textbooks use the zigzag symbol.

B.5 CURRENT AND CIRCUITS · SL + HL

A diagram is not a picture

A circuit diagram shows what is connected to what, and nothing else. Straight lines stand for wires of negligible resistance, however long or tangled the real ones are. Two diagrams that look different are the same circuit if the same components are joined at the same junctions. When a circuit confuses you, redraw it: mark every junction, then lay the branches out afresh.

Three that look alike

A variable resistor has an arrow through it. A thermistor has a line through it with a flat end. A potentiometer has an arrow that touches it from the side: a third terminal, the sliding contact.

Two symbols for the same resistor

The rectangle is the IEC international style, used in the IB data booklet and throughout this deck. The zigzag is the ANSI/US style and appears in many textbooks, especially American ones. Both represent the same fixed resistor.

Symbol comparison: DigiKey: What are schematic symbols?

The symbols are printed in the data booklet, so there is no need to memorize them, but drawing them quickly and correctly earns marks.

Quick check

Current and potential difference.

0112 C passes a point in 4.0 s. Current?

3.0 A

02Which way do electrons move, compared with conventional current?

The opposite way

0360 J is transferred when 5.0 C passes through a heater. Potential difference?

12 V

04What is one volt in terms of joules and coulombs?

1 J C⁻¹

05What makes a metal a good conductor?

Many free electrons

06Give one disadvantage of a solar cell.

No output in the dark

PRESS → TO REVEAL EACH ANSWER

Part 02

Resistance and power.

How hard it is to drive a current, and what that costs in energy

B.5 CURRENT AND CIRCUITS · SL + HL

Opposition to current

Resistance.

\(R = \dfrac{V}{I}\)

Unit: the ohm, \(1 \,\Omega = 1 \,\mathrm{V} \,\mathrm{A}^{-1}\)
The equation defines resistance, for any component

Origin: drifting electrons collide with the vibrating ions of the metal
Each collision passes energy to the ions: the wire gets hot
Hotter metal: the ions vibrate more, so more resistance

B.5 CURRENT AND CIRCUITS · SL + HL

The picture behind it

The field in the wire accelerates each free electron. Before it has gone far, it collides with an ion and loses the extra energy it had gained, then starts again. The result is a steady average drift, not a steady acceleration, and a steady transfer of energy to the ions, whose vibration is the wire’s thermal energy. Resistance measures how effectively the lattice gets in the way.

Why heating a metal raises its resistance

Ions that vibrate with a larger amplitude present a bigger target, so the electrons are scattered more often and drift more slowly for the same potential difference. A lamp filament at 2500 °C has more than ten times its resistance when cold.

A definition, not a law

\(R = \frac{V}{I}\) can be applied to anything: at each point on its graph a lamp has a resistance, the ratio of \(V\) to \(I\) there. Whether that ratio stays the same as the current changes is a separate question, answered by Ohm’s law.

Finding resistance from the gradient of a curved \(I\)–\(V\) graph. Resistance is the ratio \(\frac{V}{I}\) at the point, not the gradient of the tangent.

The story · Cologne, 1827

Background story · not assessed

A schoolteacher’s law.

Portrait of Georg Ohm, an older man with white hair in a dark coat — **Georg Simon Ohm**

Ohm taught mathematics and physics at a school in Cologne, and built his own apparatus
He tested wires of different lengths and found how the current depended on them
1827: he published his law in The Galvanic Circuit Investigated Mathematically
It was coldly received. He resigned his post
1841: the Royal Society awarded him its Copley Medal

A professor at last, in Munich, in 1852: two years before he died.

Portrait: unknown artist, public domain.

B.5 CURRENT AND CIRCUITS · SL + HL

Ohmic and non-ohmic behavior

Ohm’s law.

For a metal at constant temperature: \(I \propto V\)
Ohmic: constant resistance, a straight line through the origin
A filament lamp is non-ohmic: it heats up, and its resistance rises

OHMIC

FILAMENT LAMP

B.5 CURRENT AND CIRCUITS · SL + HL

Reading the two graphs

Both are graphs of current (up) against potential difference (across). On the left the ratio \(\frac{V}{I}\) is the same at every point. On the right the curve bends towards the \(V\) axis: each extra volt brings less extra current than the one before, so \(\frac{V}{I}\) is growing. Both graphs are symmetric, because neither a resistor nor a lamp cares which way the current goes.

The lamp does not break Ohm’s law

The law is about a metal at constant temperature, and the filament’s temperature is far from constant: it rises by more than two thousand kelvin. If the filament could be kept cool, it would be ohmic like any other piece of metal. The curve is the heating effect made visible.

Stating Ohm’s law as “\(V = IR\)”. That is the definition of resistance and is always true. Ohm’s law is the statement that \(R\) is constant, at constant temperature.

Check which way round the axes are. With \(V\) up and \(I\) across, the lamp’s curve bends the other way.

The material, not the object

Resistivity.

\(\rho = \dfrac{RA}{L}\)

So \(R = \dfrac{\rho L}{A}\): longer means more resistance, thicker means less
Unit: \(\Omega \,\mathrm{m}\)
A property of the material at a given temperature

Material at 20 °C (approximate)	ρ / Ω m
Copper	\(1.7 \times 10^{-8}\)
Aluminum	\(2.8 \times 10^{-8}\)
Nichrome (heating wire)	\(1.1 \times 10^{-6}\)
Silicon	\(\sim 10^{3}\)
Glass	\(\sim 10^{12}\)

B.5 CURRENT AND CIRCUITS · SL + HL

Why length and area

A wire twice as long is two of the original in series: the electrons meet twice as many obstacles, so the resistance doubles. A wire of twice the cross-sectional area is two of the original side by side, in parallel: there are twice as many electrons to carry the current, so the resistance halves.

Resistance and resistivity

Resistance belongs to a particular component: this wire, that lamp. Resistivity belongs to the material, whatever its shape. It plays the same part as density, which describes a material where mass describes an object.

Choosing a material

Copper is used for wiring because of its low resistivity. Overhead power lines use aluminum, which is a worse conductor but so much lighter that a thicker, lighter cable does the same job. Nichrome is chosen for heating elements because its resistivity is high and changes little as it heats.

Using the diameter as the area. \(A = \pi r^{2}\), with the radius in meters: a 1.0 mm diameter is \(r = 0.50 \times 10^{-3}\) m.

Worked example

A length of copper cable.

20 m of copper wire, diameter 1.0 mm · \(\rho = 1.7 \times 10^{-8} \,\Omega \,\mathrm{m}\) \(A = \pi r^{2} = \pi \times (0.50 \times 10^{-3})^{2}\) \(A = 7.9 \times 10^{-7} \,\mathrm{m}^{2}\) \(R = \dfrac{\rho L}{A} = \dfrac{1.7 \times 10^{-8} \times 20}{7.9 \times 10^{-7}}\) \(R = \class{res}{0.43 \,\Omega}\)

Twice the diameter: four times the area, so a quarter of the resistance
The same wire in nichrome: \(\dfrac{1.1 \times 10^{-6}}{1.7 \times 10^{-8}} = 65\) times as much, 28 Ω

B.5 CURRENT AND CIRCUITS · SL + HL

Why connecting wires can be ignored

Twenty meters of ordinary wire has less than half an ohm. The short leads in a laboratory circuit have a few hundredths of an ohm, against tens or hundreds of ohms for the components. Treating the wires as having no resistance is a very good approximation, and it is why every point along one wire in a diagram is at the same potential.

Measuring resistivity

Measure the diameter of a wire with a micrometer in several places and average it. Then measure the resistance of different lengths and plot \(R\) against \(L\). The graph is a straight line through the origin with gradient \(\frac{\rho}{A}\), so \(\rho\) = gradient × \(A\). The diameter matters most: it is squared, so its percentage uncertainty counts twice.

Keep the current small when measuring a wire’s resistance. A large current heats the wire and changes the thing you are measuring.

The heating effect

Electrical power.

\(P = IV = I^{2} R = \dfrac{V^{2}}{R}\)

The rate at which energy is transferred in a component
In a resistor, all of it becomes thermal energy

Always

\(P = IV\)

Coulombs per second × joules per coulomb

When you know the current

\(P = I^{2} R\)

Components in series; losses in cables

When you know the p.d.

\(P = \frac{V^{2}}{R}\)

Components in parallel; appliances on the mains

B.5 CURRENT AND CIRCUITS · SL + HL

Where \(IV\) comes from

\(I\) is the charge passing each second and \(V\) is the energy each coulomb transfers. Multiply them and the coulombs cancel: joules per second, which is watts. The other two forms follow by replacing \(V\) with \(IR\) or \(I\) with \(\frac{V}{R}\).

Does more resistance mean more heating, or less?

Both, depending on what is held fixed. Two resistors in series carry the same current, so by \(I^{2} R\) the larger one gets hotter. Two resistors in parallel have the same potential difference, so by \(\frac{V^{2}}{R}\) the smaller one gets hotter. A mains heater has a low resistance so that it draws a large current.

Why power is sent at high voltage

A cable of resistance \(R\) wastes \(I^{2} R\). Delivering the same power at ten times the voltage needs a tenth of the current and wastes a hundredth as much in the cable.

Energy = power × time. A 2 kW heater running for 3 hours uses 6 kWh, or \(2000 \times 3 \times 3600 = 2.2 \times 10^{7}\) J.

Worked example

A kettle and its cable.

A 2.0 kW kettle on a 230 V supply \(I = \dfrac{P}{V} = \dfrac{2.0 \times 10^{3}}{230} = \class{res}{8.7 \,\mathrm{A}}\) \(R = \dfrac{V^{2}}{P} = \dfrac{230^{2}}{2.0 \times 10^{3}} = \class{res}{26 \,\Omega}\)

Its cable has a resistance of 0.10 Ω \(P_\mathrm{cable} = I^{2} R = 8.7^{2} \times 0.10\) \(P_\mathrm{cable} = \class{res}{7.6 \,\mathrm{W}}\) Under 0.5% of the power: it stays cool

B.5 CURRENT AND CIRCUITS · SL + HL

Why the element glows and the cable does not

The same 8.7 A passes through both, one after the other. The element has 26 Ω and the cable 0.10 Ω, so by \(I^{2} R\) the element takes 260 times the power. Almost all of the 230 V is across the element and less than 1 V across the cable.

What a thin extension lead does

A long, thin lead might have 2 Ω. At 8.7 A it would dissipate \(8.7^{2} \times 2 = 150\) W along its length, enough to become dangerously hot if it is left coiled up. This is why leads carry a maximum current rating.

A fuse is a thin wire that melts when the current exceeds its rating. For this kettle a 13 A fuse is suitable: above the working current, and low enough to protect the cable.

Quick check

Resistance and power.

01A resistor carries 0.20 A with 9.0 V across it. Resistance?

45 Ω

02What does the \(I\)–\(V\) graph of an ohmic conductor look like?

A straight line through the origin

03Why does the resistance of a filament rise as it heats?

The ions vibrate more: more collisions

04A wire is stretched to twice its length with the same volume. Resistance?

Four times: twice \(L\), half \(A\)

05Power in a 10 Ω resistor carrying 3.0 A?

\(3.0^{2} \times 10 = 90 \,\mathrm{W}\)

06What is the unit of resistivity?

Ω m

PRESS → TO REVEAL EACH ANSWER

Part 03

Series and parallel.

Two ways to join components, and two conservation laws that explain them

B.5 CURRENT AND CIRCUITS · SL + HL

One path

Resistors in series.

\(I = I_{1} = I_{2} = \dots\) \(V = V_{1} + V_{2} + \dots\) \(R_\mathrm{s} = R_{1} + R_{2} + \dots\)

The total is larger than the largest resistor.

IN SERIES

B.5 CURRENT AND CIRCUITS · SL + HL

Two conservation laws

The current is the same everywhere in a single loop because charge is conserved: there is nowhere for it to pile up or leak away, so as much must leave each component every second as enters it. The potential differences add up to that of the cell because energy is conserved: each coulomb is given a certain energy by the cell and gives all of it up on the way round.

Where the resistance rule comes from

\(V = V_{1} + V_{2}\) becomes \(I R_\mathrm{s} = I R_{1} + I R_{2}\), and the current, being the same throughout, cancels.

Sharing the potential difference

With the same current in each, \(V = IR\) gives the larger resistor the larger share, in proportion to its resistance. A 2 Ω and a 4 Ω resistor in series across 6 V take 2 V and 4 V.

Old strings of decorative lights were wired in series: when one lamp failed, the single path was broken and all of them went out.

More than one path

Resistors in parallel.

\(V = V_{1} = V_{2} = \dots\) \(I = I_{1} + I_{2} + \dots\) \(\dfrac{1}{R_\mathrm{p}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dots\)

The total is smaller than the smallest resistor.

IN PARALLEL

B.5 CURRENT AND CIRCUITS · SL + HL

The same two laws

Charge is conserved at a junction: the current arriving equals the total current leaving, so the branch currents add up to the current from the cell. Energy is conserved along each route: a coulomb gives up the same energy whichever branch it takes between the same two junctions, so every branch has the same potential difference.

Why adding a resistor lowers the total

Each new branch is one more path for charge. The existing branches carry exactly what they did before, since each still has the full potential difference across it, and the new one adds its own current. More current from the same cell means less resistance overall.

Two useful short cuts

For two resistors, \(R_\mathrm{p} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}\): product over sum. For \(n\) equal resistors \(R\) in parallel, \(R_\mathrm{p} = \frac{R}{n}\).

Stopping at \(\frac{1}{R_\mathrm{p}}\). After adding the reciprocals, turn the answer upside down.

Household sockets are in parallel: each appliance gets the full supply voltage and works independently of the others.

Worked example

A combination.

The parallel pair \(\dfrac{1}{R_\mathrm{p}} = \dfrac{1}{4.0} + \dfrac{1}{12} = \dfrac{4}{12}\) so \(R_\mathrm{p} = 3.0 \,\Omega\) The whole circuit \(R = 6.0 + 3.0 = 9.0 \,\Omega\) \(I = \dfrac{V}{R} = \dfrac{9.0}{9.0} = \class{res}{1.0 \,\mathrm{A}}\) Across the pair \(V_\mathrm{p} = I R_\mathrm{p} = 1.0 \times 3.0 = \class{res}{3.0 \,\mathrm{V}}\)

THREE RESISTORS

B.5 CURRENT AND CIRCUITS · SL + HL

Finishing it off

With 3.0 V across each branch of the pair, the 4.0 Ω resistor carries \(\frac{3.0}{4.0} = 0.75\) A and the 12 Ω resistor \(\frac{3.0}{12} = 0.25\) A. They add up to the 1.0 A from the cell, as they must. The 6.0 Ω resistor has \(1.0 \times 6.0 = 6.0\) V across it, and \(6.0 + 3.0 = 9.0\) V, the emf of the cell.

The method for any network

Work inwards to find the total resistance: replace each parallel group by its single equivalent, then add the series parts. Find the current from the cell. Then work back outwards, using \(V = IR\) on each part, to find what each component gets.

Notice the split

The smaller resistor of the pair carries the larger current: three times the current in a third of the resistance.

Always finish with the two checks: currents at each junction add up, and potential differences round each loop add up to the emf.

Sharing a potential difference

The potential divider.

The same current in both resistors \(I = \dfrac{V_\mathrm{in}}{R_{1} + R_{2}}\) \(V_\mathrm{out} = I R_{2}\) \(V_\mathrm{out} = \class{res}{\dfrac{R_{2}}{R_{1} + R_{2}} V_\mathrm{in}}\)

Each resistor takes a share in proportion to its resistance.

TWO RESISTORS IN SERIES

B.5 CURRENT AND CIRCUITS · SL + HL

Why bother

A supply gives one fixed potential difference, and a circuit often needs a smaller one. Two resistors in series divide it in whatever ratio you choose. With a 12 V supply, \(R_{1} = 3\) kΩ and \(R_{2} = 1\) kΩ, the output is \(\frac{1}{4} \times 12 = 3\) V.

The idea behind sensors

If one of the two resistances changes, with temperature or with light, the way the potential difference is shared changes too, and the output follows. Part 4 uses this to turn a temperature or a light level into a voltage.

A hidden assumption

The formula is exact only if nothing draws current from the output. Connect a component across \(R_{2}\) and it forms a parallel pair with it, of lower resistance, so the output falls. The effect is small if whatever is connected has a much larger resistance than \(R_{2}\).

The potential divider equation is not in the data booklet. It is two lines from \(V = IR\), and it is quicker to reason by ratio than to remember it.

Measuring without disturbing

Ammeters and voltmeters.

Ammeter: in series, so that the current passes through it
Ideal ammeter: zero resistance
Voltmeter: in parallel, across the component
Ideal voltmeter: infinite resistance

WHERE THE METERS GO

B.5 CURRENT AND CIRCUITS · SL + HL

Why those ideals

A meter should report what was happening before it was connected. An ammeter is added in series, so any resistance of its own would add to the circuit’s and reduce the very current it is measuring: it should have none. A voltmeter is added in parallel, so it offers the charge an extra path round the component: it should carry no current, which means an infinite resistance.

A real voltmeter

Suppose a voltmeter of resistance 10 kΩ is placed across a 10 kΩ resistor that is in series with another 10 kΩ resistor on a 6.0 V supply. Before, each resistor had 3.0 V. The voltmeter and its resistor now form a parallel pair of 5.0 kΩ, so the reading is \(\frac{5.0}{15} \times 6.0 = 2.0\) V. The meter has changed what it set out to measure. A modern digital voltmeter has about 10 MΩ, and the same reading would be 3.0 V.

Meters are ideal unless a question says otherwise. If one is not, treat it as an ordinary resistor of constant resistance and include it in the circuit.

Quick check

Series and parallel.

0110 Ω and 15 Ω in series. Total?

25 Ω

02Two 10 Ω resistors in parallel. Total?

5.0 Ω

033.0 Ω and 6.0 Ω in parallel. Total?

\(\frac{3.0 \times 6.0}{9.0} = 2.0 \,\Omega\)

04Which conservation law explains why currents add at a junction?

Conservation of charge

052.0 kΩ above 4.0 kΩ across 12 V. P.d. across the 4.0 kΩ?

\(\frac{4.0}{6.0} \times 12 = 8.0 \,\mathrm{V}\)

06How is a voltmeter connected, and what is its ideal resistance?

In parallel; infinite

PRESS → TO REVEAL EACH ANSWER

Part 04

Real cells and sensors.

A cell with resistance of its own, and resistors that respond to the world

B.5 CURRENT AND CIRCUITS · SL + HL

A real cell

Emf and internal resistance.

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

\(r\): the resistance of the cell itself
Terminal p.d.: \(V = \varepsilon - Ir\)
\(Ir\): the “lost volts”, dissipated inside the cell

AN IDEAL CELL IN SERIES WITH r

B.5 CURRENT AND CIRCUITS · SL + HL

Energy for each coulomb

The cell gives each coulomb \(\varepsilon\) joules. The coulomb gives up \(Ir\) of it in getting through the cell’s own materials and the rest, \(IR\), in the external circuit: \(\varepsilon = IR + Ir\). The equation in the data booklet is that energy balance.

What you notice

The potential difference at the terminals depends on the current. With nothing connected, \(I = 0\), no volts are lost and a voltmeter across the cell reads the emf. Draw a large current and the reading drops. Car headlamps dim for a moment when the starter motor runs, because the motor draws hundreds of amperes and the lost volts inside the battery become large.

Why a flat battery still “has its voltage”

As a cell runs down, its internal resistance rises. Measured with a voltmeter, which draws almost no current, it may still read close to its emf, yet it cannot supply a useful current without most of that being lost inside it.

Treat \(r\) as one more resistor in series with the rest of the circuit. Nothing else about the method changes.

Worked example

A cell and a lamp.

\(\varepsilon = 1.5 \,\mathrm{V}\) · \(r = 0.50 \,\Omega\) · lamp \(R = 2.5 \,\Omega\) \(\varepsilon = I(R + r)\) so \(I = \dfrac{\varepsilon}{R + r}\) \(I = \dfrac{1.5}{2.5 + 0.50} = \class{res}{0.50 \,\mathrm{A}}\) \(V = IR = 0.50 \times 2.5 = \class{res}{1.25 \,\mathrm{V}}\) Lost volts \(= Ir = 0.50 \times 0.50 = 0.25 \,\mathrm{V}\)

Where the power goes Cell supplies: \(\varepsilon I = 1.5 \times 0.50 = 0.75 \,\mathrm{W}\) Lamp: \(I^{2} R = 0.50^{2} \times 2.5 = 0.625 \,\mathrm{W}\) Inside the cell: \(I^{2} r = 0.50^{2} \times 0.50 = 0.125 \,\mathrm{W}\)

B.5 CURRENT AND CIRCUITS · SL + HL

The energy check

\(0.625 + 0.125 = 0.75\) W: everything the cell supplies is accounted for. Five sixths reaches the lamp and one sixth warms the cell, in the same ratio as the two resistances, 2.5 to 0.50.

Short circuit

Join the terminals with a thick wire and \(R\) is nearly zero. The current is limited only by the internal resistance: \(\frac{1.5}{0.50} = 3.0\) A, and the whole 4.5 W heats the cell. For a car battery, with \(r\) of about 0.01 Ω and an emf of 12 V, the short-circuit current is over a thousand amperes, which is why a spanner dropped across its terminals is dangerous.

A source delivers the most power to a load whose resistance equals its internal resistance, but then half the energy is wasted inside the source.

From a graph

Measuring emf and internal resistance.

\(V = \varepsilon - Ir\)

Vary the external resistance; record \(V\) and \(I\)
A straight line: compare with \(y = c + mx\)
Intercept = \(\varepsilon\)
Gradient = \(-r\)

TERMINAL P.D. AGAINST CURRENT

B.5 CURRENT AND CIRCUITS · SL + HL

The circuit

Connect a variable resistor across the cell, with an ammeter in series and a voltmeter across the cell’s terminals. For each setting of the variable resistor, read both meters. Open the switch between readings, so that the cell does not run down or warm up and change its internal resistance during the experiment.

Why a graph beats one pair of readings

The emf is the reading you would get at zero current, which you cannot quite reach with a real voltmeter. Extending the line back to the axis finds it. And the gradient uses every reading, not just two.

The other intercept

Where the line would meet the current axis, \(V = 0\): the cell is short-circuited and the current is \(\frac{\varepsilon}{r}\). Nobody takes that reading, but it can be read from the extended line.

Fruit and vegetable “batteries” have internal resistances of thousands of ohms. They show their emf on a voltmeter but cannot light a lamp.

Resistance that changes

Thermistors and LDRs.

Thermistor: hotter, lower resistance
LDR: brighter, lower resistance
Both are semiconductors: energy frees more charge carriers
Neither is ohmic

TWO SENSORS

B.5 CURRENT AND CIRCUITS · SL + HL

The opposite of a metal

Heating a metal makes its ions vibrate more, which gets in the way of the electrons: resistance rises. In a semiconductor that still happens, but a far bigger effect swamps it. Most of its electrons are bound, and warming the material shakes more of them free. The number of carriers can rise many times over for a few tens of kelvin, so the resistance falls steeply. In an LDR the energy to free the carriers comes from photons of light.

The sizes involved

A typical thermistor might fall from 10 kΩ at 0 °C to a few hundred ohms at 100 °C. A typical LDR might go from about 1 MΩ in the dark to a few hundred ohms in bright light.

The thermistors of this course are the NTC type: negative temperature coefficient, meaning resistance falls as temperature rises.

The variable resistors required are the thermistor, the LDR and the potentiometer.

Worked example

A thermistor in a potential divider.

\(V_\mathrm{in} = 6.0 \,\mathrm{V}\) · \(R_{2} = 10 \,\mathrm{k}\Omega\) · thermistor \(R_{1}\) \(V_\mathrm{out} = \dfrac{R_{2}}{R_{1} + R_{2}} V_\mathrm{in}\) Cold: \(R_{1} = 5.0 \,\mathrm{k}\Omega\) \(V_\mathrm{out} = \dfrac{10}{5.0 + 10} \times 6.0 = \class{res}{4.0 \,\mathrm{V}}\) Hot: \(R_{1} = 1.0 \,\mathrm{k}\Omega\) \(V_\mathrm{out} = \dfrac{10}{1.0 + 10} \times 6.0 = \class{res}{5.5 \,\mathrm{V}}\)

THERMISTOR ABOVE, FIXED RESISTOR BELOW

B.5 CURRENT AND CIRCUITS · SL + HL

Reasoning without the formula

The two components share 6.0 V in proportion to their resistances. When the thermistor warms, its resistance falls, so its share falls. Whatever it gives up, the fixed resistor gains, because the two shares must still add up to 6.0 V. So the output, taken across the fixed resistor, rises with temperature.

Turning it round

Take the output across the thermistor instead and it falls as the temperature rises: 2.0 V cold and 0.5 V hot. Which arrangement you choose depends on what the output has to switch on: a heater when it is cold, or a fan when it is hot.

Light sensors work the same way

Replace the thermistor with an LDR and the output across the fixed resistor rises as the light gets brighter. Street lamps that switch on at dusk use the other arrangement.

Deciding which way the output moves by guessing. Ask first what happens to the sensor’s resistance, then to its share of the potential difference, then to the other component’s share.

A variable potential divider

The potentiometer.

One resistor with a sliding contact
The contact divides it into two parts, \(R_{1}\) and \(R_{2}\)
Moving it changes \(V_\mathrm{out}\) smoothly, from 0 to \(V_\mathrm{in}\)

Volume controls and dimmers are potentiometers.

THREE TERMINALS

B.5 CURRENT AND CIRCUITS · SL + HL

Potentiometer or variable resistor?

The same component can be wired in two ways. Use two of its terminals, one end and the slider, and it is a variable resistor in series with the load: it changes the current, but can never reduce it to zero, because the load itself still has the supply across it. Use all three terminals as on the slide and it is a potential divider: the output can be taken all the way down to 0 V.

Why that matters in the laboratory

To plot the full \(I\)–\(V\) graph of a lamp, starting from the origin, you need potential differences right down to zero. A variable resistor in series cannot provide them. A potentiometer can, which is why it is the standard circuit for that experiment.

With a uniform track the output is proportional to the position of the slider: half way along gives half the input, provided little current is drawn from the output.

In exam diagrams, look for the third connection. An arrow touching the side of a resistor is a potentiometer; an arrow drawn through it is a variable resistor.

Quick check

Real cells and sensors.

01A cell of emf 6.0 V and \(r\) = 1.0 Ω drives 0.50 A. Terminal p.d.?

\(6.0 - 0.50 \times 1.0 = 5.5 \,\mathrm{V}\)

02When does the terminal p.d. equal the emf?

When no current flows

03On a graph of \(V\) against \(I\) for a cell, what is the gradient?

\(-r\)

04A thermistor is warmed. Its resistance?

It falls

05An LDR is the upper resistor of a divider. It gets dark. The output across the lower resistor?

It falls

06How many terminals of a potentiometer are used in a potential divider?

Three

PRESS → TO REVEAL EACH ANSWER

Summary

Current and circuits on one slide.

Current and p.d.

\(I = \frac{\Delta q}{\Delta t}\) · \(V = \frac{W}{q}\)

Coulombs per second; joules per coulomb

Resistance

\(R = \frac{V}{I}\) · \(\rho = \frac{RA}{L}\)

Ohmic: \(R\) constant at constant temperature

Power

\(P = IV = I^{2} R = \frac{V^{2}}{R}\)

The heating effect of a current

Series

\(R_\mathrm{s} = R_{1} + R_{2}\)

Same current; potential differences add

Parallel

\(\frac{1}{R_\mathrm{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}\)

Same potential difference; currents add

Real cell

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

Terminal p.d. \(V = \varepsilon - Ir\)

B.5 CURRENT AND CIRCUITS · SL + HL

Summary

Key words.

CurrentRate of flow of charge

Potential differenceWork done per unit charge between two points

EmfEnergy supplied per unit charge by a source

ResistanceRatio of potential difference to current

ResistivityThe resistance property of a material

OhmicCurrent proportional to potential difference

Internal resistanceThe resistance of the cell itself

Terminal p.d.What the cell delivers to the circuit: \(\varepsilon - Ir\)

Potential dividerResistors in series sharing a potential difference

B.5 CURRENT AND CIRCUITS · SL + HL