Circuits and Ohm's law. Attnom Academy

S4.12.1 Chapter 1

Charge and current

Amina presses the switch of a torch and the lamp lights at once. What is moving in the wires, and how can we count it?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A Tiny particles that were already in the wire all start to move together.
You picked A. Read on. At the end, the chapter comes back to your guess.
B One electron races from the cell to the lamp, faster than light.
You picked B. Read on. At the end, the chapter comes back to your guess.
C The lamp uses up something from the wire, so less comes back to the cell.
You picked C. Read on. At the end, the chapter comes back to your guess.
D Light runs along the wire from the cell to the lamp.
You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

Calculate charge from current and time
Explain current in metals as a flow of electrons

Press a torch switch and the lamp lights at once, even when the wires are long. So something must already be in the wire, ready to move. That something is electric charge.

Some particles carry charge. An electron carries a negative charge. A metal is made of atoms packed together. Each atom keeps most of its electrons, but one or two are loose. These free electrons wander between the atoms. The rest of each atom is a positive ion, and it stays in its place.

Inside a metal wire. The ions stay still. The free electrons move. The dashed line is an imagined cut across the wire.

With no cell joined, the free electrons wander in every direction, so nothing flows along the wire. Join a cell, and its push acts on every free electron at once. All of them now drift the same way along the wire. A flow of charge is an electric current.

The drift is slow. In an ordinary copper wire carrying 1 A, the electrons move at well under 1 mm each second. The lamp lights at once because the wire is already full of electrons. The cell pushes one end, and all the electrons start to move together, like beads in a full tube.

To count a current, count the charge that passes. Charge is measured in coulombs, written C. Current is the charge that passes a point each second. It is measured in amperes, called amps for short and written A. A current of 1 A means that 1 C of charge passes each second.

See it three ways

As a picture

In words

Current is the charge that passes each second. A bigger current means more charge each second.

In symbols

I = Q ÷ t. Charge Q is in C, time t is in s and current I is in A. So 1 A = 1 C ÷ 1 s.

A steady current gives the same charge every second. After 2 s, twice as much has passed. After 3 s, three times as much. So a graph of charge against time is a straight line through the origin. Look at a current of 2 A.

Draw it first

Charge against time for a steady current

1 Draw the line. A current of 2 A passes 2 C each second, so the line climbs 2 C for every 1 s.
2 At 3 s the line is at 6 C. The charge is the current times the time: 2 A × 3 s = 6 C.
3 At 5 s the line is at 10 C. The steeper the line, the bigger the current. The gradient of the line is the current.

Turn the equation round to find the charge: Q = It. Charge in coulombs is current in amps times time in seconds. The units agree: A × s = (C ÷ s) × s = C. You can also find the time: t = Q ÷ I.

The lamp loop of the next worked example. The ammeter reads 0.5 A.

Worked example

The ammeter in the loop reads 0.5 A. How much charge passes through the lamp in 20 s?

Write the equation: Q = It.
Put in the numbers: Q = 0.5 A × 20 s.
Q = 10 C.

Answer: 10 C

Worked example

A steady current of 0.4 A flows through a lamp. How long does it take 72 C of charge to pass through it? Give the time in minutes.

We want the time. Turn Q = It round to t = Q ÷ I.
Put in the numbers: t = 72 C ÷ 0.4 A = 180 s.
Change to minutes: 180 s ÷ 60 = 3 min.

Answer: 3 min

Before you put numbers in, change them to the base units: amps and seconds. A milliamp, mA, is one thousandth of an amp, so 1 mA = 0.001 A. A minute is 60 s. Skip this step and the answer is wrong by a factor of 1 000 or 60.

Worked example

A phone charger gives 250 mA for 3 minutes. How much charge passes through it?

Change the current to amps: 250 mA = 250 ÷ 1 000 = 0.25 A.
Change the time to seconds: 3 min = 3 × 60 = 180 s.
Put in the numbers: Q = 0.25 A × 180 s.
Q = 45 C.

Answer: 45 C

How big is a coulomb? One electron carries a charge of only 1.6 × 10⁻¹⁹ C. So 1 C is the charge of about 6.25 × 10¹⁸ electrons. That is about 6 million million million electrons. A small current moves a huge number of them.

Now the direction. Early scientists guessed that current was a flow of positive charge from the + end of a cell round to the − end. Later we found that the moving charges in a metal are electrons, and they go the other way. The old rule stayed. Conventional current is drawn from + round to −. Every rule and equation in this book uses it.

Conventional current leaves the + end of the battery and goes round to the − end.

The electrons go the other way, from the − end round to the + end. The same loop, the same lamp.

An ammeter measures the current. It goes in the loop, one part after another, which is called being in series. That way all the charge that goes round passes through it.

One ammeter on each side of the lamp. Both read 0.3 A. The lamp does not use up the current.

The current is the same everywhere in a loop with one path. The lamp does not use up charge. What the lamp changes is energy. The cell gives energy to the charges, and the lamp transfers it to its surroundings as heat and light. The next chapter shows how to measure that.

Try it

A tube of peas

You need: a clear plastic tube or a long drinking straw, about 30 cm, dried peas, lentils or small beads that just slide inside it, a thin stick or a pencil.

Be safe: Dried peas and beads are a choking danger. Keep them away from young children, and never put them in the mouth. Do this on a table, and pick up every pea afterwards.

Fill the tube with peas, so that they touch in a row from one end to the other.
Hold the tube level. Push one more pea into the left end with the stick.
Watch the right end. Notice how soon a pea comes out.
Now take out most of the peas and leave only two, far apart. Push one in at the left end and watch again.

What you will see: With a full tube, a pea falls out of the right end at the moment you push at the left. Each pea moved only a little. The push went along the whole row at once. A full wire of free electrons does the same, so the lamp lights at once.

If you cannot try it, this is what happens.

A wrong idea many people have

The lamp uses up the current, so less current comes back to the cell.

Why it feels right: The lamp gives out heat and light, so it seems to take something away. A battery goes flat, so it seems that something is used up.

The test that breaks it: Put an ammeter on each side of the lamp, as in the picture. Both read 0.3 A. If the lamp used up current, the second would read less.

What is true: Current is not used up. The same charge flows out of the lamp as flows in. The lamp changes energy: it takes energy from the charges and gives out heat and light.

A wrong idea many people have

The lamp lights at once because electrons race along the wire at the speed of light.

Why it feels right: Light and electricity both seem instant. Pictures often show an electron dashing round the loop.

The test that breaks it: Work out the drift speed. In a copper wire carrying 1 A, the electrons drift well under 1 mm each second. A 1 m wire would take hours for one electron. The lamp does not wait for that.

What is true: The wire is already full of free electrons. The push of the cell moves all of them together, so the current starts everywhere at once. Each electron drifts slowly.

A wrong idea many people have

Electrons flow from the + end of the cell to the − end, the same way as the arrows in a circuit drawing.

Why it feels right: We draw the current arrow from + to −, and we tend to think the moving things follow the arrow.

The test that breaks it: An electron has a negative charge. Opposite charges attract, so the + end of the cell pulls electrons towards it and the − end pushes them away. Look at the second loop picture.

What is true: Electrons move from the − end round to the + end. The arrows we draw show conventional current, which goes the other way. The maths gives the same answers either way.

Key words

charge: a property of some particles. It makes them feel electric forces. It is measured in coulombs.
electron: a tiny particle with a negative charge. Free electrons in a metal carry the current.
current: the flow of charge. It is the charge that passes a point each second. Its symbol is I.
coulombs: the unit of charge, written C.
amperes: the unit of current, written A. One ampere is one coulomb each second.
conventional current: current drawn from the + end of a cell round to the − end. The electrons go the other way.
ammeter: a meter that measures current. It goes in the loop, in series.
milliamp: one thousandth of an amp, written mA.

Back to your guess

The answer is A: Tiny particles that were already in the wire all start to move together.

A metal wire is full of free electrons. The cell pushes all of them along together, so the lamp lights at once. They drift slowly, and none is used up: the same charge passes every point of the loop each second.

Keep it

Copy this into your exercise book.

Current is charge each second. I = Q ÷ t, so Q = It. Charge in C, current in A, time in s. Conventional current goes + to −. Electrons go − to +. Current is not used up.

Draw: A loop with a battery, a lamp and an ammeter. Draw the current arrow from + round to −, and a second arrow for the electrons the other way. Write Q = It.

S4.12.2 Chapter 2

Potential difference

Ana's torch cell is marked 1.5 V and her radio battery is marked 9 V. What are those volts measuring?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A How much energy each coulomb of charge gets from the battery.
You picked A. Read on. At the end, the chapter comes back to your guess.
B How much current the battery pushes out each second.
You picked B. Read on. At the end, the chapter comes back to your guess.
C How much charge is stored inside the battery.
You picked C. Read on. At the end, the chapter comes back to your guess.
D How fast the electrons move along the wire.
You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

Explain potential difference as energy transferred per unit of charge
Connect a voltmeter in parallel and an ammeter in series

A torch cell is marked 1.5 V. A radio battery is marked 9 V. Both push charge round a loop. What do the volts measure? Not the current, and not the charge stored inside. They measure the energy that each coulomb of charge gets.

When charge goes through a cell, the cell does work on it and the charge gains energy from the cell's chemical store. When charge goes through a lamp, it hands that energy to the surroundings as heat and light. The potential difference between two points is the energy transferred for each coulomb that passes from one to the other. We write p.d. for short.

Potential difference is measured in volts, written V. One volt is one joule for each coulomb: 1 V = 1 J/C. In symbols, V = E ÷ Q. E is the energy transferred, in joules (J). Q is the charge, in coulombs (C). Watch the loop of a 3 V battery and a lamp.

Draw it first

Energy for each coulomb

1 A 3 V battery and a lamp in one loop. The charge goes round it.
2 Each coulomb that goes through the battery gains 3 J of energy.
3 Each coulomb that goes through the lamp hands 3 J to the surroundings as heat and light. It comes back to the battery with no spare energy, ready to gain 3 J again.

The p.d. tells you the energy for one coulomb, however much charge passes. A 9 V battery gives 9 J to every coulomb. If 100 C passes, the battery transfers 100 times as much energy, so E = QV. Rearranged, V = E ÷ Q, and Q = E ÷ V.

See it three ways

As a picture

In words

The p.d. is the energy given to each coulomb. A bigger p.d. gives more energy to each coulomb.

In symbols

V = E ÷ Q, E = QV, Q = E ÷ V. V in volts, E in joules, Q in coulombs.

Worked example

A 3 V battery transfers energy to a lamp. How much energy does it give to 8 C of charge?

Write the equation: E = QV.
Put in the numbers: E = 8 C × 3 V.
E = 24 J.

Answer: 24 J

Worked example

A lamp transfers 0.27 kJ of energy as 30 C of charge passes through it. What is the p.d. across the lamp?

Change the energy to joules: 0.27 kJ = 0.27 × 1 000 = 270 J.
We want V. Turn E = QV round to V = E ÷ Q.
Put in the numbers: V = 270 J ÷ 30 C = 9 V.

Answer: 9 V

To measure a p.d. you need a voltmeter. It reads the energy for each coulomb between the two points it touches. So it is joined across the part, with its two leads on the two ends of the part. We say it is in parallel with the part. First, the symbols for the parts you meet in circuit diagrams.

Circuit symbols. The long plate of a cell is its + end. A battery is two or more cells. In circuit drawings the number of plates need not match the cells, so read the label for the p.d.

An ammeter reads the current that goes through a part, so it must be in the loop, in series. A good ammeter has a very small resistance, so it hardly changes the current. A voltmeter is joined in parallel. A good voltmeter has a very large resistance, so it takes almost no current from the loop.

The ammeter is in the loop. The voltmeter is across the lamp, side by side with it.

Here is how to tell them apart. The current goes through a part: use an ammeter in series. The p.d. is across a part: use a voltmeter in parallel. This lamp has 3 V across it. That means each coulomb that goes through it hands over 3 J.

Charge and energy often come together in one sum. A current gives the charge, Q = It. A p.d. gives the energy for each coulomb, E = QV. So you can find the energy from a current, a time and a p.d.

Worked example

A 6 V battery drives a current of 500 mA through a motor for 2 minutes. How much energy does it transfer?

Change the current to amps: 500 mA = 500 ÷ 1 000 = 0.5 A.
Change the time to seconds: 2 min = 2 × 60 = 120 s.
Find the charge: Q = It = 0.5 A × 120 s = 60 C.
Find the energy: E = QV = 60 C × 6 V = 360 J.

Answer: 360 J

Two mistakes are common with meters. Neither is harmless. A voltmeter in the loop blocks the current, because of its very large resistance. An ammeter joined across a part gives the current a path with almost no resistance. A very large current can flow, and it can damage the ammeter or the cell. Look at the first mistake.

A voltmeter put in the loop. The lamp stays dark, and the voltmeter reads nearly the whole battery p.d.

Try it

Cells that add and cells that cancel

You need: two torch cells of 1.5 V each, a small torch bulb, two pieces of bare wire or two strips of kitchen foil, sticky tape.

Be safe: Use only torch cells. Never join a wire straight from one end of a cell to the other. The wire and the cell get hot. Switch the bulb off by lifting a wire when you have looked.

Tape a wire from the bottom of the bulb to the + end of one cell. Touch another wire from the side of the bulb to the − end. Note how bright the bulb is.
Now put the two cells end to end in a line, + of one touching − of the other. Join the bulb across the two ends of the line. Note the brightness.
Turn one cell round, so that the two + ends touch. Join the bulb across the two free ends. Note the brightness.

What you will see: One cell lights the bulb dimly. Two cells the same way round make it brighter. Their p.d.s add: 1.5 V + 1.5 V = 3 V, so each coulomb gets 3 J. Two cells facing opposite ways push against each other, and the bulb stays dark or very dim.

If you cannot try it, this is what happens.

A wrong idea many people have

A voltmeter goes in the loop, one part after another, just like an ammeter.

Why it feels right: Both are meters with a letter in a circle, and an ammeter does go in the loop. It seems that all meters should be joined the same way.

The test that breaks it: Look at the picture of the voltmeter in the loop. Its very large resistance leaves almost no current, so the lamp is dark. The reading is close to the whole battery p.d., which tells you nothing about the lamp.

What is true: A p.d. is measured across two points, so a voltmeter goes in parallel with the part. Current is measured through a point, so an ammeter goes in series.

A wrong idea many people have

A battery marked with more volts holds more charge, so it lasts longer.

Why it feels right: Bigger numbers on a label seem to mean a bigger battery.

The test that breaks it: A large 1.5 V torch cell can supply charge for much longer than a tiny 9 V battery. The small battery has more volts but holds much less charge.

What is true: Volts tell the energy each coulomb gets. How much charge a battery can supply is a different thing, called its capacity.

Key words

potential difference: the energy transferred for each coulomb of charge that passes between two points. Its short name is p.d.
volts: the unit of potential difference, written V. One volt is one joule for each coulomb.
voltmeter: a meter that reads the p.d. across a part. It goes in parallel with the part.
in parallel: joined across a part, with its two ends on the two ends of the part.

Back to your guess

The answer is A: How much energy each coulomb of charge gets from the battery.

Volts measure energy for each coulomb of charge: 1 V is 1 J for each C. A 9 V battery gives 9 J to every coulomb that goes through it. The current, the charge stored and the drift speed are different things.

Keep it

Copy this into your exercise book.

p.d. is energy for each coulomb. V = E ÷ Q, so E = QV. 1 V = 1 J/C. Ammeter in series, voltmeter in parallel, across the part.

Draw: A loop with a battery and a lamp. Draw an ammeter in the loop and a voltmeter across the lamp, with the wires to its two ends. Write V = E ÷ Q.

S4.12.3 Chapter 3

Ohm's law

Priya joins a resistor to a 3 V battery and measures the current. Then she doubles the p.d. to 6 V. What happens to the current?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A The current stays the same, because the resistor sets it.
You picked A. Read on. At the end, the chapter comes back to your guess.
B The current doubles.
You picked B. Read on. At the end, the chapter comes back to your guess.
C The current goes up by 3 A.
You picked C. Read on. At the end, the chapter comes back to your guess.
D The current halves.
You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

Use V equals I times R in calculations
Plot a graph of current against voltage for a resistor

A bigger p.d. pushes a bigger current through the same part. How much bigger? The answer has a number in it, and it was found by measuring. This chapter shows how the three quantities fit together: p.d., current and resistance.

The resistance of a part tells how hard it makes it for charge to flow. It is the p.d. across the part divided by the current through it: R = V ÷ I. It is measured in ohms, written Ω. A part has a resistance of 1 Ω when a p.d. of 1 V drives a current of 1 A through it. So 1 Ω = 1 V/A.

A resistor is a part made to have a certain resistance. A variable resistor has a resistance you can change, with a sliding contact. To see how current depends on p.d., set up a circuit that lets you change the p.d. across one resistor.

The symbols for the two kinds of resistor.

The test circuit. The ammeter reads the current through the resistor labelled R. The voltmeter, with the reading 3 V under it, is across that resistor. The variable resistor changes both.

Slide the variable resistor to a new place, then write down the voltmeter and the ammeter readings. Do this for six settings. Here are the readings for a resistor called R.

Six readings. Look at the last column. The p.d. divided by the current is the same every time.

Double the p.d. and the current doubles. Treble it and the current trebles. That pattern is called direct proportion. Plot the readings on a graph with the p.d. along the bottom and the current up the side.

Draw it first

The graph of current against p.d.

1 Plot each reading as a dot: p.d. along, current up.
2 The dots lie on a straight line, and the line goes through the origin. No p.d. means no current.
3 Read a point on the line, such as 4 V and 0.4 A. The p.d. divided by the current is 4 ÷ 0.4 = 10 Ω, as in the table.

A straight line through the origin means the current is proportional to the p.d. This is Ohm's law. For a part like this, current is directly proportional to p.d., as long as its temperature stays the same. A part that follows it is an ohmic conductor. A metal wire or a resistor at a steady temperature is one.

Because V ÷ I stays the same, we can write V = IR. Look closely at what this says. R = V ÷ I is the meaning of resistance, and it holds for every part. Ohm's law is the extra fact that R does not change for an ohmic conductor. For other parts, such as a lamp, V = IR still holds at each moment, but R changes.

See it three ways

As a picture

In words

The p.d. is the current times the resistance. The current is the p.d. divided by the resistance. The resistance is the p.d. divided by the current.

In symbols

V = IR, so I = V ÷ R and R = V ÷ I. The units agree: V = A × Ω, A = V ÷ Ω and Ω = V ÷ A.

You do not need to remember three equations. Start from V = IR and do the same to both sides. To get I alone, divide both sides by R: V ÷ R = I. To get R alone, divide both sides by I: V ÷ I = R. The letter you want ends up alone on one side.

Worked example

A resistor of 20 Ω carries a current of 0.3 A. What is the p.d. across it?

Write the equation: V = IR.
Put in the numbers: V = 0.3 A × 20 Ω.
V = 6 V.

Answer: 6 V

Worked example

A 12 V battery is joined across a resistor of 40 Ω. What current flows? Give the answer in milliamps.

We want I. Turn V = IR round to I = V ÷ R.
Put in the numbers: I = 12 V ÷ 40 Ω = 0.3 A.
Change to milliamps: 0.3 A × 1 000 = 300 mA.

Answer: 300 mA

Worked example

A lamp has a p.d. of 4.5 V across it and a current of 150 mA through it. What is its resistance at that moment?

Change the current to amps: 150 mA = 150 ÷ 1 000 = 0.15 A.
We want R. Turn V = IR round to R = V ÷ I.
Put in the numbers: R = 4.5 V ÷ 0.15 A = 30 Ω.
Check: 0.15 A × 30 Ω = 4.5 V, which is the p.d. we started with.

Answer: 30 Ω

Now compare different resistors on one graph. A bigger resistance lets less current through at the same p.d. So its line is lower, and shallower. The gradient of an I-V graph is I ÷ V, which is 1 ÷ R. A steep line means a small resistance.

Three resistors. The steepest line is the smallest resistance. The gradient of the line is 1 ÷ R.

Try it

A longer pencil, a bigger resistance

You need: two torch cells of 1.5 V each, a small torch bulb, two pencils, one short and one long, soft ones marked B if you have them, three pieces of bare wire or kitchen foil strips, sticky tape.

Be safe: Use only torch cells. The lead of a pencil is safe to touch. Switch off by lifting a wire when you have looked, because the lead can get warm. Use a pencil sharpener, never a knife.

Sharpen both ends of each pencil with a sharpener, so that the dark lead shows at both points. Do not use a knife.
Put the two cells end to end in a line, and tape a wire to each free end. Tape one cell wire to the tip at the bottom of the bulb. Tape a third wire to the bulb's metal side. This leaves one gap, between the third wire and the other cell wire.
Close the gap with the short pencil. Press the free end of the cell wire firmly on one point. Press the free end of the third wire on the other. Note how bright the bulb is.
Do the same with the long pencil, and compare the brightness.

What you will see: The long pencil gives the dimmer bulb. The graphite lead inside is longer, so it has a larger resistance. The p.d. of the cells stays the same, so V = IR gives a smaller current. The bulb gets less current and shines less. A soft, dark pencil works best.

If you cannot try it, this is what happens.

A wrong idea many people have

The resistance of a resistor gets bigger when you increase the p.d. across it.

Why it feels right: A bigger p.d. gives a bigger current, and a bigger current seems to mean more trouble for the charge.

The test that breaks it: Look at the table. The p.d. goes from 1 V to 6 V, and the last column stays at 10 Ω all the way down.

What is true: For an ohmic conductor at a steady temperature, the resistance does not change. A bigger p.d. gives a bigger current, but V ÷ I stays the same.

A wrong idea many people have

R = I ÷ V, or R = V × I.

Why it feels right: The three letters V, I and R can be put together in many ways. People often pick the wrong one.

The test that breaks it: Check the units. Ω is V per A, so R must be V ÷ I. For 4 V and 0.4 A, V × I gives 1.6 and I ÷ V gives 0.1. Neither is the 10 Ω of the graph.

What is true: Start from V = IR, and divide both sides by what you want to remove. R = V ÷ I, and I = V ÷ R. Check that the units of your answer agree.

A wrong idea many people have

The resistance is the gradient of the graph of current against p.d.

Why it feels right: The gradient of a graph is often the quantity of interest, and a steeper line looks like a bigger something.

The test that breaks it: Look at the graph of the three resistors. The steepest line belongs to the smallest resistance, 5 Ω. The gradient of an I-V line is 1 ÷ R.

What is true: On a graph of current against p.d., the gradient is 1 ÷ R. A steeper line means a smaller resistance. To find R, divide a p.d. by its current.

Key words

resistance: how hard a part makes it for charge to flow. R = V ÷ I.
ohm: the unit of resistance, written Ω. One ohm is one volt for each amp.
resistor: a part made to have a certain resistance.
variable resistor: a resistor whose resistance can be changed with a sliding contact.
Ohm's law: the current through an ohmic conductor is directly proportional to the p.d. across it, at a steady temperature.
ohmic conductor: a part that follows Ohm's law, so its resistance does not change.

Back to your guess

The answer is B: The current doubles.

For a resistor at a steady temperature, the current is proportional to the p.d. Double the p.d. and the current doubles. The resistance, which is the p.d. divided by the current, does not change.

Keep it

Copy this into your exercise book.

R = V ÷ I, in ohms. V = IR, I = V ÷ R. An ohmic conductor at a steady temperature gives a straight I-V line through the origin. Steeper means smaller R.

Draw: Axes with p.d. along and current up. Draw two straight lines through the origin, a steep one and a shallow one. Label the steep one small R and the other big R.

S4.12.4 Chapter 4

Different components

A resistor gives a straight line on a graph of current against p.d. Does a lamp do the same? What about a diode?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A Yes, every part gives a straight line, because V = IR holds for all of them.
You picked A. Read on. At the end, the chapter comes back to your guess.
B No, a lamp gives a straight line but a diode does not.
You picked B. Read on. At the end, the chapter comes back to your guess.
C No, a lamp and a diode both give curves, and a diode lets current through one way only.
You picked C. Read on. At the end, the chapter comes back to your guess.
D No, both give a straight line, but with different gradients.
You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

Compare current and voltage graphs for a resistor, a lamp and a diode
Explain how thermistors and light dependent resistors work in sensors

A resistor at a steady temperature follows Ohm's law, and its graph is a straight line through the origin. Many useful parts do not. Their graphs curve or bend, and the shape of the graph tells you what the part does.

Start with a filament lamp. Its filament is a thin coil of metal wire. As the current grows, the filament gets hotter. Its atoms vibrate more, and the moving electrons bump into them more often. So the resistance rises as the filament gets hot. Ohm's law needs a steady temperature, and here the temperature changes.

Look at the graph of a lamp beside a graph of an 8 Ω resistor. We draw only the part for a battery joined the usual way. Reverse the battery and the current reverses. The curve repeats upside down and back to front: at −3 V the current is −0.5 A.

A resistor of 8 Ω and a filament lamp. The dashed lines mark the lamp at 3 V, with 0.5 A, and at 6 V, with 0.75 A. At 6 V the lamp has 8 Ω, the same as the resistor, so the two meet. The curve bends over, so each extra volt gives less extra current.

Find the resistance of the lamp at each point by dividing the p.d. by the current at that point. Do not use the gradient of the curve. R = V ÷ I works at any point, for any part.

See it three ways

As a picture

In words

At 3 V the lamp has a resistance of 6 Ω. At 6 V it has 8 Ω. The hotter filament has the larger resistance.

In symbols

R = V ÷ I. At 3 V: 3 ÷ 0.5 = 6 Ω. At 6 V: 6 ÷ 0.75 = 8 Ω.

A real lamp filament, white hot, can have a resistance ten times its cold value. This lamp rises by much less, so that the numbers stay small.

Next, the diode. A diode lets current through in one direction only. Its symbol is a triangle and a bar. Current can go the way the triangle points. Joined that way round, a silicon diode lets almost no current through until the p.d. across it is about 0.6 V. Above that, the current rises very steeply.

Joined the other way round, a diode lets almost no current through, however big the p.d. A p.d. far too big will damage it, but within its limits the current is close to zero.

A diode. To the left of the current axis it is joined the other way round, and the current is almost zero. To the right, there is almost none until about 0.6 V, and then the current climbs steeply.

To see which way a diode lets current through, follow the conventional current round the loop. In a loop with a battery, the current leaves the + end. If it goes the way the triangle points, the lamp lights. If the diode is turned round, the loop is blocked.

Above, the diode points the way the current goes and the lamp lights. Below, it is turned round and the lamp is dark.

Try it

A diode in a torch

You need: a plain LED torch that takes two or three torch cells, such as AA cells.

Be safe: Use only ordinary torch cells. Leave the cells reversed for a few seconds only. Do not do this with rechargeable cells, or with cells that are damaged or hot. If the turned round cells do not touch the contacts, stop and do not force them. Skip this with a torch you cannot do without. Never open anything that plugs in to a socket.

Switch on the torch. It lights. An LED is a diode that gives out light, so it works one way round only. This works best with a plain torch: cells, a switch and the LED.
Switch off. Open the torch and take out the cells. Look at the + marks.
Put all the cells back the wrong way round, so that every + end is where a − end was. Switch on and look.
Switch off. Take the cells out and put them back the right way round.

What you will see: With the cells the right way round the LED lights. With every cell reversed the LED is joined the wrong way round, so no current goes through it and it stays dark. This is the diode working. A torch with a small circuit board inside may stay dark for another reason. Then the test does not show that the diode did it.

If you cannot try it, this is what happens.

Now two parts whose resistance changes with the world around them. A thermistor is a resistor whose resistance depends on its temperature. The kind in this chapter is called NTC. As it gets hotter, its resistance falls. In this material, heating frees more charge carriers, so more can carry the current. A metal filament does the opposite.

An NTC thermistor. This one halves its resistance for every 20 °C. Other thermistors have other numbers, but the same shape.

Put a thermistor in a loop with a battery and an ammeter. At 20 °C its resistance is 200 Ω. The current is 6 V ÷ 200 Ω = 0.03 A. At 60 °C its resistance is 50 Ω, so the current is 6 V ÷ 50 Ω = 0.12 A. The ammeter reading follows the temperature.

The same thermistor in a loop with a 6 V battery and an ammeter. Hotter means less resistance and more current.

The current shows the temperature. A fire alarm or a thermostat can use this. Electronic parts watch the current or a p.d. When it passes a set value, they switch on a buzzer or a heater. The thermistor does the sensing. The electronics do the switching.

A light dependent resistor, or LDR, works the same way with light. When bright light falls on it, its resistance falls. In the dark, its resistance is very high, often about a million ohms (1 MΩ) or more. Light is measured in lux. A bright room has some hundreds of lux.

An example LDR. More light, less resistance.

One kilohm, kΩ, is 1 000 Ω. A p.d. in volts divided by a resistance in kilohms gives a current in milliamps, mA. For example 6 V ÷ 100 kΩ = 0.06 mA. Use this, or change to ohms and amps first.

The same LDR in a loop with a 6 V battery and an ammeter. Brighter light means less resistance and more current.

The four symbols of this chapter. The two arrows on the LDR are the light falling on it.

Worked example

A thermistor has a resistance of 200 Ω. It is joined to a 6 V battery. What current flows?

Write the equation for the current: I = V ÷ R.
Put in the numbers: I = 6 V ÷ 200 Ω = 0.03 A.
The current is 0.03 A.

Answer: 0.03 A

Worked example

An LDR has a resistance of 30 kΩ in bright light. It is joined to a 6 V battery. What current flows, in milliamps?

Change to ohms: 30 kΩ = 30 × 1 000 = 30 000 Ω.
Write the equation for the current: I = V ÷ R.
Put in the numbers: I = 6 V ÷ 30 000 Ω = 0.0002 A.
Change to milliamps: 0.0002 A × 1 000 = 0.2 mA.

Answer: 0.2 mA

A wrong idea many people have

Every part that has a resistance follows Ohm's law, so every I-V graph is a straight line.

Why it feels right: We write V = IR for everything, and a resistor gives a straight line, so it seems that all parts should.

The test that breaks it: Look at the lamp curve. At 3 V the lamp had 6 Ω and at 6 V it had 8 Ω. V ÷ I changes, so the graph cannot be a straight line through the origin.

What is true: V = IR gives the resistance of any part at any moment. Ohm's law is the extra fact that R stays the same. A resistor at a steady temperature follows it. A lamp, a diode and a thermistor do not.

A wrong idea many people have

A thermistor gets a bigger resistance when it gets hotter, like the filament of a lamp.

Why it feels right: We know that hot metal has more resistance, and a thermistor is also a kind of resistor.

The test that breaks it: Look at the thermistor graph. At 20 °C it has 200 Ω and at 60 °C only 50 Ω. The curve falls as the temperature rises.

What is true: In an NTC thermistor, heating frees more charge carriers, so the resistance falls. In a metal, heating makes the atoms vibrate more, so the resistance rises. The two parts behave differently.

A wrong idea many people have

The resistance of the lamp at a point is the gradient of its curve at that point.

Why it feels right: On many graphs the gradient is the quantity of interest.

The test that breaks it: Take the point 6 V and 0.75 A. V ÷ I gives 8 Ω. The slope of the curve there is about 0.06 A for each volt, and 1 ÷ 0.0625 is 16 Ω, a different number.

What is true: The resistance at a point is the p.d. there divided by the current there. This is true whatever the shape of the curve.

Key words

filament lamp: a lamp with a thin coil of wire that glows when hot. Its resistance rises as it heats.
diode: a part that lets current through one way only, above about 0.6 V for a silicon diode.
thermistor: a resistor whose resistance changes with temperature. An NTC thermistor falls in resistance as it gets hotter.
light dependent resistor: a resistor whose resistance falls as the light on it gets brighter. It is called an LDR for short.
kilohm: one thousand ohms, written kΩ.

Back to your guess

The answer is C: No, a lamp and a diode both give curves, and a diode lets current through one way only.

Only an ohmic conductor at a steady temperature gives a straight line through the origin. A filament lamp gives a curve, because its resistance rises as it heats. A diode lets current through one way and almost none the other way.

Keep it

Copy this into your exercise book.

Resistor: straight I-V line. Lamp: curve, resistance rises as it heats. Diode: one way only, above 0.6 V. Thermistor: R falls as hotter. LDR: R falls as brighter.

Draw: Axes of current and p.d. Sketch a straight line for a resistor, a bending curve for a lamp and a flat then steep curve for a diode.

S4.12.5 Chapter 5

Series circuits

Tariq puts a second lamp in the same loop, and both lamps glow more dimly than one did. Why?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A The first lamp uses up some current, so the second lamp gets less.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Each lamp gets the whole p.d. of the battery but only half the current.
You picked B. Read on. At the end, the chapter comes back to your guess.
C The battery gets weaker as soon as a second lamp is joined.
You picked C. Read on. At the end, the chapter comes back to your guess.
D The lamps share the battery p.d. and add their resistances, so the current falls.
You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

Calculate total resistance in series
Explain how voltage is shared in a series circuit

In a series circuit the parts are joined one after another in a single loop. The charge has only one path to take. Three rules follow from that, and each one comes from something you already know.

Rule 1: the current is the same everywhere in the loop. Charge cannot pile up or vanish, so the same charge passes every point each second. You saw this with the ammeters in the first chapter.

Rule 2: the resistances add up. The charge goes through each resistor in turn, and each one resists. Two resistors end to end are like one longer one. The total resistance is R = R₁ + R₂, and for three resistors R = R₁ + R₂ + R₃.

Adding a resistor in series always raises the total resistance. With the same battery, the current then falls, because I = V ÷ R. Watch it on one battery of 6 V.

Draw it first

More resistors in series, less current

1 One resistor of 4 Ω. The total resistance is 4 Ω, so I = 6 V ÷ 4 Ω = 1.5 A.
2 Add a 6 Ω resistor in series. The total is 4 Ω + 6 Ω = 10 Ω, so I = 6 V ÷ 10 Ω = 0.6 A.
3 Add a 10 Ω resistor. The total is 20 Ω, so I = 6 V ÷ 20 Ω = 0.3 A. Each resistor made the current smaller.

Lamps are resistors that glow. Two lamps in series have about twice the resistance of one. It is a little less, because each dimmer filament is cooler. So the current falls. Each lamp now has less current, and it also has only part of the battery p.d. Both glow less. Look at one lamp and two lamps on a 3 V battery.

One lamp, then two lamps in series on the same battery. The rays show how brightly each lamp glows. Each of two lamps glows less than one lamp alone.

Rule 3: the p.d.s across the parts add up to the p.d. of the supply. Each coulomb gains its energy from the battery, and it gives that energy away in steps, a little in each resistor. So V = V₁ + V₂. Look at the 6 V battery with a 4 Ω and a 6 Ω resistor.

The current is 0.6 A everywhere. Each resistor has its label under its voltmeter reading. The p.d.s are 2.4 V across the 4 Ω resistor and 3.6 V across the 6 Ω one. Together they make the 6 V of the battery.

See the same thing as a stack. The bar is as high as the battery p.d. It is cut into the p.d. across each resistor. The bigger resistor has the bigger part of the bar.

The p.d. of the battery is shared between the resistors. The 6 Ω resistor has 1.5 times the resistance of the 4 Ω resistor, and 1.5 times the p.d.

In a series circuit the p.d. is shared in proportion to the resistance. A resistor with twice the resistance of the other takes twice the p.d. This follows from V = IR, because the current I is the same for both.

See it three ways

As a picture

In words

The same current goes through every part. The p.d.s of the parts add up to the p.d. of the battery. The resistances add up to the total resistance.

In symbols

I = I₁ = I₂. V = V₁ + V₂. R = R₁ + R₂.

Worked example

Two resistors of 15 Ω and 25 Ω are joined in series. What is the total resistance?

Write the rule: R = R₁ + R₂.
Put in the numbers: R = 15 Ω + 25 Ω.
R = 40 Ω.

Answer: 40 Ω

Worked example

A 12 V battery is joined to a 20 Ω resistor and a 40 Ω resistor in series. What current flows?

Find the total resistance: R = 20 Ω + 40 Ω = 60 Ω.
Turn V = IR round to I = V ÷ R.
Put in the numbers: I = 12 V ÷ 60 Ω = 0.2 A.

Answer: 0.2 A

The loop of the next worked example, a 6 V battery with a 10 Ω and a 20 Ω resistor.

Worked example

A 6 V battery is joined to a 10 Ω resistor and a 20 Ω resistor in series. Find the p.d. across each resistor.

Find the total resistance: R = 10 Ω + 20 Ω = 30 Ω.
Find the current: I = V ÷ R = 6 V ÷ 30 Ω = 0.2 A.
The p.d. across the 10 Ω resistor: V = IR = 0.2 A × 10 Ω = 2 V.
The p.d. across the 20 Ω resistor: V = IR = 0.2 A × 20 Ω = 4 V.
Check: 2 V + 4 V = 6 V, which is the p.d. of the battery.

Answer: 2 V and 4 V

A series circuit has a weak point. If any one part is taken out, or a lamp burns out, the loop is broken and the current stops everywhere. This is why a switch goes in series with what it controls.

Try it

One bulb, two bulbs, one removed

You need: two torch cells of 1.5 V each, two small torch bulbs of the same kind, bare wire or kitchen foil strips, sticky tape.

Be safe: Use only torch cells. Do not leave the loop joined for long, because the wires and the bulb can get warm. Lift a wire to switch off.

Put the two cells end to end, + touching −. Join one bulb across the two ends, with wire or foil. Look at how bright it is.
Now put the second bulb in the same loop, one after the other, with a wire between them. Look at both bulbs.
Lift one bulb out of its place, so the loop is broken. Look at the other bulb.

What you will see: One bulb on its own is bright. With two bulbs in the same loop, each is dimmer, because they share the p.d. and the current falls. Take one away and the other goes out too, because the loop is broken.

If you cannot try it, this is what happens.

A wrong idea many people have

In a series circuit the current is shared between the resistors, just like the p.d.

Why it feels right: We say the p.d. is shared, and we tend to think that everything is shared.

The test that breaks it: Remember the two ammeters in chapter 1, or look at the main loop. The ammeter reads 0.6 A. The same charge must go through both resistors, one after the other.

What is true: The current is the same in every part of a series loop. It is the p.d. that is shared. The current is not used up and not split.

A wrong idea many people have

Each resistor in series has the whole battery p.d. across it.

Why it feels right: One resistor alone on the battery does have the whole p.d., so it seems that every resistor must.

The test that breaks it: Look at the main loop. The battery is 6 V, but the voltmeters read 2.4 V and 3.6 V. If each had 6 V, the charge would give away 12 J for each coulomb, and it only gained 6 J.

What is true: The p.d.s across the resistors in series add up to the battery p.d. Each coulomb gives away its energy in steps.

A wrong idea many people have

Adding a resistor in series makes it easier for the current to flow, so the current grows.

Why it feels right: More parts seem to mean more paths. Parallel circuits do give more paths.

The test that breaks it: Look at the strip. The current goes 1.5 A, then 0.6 A, then 0.3 A as resistors are added in series. A resistor in series makes the total resistance bigger, so the current falls.

What is true: In series, the resistances add up, so the total is bigger than any one resistor. The current is smaller. More paths come only with a parallel circuit.

Key words

series circuit: a circuit with the parts in one loop, one after another, so the current has only one path.
total resistance: the one resistance that would give the same current as all the resistors together. In series, R = R₁ + R₂ + R₃.

Back to your guess

The answer is D: The lamps share the battery p.d. and add their resistances, so the current falls.

In a series loop the resistances add, so the current is smaller. The p.d. of the battery is shared between the lamps, so each gets only part of it. Each lamp has less current and less p.d., so it glows less.

Keep it

Copy this into your exercise book.

Series: one path. Same current everywhere. R = R₁ + R₂. The p.d.s add up to the supply. The bigger resistor takes the bigger share. One break stops all.

Draw: A loop with a battery and two resistors in series. Write the same I on both, V₁ and V₂ across them, and V = V₁ + V₂.

S4.12.6 Chapter 6

Parallel circuits

Ana joins two lamps side by side across one battery. Both glow as brightly as one lamp, and she can unscrew one while the other stays lit. How is this loop different?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A The current is shared equally, so each lamp gets half the p.d.
You picked A. Read on. At the end, the chapter comes back to your guess.
B The lamps are in series, but the battery is stronger.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Each lamp has its own path, and each path has the full p.d. of the battery.
You picked C. Read on. At the end, the chapter comes back to your guess.
D The current goes through one lamp only, and the other is lit by its light.
You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

Explain how current splits between parallel branches
Explain why adding resistors in parallel lowers the total resistance

In a parallel circuit the current has more than one path. Each path is a branch. The branches start at one junction and meet again at another, and each branch is joined straight across the supply.

Take a 6 V battery with two branches. One branch has a 6 Ω resistor and the other has a 12 Ω resistor. An ammeter is in each branch, one more is on the wire from the battery, and a voltmeter is across each resistor.

A parallel circuit. Both voltmeters read 6 V. The branch currents are 1 A and 0.5 A, and the battery gives 1.5 A.

Rule 1: every branch has the same p.d. as the supply. Each branch is joined to the same two points. Each coulomb that takes a branch gains the battery's energy, and gives all of it away in that branch.

Rule 2: the currents in the branches add up to the current from the battery. Charge is not lost at a junction, so what goes in comes out. I = I₁ + I₂. Here 1 A + 0.5 A = 1.5 A.

Each branch is a loop of its own with the battery, so each follows V = IR. The branch with 6 Ω has I = 6 V ÷ 6 Ω = 1 A. The branch with 12 Ω has I = 6 V ÷ 12 Ω = 0.5 A. The smaller resistance takes the bigger current, in inverse proportion to the resistance.

See it three ways

As a picture

In words

Each branch has the full p.d. The branch currents add up to the current from the battery. The total resistance is less than the smallest branch.

In symbols

V = V₁ = V₂. I = I₁ + I₂. R is less than R₁ and less than R₂.

Now the surprising part. The total resistance of the two branches is smaller than either resistor. Think of paths. Each new branch gives the charge another way through. The p.d. is unchanged, so the old branches carry the same current as before, and the new branch adds its own. Watch it with 6 Ω resistors on a 6 V battery.

Draw it first

More branches, more current

1 One 6 Ω resistor. The battery gives I = 6 V ÷ 6 Ω = 1 A.
2 A second branch takes another 1 A. The battery gives 2 A, so the total resistance is 6 V ÷ 2 A = 3 Ω.
3 A third branch takes another 1 A. The battery gives 3 A, so the total resistance is 6 V ÷ 3 A = 2 Ω. Each new branch lowers the total resistance.

Back to the first loop. The battery gives 1.5 A at 6 V, so the total resistance is 6 V ÷ 1.5 A = 4 Ω. That is less than 6 Ω, the smallest branch. The 12 Ω branch is an extra path, and it can only add current, never take any away. So the total is always less than the smallest branch.

There is a rule for the total resistance R of resistors in parallel. It adds the reciprocals: 1 ÷ R = 1 ÷ R₁ + 1 ÷ R₂ + 1 ÷ R₃. For just two resistors it simplifies to the product over the sum: R = (R₁ × R₂) ÷ (R₁ + R₂).

Worked example

Two resistors of 6 Ω and 12 Ω are joined in parallel. What is the total resistance?

Write the rule for two resistors: R = (R₁ × R₂) ÷ (R₁ + R₂).
Multiply: 6 × 12 = 72. Add: 6 + 12 = 18.
Divide: R = 72 ÷ 18 = 4 Ω.

Answer: 4 Ω

Worked example

Three resistors of 10 Ω, 20 Ω and 20 Ω are joined in parallel. What is the total resistance?

Write the rule: 1 ÷ R = 1 ÷ R₁ + 1 ÷ R₂ + 1 ÷ R₃.
Put in the numbers: 1 ÷ R = 1 ÷ 10 + 1 ÷ 20 + 1 ÷ 20 = 0.1 + 0.05 + 0.05 = 0.2.
Turn it over: R = 1 ÷ 0.2 = 5 Ω.

Answer: 5 Ω

The loop of the next worked example, a 6 V battery with a 20 Ω branch and a 30 Ω branch.

Worked example

A 6 V battery is joined across a 20 Ω resistor and a 30 Ω resistor in parallel. Find the current in each branch and the current from the battery.

Each branch has the full 6 V across it.
The 20 Ω branch: I = V ÷ R = 6 V ÷ 20 Ω = 0.3 A.
The 30 Ω branch: I = V ÷ R = 6 V ÷ 30 Ω = 0.2 A.
The current from the battery is the sum: 0.3 A + 0.2 A = 0.5 A.
Check with the total resistance: R = (20 × 30) ÷ (20 + 30) = 12 Ω, and 6 V ÷ 12 Ω = 0.5 A.

Answer: 0.3 A, 0.2 A and 0.5 A

Homes are wired in parallel. Each lamp and each socket is on its own branch, with the full p.d. of the supply. Each works at full strength, and a switch in one branch turns off only that branch. Look at two lamps with a switch in one branch.

The switch in the second branch is open, so lamp 2 is off. Lamp 1 is on its own branch, and it stays lit at full brightness.

Try it

Two bulbs side by side

You need: two torch cells of 1.5 V each, two small torch bulbs of the same kind, bare wire or kitchen foil strips, sticky tape.

Be safe: Use only torch cells. Do not leave the loop joined for long, because the wires and the cells can get warm. Lift a wire to switch off.

Put the two cells end to end, + touching −. Join one bulb across the two ends. Look at how bright it is.
Now join the second bulb across the same two ends, side by side with the first. Each bulb needs its own wires to the cells. Look at both bulbs.
Lift one bulb out of its place. Look at the other bulb.

What you will see: The two bulbs side by side are as bright as one on its own. Each has its own path, and each has the full p.d. of the cells. Take one away and the other stays lit, because its branch is still complete. The cells give twice the current, so they run down faster.

If you cannot try it, this is what happens.

A wrong idea many people have

Resistors in parallel add up: R = R₁ + R₂.

Why it feels right: Resistors in series do add, and "total resistance" sounds like a sum.

The test that breaks it: Add 6 Ω and 12 Ω and you get 18 Ω, which is more than either. But the second branch is an extra path, and the loop gave 1.5 A from 6 V, which is 4 Ω.

What is true: In parallel the total is smaller than the smallest branch. Use the product over the sum for two resistors, and then check that your answer is smaller than both.

A wrong idea many people have

In a parallel circuit the p.d. is shared between the branches, as it is in series.

Why it feels right: We say the p.d. is shared in a series circuit, and it is natural to carry the idea over.

The test that breaks it: Look at the two voltmeters in the main picture. Each reads the full 6 V of the battery. No branch has only a part of it.

What is true: Each branch is joined across the supply, so each has the full p.d. In parallel it is the current that is shared out.

A wrong idea many people have

The current splits equally between the branches.

Why it feels right: Two paths look like two halves.

The test that breaks it: In the main picture the branches carry 1 A and 0.5 A. The branch with the smaller resistance took twice as much.

What is true: The current splits in inverse proportion to the resistances. With the same p.d., I = V ÷ R, so the smaller resistance takes the bigger current. The shares are equal only when the resistances are equal.

Key words

parallel circuit: a circuit with more than one path, where each branch is joined straight across the supply.
branch: one of the paths in a parallel circuit.

Back to your guess

The answer is C: Each lamp has its own path, and each path has the full p.d. of the battery.

In a parallel circuit each lamp is joined straight across the battery. Each has the full p.d., so each glows as brightly as it would alone. A break in one branch leaves the other branch complete.

Keep it

Copy this into your exercise book.

Parallel: branches across the supply. Same p.d. in each. Branch currents add. Smaller R takes more current. Total R is less than the smallest. Two resistors: R = R₁R₂ ÷ (R₁ + R₂).

Draw: A battery with two branches, each with a resistor. Write the same V on both, I₁ and I₂ on the branches and I = I₁ + I₂ at the junction.

S4.12.7 Chapter 7

Power in circuits

Amina's heater is marked 920 W and her lamp is marked 40 W. What does a watt measure in a circuit, and why does the heater need a bigger fuse?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A The energy transferred each second, which is current times p.d. The heater takes a bigger current.
You picked A. Read on. At the end, the chapter comes back to your guess.
B The charge stored in the appliance. The heater stores more charge.
You picked B. Read on. At the end, the chapter comes back to your guess.
C The p.d. across the appliance. The heater has a bigger p.d.
You picked C. Read on. At the end, the chapter comes back to your guess.
D The resistance of the appliance. The heater has a bigger resistance.
You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

Calculate power using P equals I times V and I squared times R
Choose the right fuse for an appliance

You know that power is the energy transferred each second: P = E ÷ t, in watts. One watt is one joule each second. A lamp marked 40 W transfers 40 J every second. This chapter finds the power of a circuit part from its current and p.d.

Start with what you know. A p.d. of V volts means V joules for each coulomb. A current of I amps means I coulombs each second. So in one second, I coulombs each hand over V joules. The energy each second is I × V.

In symbols: E = QV and Q = It, so E = ItV. Divide both sides by t, and the power is P = E ÷ t = IV. The power in watts is the current in amps times the p.d. in volts. The units agree: A × V = (C ÷ s) × (J ÷ C) = J ÷ s = W.

A lamp with an ammeter and a voltmeter. Multiply the two readings to find the power.

Worked example

The ammeter reads 0.5 A and the voltmeter reads 3 V. What power does the lamp transfer?

Write the equation: P = IV.
Put in the numbers: P = 0.5 A × 3 V.
P = 1.5 W.

Answer: 1.5 W

Why P = IV. Each second 0.5 C goes round, and each coulomb hands over 3 J, which is 3 J/C. That is 1.5 J each second, which is 1.5 W.

Often you know the current and the resistance but not the p.d. Put V = IR into P = IV. Then P = I × IR, which is P = I²R. For a resistor, the power is the square of the current times the resistance. This is the energy turned to heat each second.

Because the current is squared, a small rise in current makes a big rise in power. Double the current and the power is four times as big. Treble it and the power is nine times as big. Look at the graph for a 4 Ω resistor.

P = I²R for a 4 Ω resistor. At 1 A the power is 4 W. At 2 A it is 16 W, four times as much.

Worked example

A current of 2 A flows through a heating element of 5 Ω. What power does it transfer?

Write the equation: P = I²R.
Square the current: 2 A × 2 A = 4.
Put in the numbers: P = 4 × 5 Ω = 20 W.

Answer: 20 W

Energy follows from power. In a time t, the energy transferred is E = Pt. With P in watts and t in seconds, E is in joules. A 60 W lamp left on for 10 s transfers 60 W × 10 s = 600 J.

For big amounts of energy, an electricity meter uses the kilowatt-hour, kWh. It is the energy of a 1 kW appliance running for 1 hour. With P in kilowatts and t in hours, E = Pt gives kWh. Since 1 kW = 1 000 W and 1 h = 3 600 s, 1 kWh = 3.6 million J.

Mains voltage differs between countries. These examples use 230 V. At 120 V, an appliance needs about twice the current for the same power. The table shows the current I = P ÷ V for some devices at 230 V.

See it three ways

As a picture

In words

The more power a device has, the more current it takes from the same p.d. A kettle takes about sixty times the current of a lamp.

In symbols

I = P ÷ V. For the kettle: I = 2 300 W ÷ 230 V = 10 A.

A large current heats the wires. A fault can make a very large current flow, and wires that get too hot can start a fire. A fuse is a thin wire in a small cartridge, joined in series in the live wire. If the current goes above its rating, the thin wire gets hot and melts. That breaks the circuit.

A fuse in a heater circuit. Above, the fuse wire is whole and the heater works. Below, it has melted and the circuit is broken.

A fuse is chosen by its rating, in amps. The rating must be a little above the normal current of the appliance. Too low a rating, and the fuse melts in normal use. Too high a rating, and a fault can send a dangerous current through the wires without melting it.

Find the normal current from I = P ÷ V. Then pick the smallest rating above it. Fuses come in set sizes, and the sizes differ from place to place. In the examples here they are 3 A, 5 A and 13 A.

Worked example

A heater is marked 920 W and runs from 230 V. The fuses available are 3 A, 5 A and 13 A. Which fuse should it have?

Find the normal current: I = P ÷ V = 920 W ÷ 230 V = 4 A.
The fuse must be a little above 4 A. The smallest rating above 4 A is 5 A.
A 3 A fuse is below 4 A and would melt in normal use. A 13 A fuse is far above 4 A.
Check the margin: 5 A − 4 A = 1 A, which is small.
Choose the 5 A fuse.

Answer: The 5 A fuse

Try it

The power of a torch bulb

You need: a small torch bulb with its markings, for example 2.5 V 0.3 A, a pencil and paper.

Be safe: Only read the markings on a bulb that is not in a circuit. Do not touch, plug in or switch on any mains lamp or appliance. Keep the glass of the bulb whole.

Look at the metal base of the bulb or at its glass. The p.d. in V and the current in A are marked on it, in small print.
Write them down. If only the power in W is marked, find the current from I = P ÷ V.
Work out the power: P = IV. Then compare it with the 40 W of a room lamp, a lamp you do not touch.

What you will see: A torch bulb marked 2.5 V and 0.3 A has a power of 2.5 V × 0.3 A = 0.75 W. A 40 W lamp transfers about fifty times as much energy each second. That is why the room lamp needs mains wires and a fuse, and the torch needs only a cell.

If you cannot try it, this is what happens.

A wrong idea many people have

Doubling the current through a resistor doubles the power.

Why it feels right: Many things double when the current doubles, such as the charge in a given time. It seems that power should too.

The test that breaks it: Look at the graph of P = I²R for 4 Ω. At 1 A the power is 4 W. At 2 A it is 16 W, which is four times as much, not twice.

What is true: The power depends on the square of the current. Double the current and the power is four times as big. Use this to see why thin wires carrying large currents get hot.

A wrong idea many people have

A bigger resistance always means a bigger power.

Why it feels right: P = I²R has R in it, so it seems that a larger R must give a larger P.

The test that breaks it: Join 6 Ω and then 12 Ω across the same 12 V battery. The 6 Ω resistor takes 2 A, so P = 2 A × 12 V = 24 W. The 12 Ω resistor takes 1 A, so P = 1 A × 12 V = 12 W. The bigger resistance gave the smaller power.

What is true: The answer depends on what stays the same. With the current fixed, a bigger R gives more power. With the p.d. fixed, as with a battery or the mains, a bigger R takes a smaller current and gives less power.

A wrong idea many people have

A fuse of a bigger rating is safer, so always choose the biggest.

Why it feels right: A big fuse never melts in normal use, so it seems to be the more reliable choice.

The test that breaks it: A 4 A kettle in a loop with a 13 A fuse could carry a fault of 12 A. That heats the wires a great deal, and the fuse does not melt. A 5 A fuse would melt long before.

What is true: A fuse must melt before the wires get dangerously hot. Choose a rating just above the normal current: low enough to protect, high enough not to melt in normal use.

Key words

kilowatt-hour: a unit of energy, written kWh. It is the energy of a 1 kW appliance running for 1 hour. 1 kWh is 3.6 million J.
fuse: a thin wire in the live wire that melts and breaks the circuit when the current goes above its rating.

Back to your guess

The answer is A: The energy transferred each second, which is current times p.d. The heater takes a bigger current.

Power is the energy transferred each second. In a circuit P = IV. On the same mains, a 920 W heater takes a much bigger current than a 40 W lamp. So it needs a stronger fuse. A heater has a smaller resistance than a lamp.

Keep it

Copy this into your exercise book.

P = IV, and P = I²R. Watts are joules each second. E = Pt. Normal current I = P ÷ V. A fuse goes in the live wire, rated just above I. Double I, four times the power.

Draw: A loop with a lamp, an ammeter and a voltmeter. Write P = IV beside it. Add a fuse in the live wire and its rating just above the normal current.

S4.12.8 Chapter 8

Mains electricity and safety

Hana can take a torch apart, but she must never take apart a plugged in lamp. What is different about the mains?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A The mains alternates and has more energy, but a torch cell is just as safe.
You picked A. Read on. At the end, the chapter comes back to your guess.
B The wall socket gives no current until a lamp is plugged in, so the two are alike.
You picked B. Read on. At the end, the chapter comes back to your guess.
C A torch cell gives a few volts. The mains gives about 230 V, which can push a deadly current through the body.
You picked C. Read on. At the end, the chapter comes back to your guess.
D The mains is direct current and a torch cell is alternating current.
You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

Explain the difference between direct and alternating current
Explain the jobs of the live, neutral and earth wires and the fuse
Explain why mains electricity is never used for home experiments

A torch cell pushes charge round the loop the same way all the time. This is direct current, d.c. The p.d. of the cell stays at about 1.5 V, and the current keeps its direction. Look at a graph of p.d. against time.

Direct current. The p.d. of a cell stays the same and never changes sign.

The mains is different. It gives alternating current, a.c. The p.d. swings up, falls through zero, swings the other way, and returns. The current reverses direction again and again. The electrons do not travel round the loop. They shuffle to and fro about their places, and the energy still reaches the appliance.

Alternating current from a 230 V, 50 Hz supply. One whole swing takes 20 ms. The peaks, at 5 ms and 25 ms, are 20 ms apart.

Read the graph like a wave. The time for one whole swing is the period T. The number of swings each second is the frequency f. They are linked by f = 1 ÷ T. A supply with a frequency of 50 Hz swings 50 times each second. Each swing takes 1 ÷ 50 = 0.02 s, which is 20 ms.

The 230 V we quote is not the top of the swing. It is the root mean square, or r.m.s., value: the steady p.d. that would give the same heating. For a smooth swing like this, the peak is always about 1.41 times the r.m.s. value. That 1.41 is the square root of 2. So the peak is 230 V × 1.41, about 325 V.

The current in the loop in one half swing, above, and in the next half swing, below. The direction reverses, and the lamp stays lit both ways.

Worked example

The p.d. of a supply repeats every 0.02 s. What is the frequency?

Write the equation: f = 1 ÷ T.
Put in the numbers: f = 1 ÷ 0.02 s.
f = 50 Hz.

Answer: 50 Hz

The mains p.d. and frequency are not the same everywhere. Many places use 230 V and 50 Hz. Others use 120 V and 60 Hz. Some use other values. The label on an appliance says which supply it is made for.

Examples of mains supplies. The values differ from place to place.

A mains cable has three wires, each with insulation. The live wire carries the alternating p.d. of the supply. The neutral wire completes the circuit, so the current can return. The earth wire is only for safety. It does not carry current in normal use.

Inside a mains cable. The dots, hatching and cross hatching tell the three wires apart on a black and white copy.

The wires are told apart by the colour of the insulation. One common scheme is brown for live, blue for neutral and green with a yellow stripe for earth. Colours vary from place to place, for example black, white and green in North America. Always follow the rule of your own place, and leave wiring to a trained person.

The jobs of the three wires. In the common scheme, live is brown, neutral is blue and earth is green with a yellow stripe.

Why a safety wire? Many appliances, such as a kettle or a toaster, have a metal case. Suppose the live wire inside comes loose and touches the case. The case is then live, and anyone who touches it gets a path to the ground through their body.

The earth wire joins the case to the ground. Now the fault has a path of very low resistance. A large current flows at once through the earth wire, and the fuse in the live wire melts. The appliance is cut off from the live supply, and the case is safe.

Above, with an earth wire, the fault current melts the fuse. Below, with no earth wire, the case stays live and the only way to the ground is through a person.

Appliances with a plastic case, called double insulated, have no earth wire. No metal part can become live, so there is nothing to earth.

The fuse goes in the live wire. When it melts, the appliance is cut off from the live supply. A fuse in the neutral wire would leave the appliance joined to the live wire after it melted. On its own, a fuse does not stop a shock to a person who touches a live wire. 13 A is far more than can kill, and a fuse takes time to melt. With an earth wire, it cuts off a faulty metal case before anyone touches it.

Many homes also have a residual current device. It cuts off the supply very fast when a little current leaks to earth, for example through a person. A trained person fits and tests it.

Now see why the mains is never used for experiments. The current through a person is I = V ÷ R. Here R is the resistance of the body between the two touched points. R varies a lot. Dry skin can have 100 000 Ω or more. Wet skin or a cut can have 2 000 Ω or less. These are rough values.

Worked example

A person touches both ends of a 1.5 V cell. The resistance of the body between the touched points is 5 000 Ω. What is the current, in milliamps?

Write the equation: I = V ÷ R.
Put in the numbers: I = 1.5 V ÷ 5 000 Ω = 0.0003 A.
Change to milliamps: 0.0003 A × 1 000 = 0.3 mA.

Answer: 0.3 mA

Worked example

A person with wet skin touches a live mains wire at 230 V. The resistance of the body is 2 kΩ. What is the current, and how does it compare with the 0.05 A that can kill?

Change to ohms: 2 kΩ = 2 × 1 000 = 2 000 Ω.
Write the equation: I = V ÷ R.
Put in the numbers: I = 230 V ÷ 2 000 Ω = 0.115 A.
Compare: 0.115 A ÷ 0.05 A = 2.3, more than twice the current that can kill.

Answer: 0.115 A

The cell pushes less than a thousandth of an amp through the body, too little to feel. The mains pushes a current that can stop the heart. Rough guides for the current through the body are in the table. They depend on the path through the body and on how long the current lasts.

Rough guides to what a current through the body does. A mains p.d. can drive far more than 50 mA through wet skin.

A phone charger turns the mains into a few volts of d.c. inside a sealed case. The sealed case is the safe part. Never open a charger or any plugged in appliance. Any lamp, motor or sensor you build for a test should run from cells, and never from the wall socket.

Try it

Alternating current by hand

You need: one torch cell of 1.5 V, a small torch bulb, two pieces of bare wire or kitchen foil strips, sticky tape.

Be safe: Use only a torch cell. Do not leave the wires on the cell for long, because they and the bulb get warm. Never try this with anything that plugs in to a socket.

Tape one wire to the tip at the bottom of the bulb and the other to its metal side. Touch the free ends to the two ends of the cell. The bulb lights.
Swap the wires, so that each wire touches the other end of the cell. The bulb lights again. The current now goes the other way round the loop.
Swap the wires back and forth, about once each second, and say "one, two" aloud. You are making an alternating current of about 0.5 Hz.

What you will see: The bulb lights whichever way the current goes. A lamp does not care about the direction. A 50 Hz mains reverses the current 100 times each second, once every hundredth of a second.

If you cannot try it, this is what happens.

A wrong idea many people have

The neutral wire is at about zero p.d., so it is safe to touch.

Why it feels right: The neutral wire is labelled as the return wire and sits close to zero p.d. in a working circuit.

The test that breaks it: A fault, or a break in the neutral wire, can leave the neutral wire at a high p.d. A wire you cannot see is working may carry the live p.d. You cannot tell by looking.

What is true: Treat every wire in a mains circuit as live until a trained person has made it safe. Never touch exposed wire in any socket, plug or appliance.

A wrong idea many people have

A fuse stops a person who touches a live wire from getting a shock.

Why it feels right: A fuse is a safety part, and with an earth wire it does cut off a faulty metal case.

The test that breaks it: A 13 A fuse lets 13 A flow before it melts. That is more than 200 times the 0.05 A that can kill, and a fuse can take a moment to melt.

What is true: A fuse melts at currents far above what can kill. It protects the wires from overheating, and with the earth wire it cuts off a faulty metal case. A residual current device, and keeping away from live parts, protect a person who touches a live wire.

A wrong idea many people have

In alternating current the electrons travel from the power station to the lamp.

Why it feels right: We say that electricity is sent from the power station, and we picture the electrons going the whole way.

The test that breaks it: With a 50 Hz supply the current reverses 100 times each second. An electron that moves forward for a hundredth of a second is moved back again for the next hundredth. It ends where it started.

What is true: The electrons shuffle to and fro about their places. The energy travels along the wire in the changing p.d. and reaches the appliance.

Key words

r.m.s.: root mean square. The r.m.s. value of an alternating p.d. is the steady p.d. that would give the same heating. The quoted mains p.d. is an r.m.s. value.
direct current: current that flows one way only, such as from a cell. It is written d.c.
alternating current: current that keeps reversing direction, such as from the mains. It is written a.c.
live: the wire that carries the alternating p.d. of the mains supply.
neutral: the wire that completes the circuit so the current can return.
earth: the safety wire that joins a metal case to the ground. It carries no current in normal use.

Back to your guess

The answer is C: A torch cell gives a few volts. The mains gives about 230 V, which can push a deadly current through the body.

A torch cell gives a few volts and cannot push a harmful current through dry skin. The mains gives hundreds of volts, and the current it can drive through the body can kill. The mains is also alternating, but it is the p.d. and the current that make it dangerous.

Keep it

Copy this into your exercise book.

d.c. one way, a.c. keeps reversing. f = 1 ÷ T. Live: supply. Neutral: return. Earth: safety. Fuse in the live wire. Mains can kill: use cells only.

Draw: A metal cased kettle with a live, a neutral and an earth wire and a fuse in the live wire. Mark where a fault would put the live wire on the case.

Charge against time for a steady current

A tube of peas

Energy for each coulomb

Cells that add and cells that cancel

The graph of current against p.d.

A longer pencil, a bigger resistance

A diode in a torch

More resistors in series, less current

One bulb, two bulbs, one removed

More branches, more current

Two bulbs side by side

The power of a torch bulb

Alternating current by hand

Key words, A to Z