Chapter 2 of 8

All 8 chapters
  1. 1Charge and current, readunderstood
  2. 2Potential difference, readunderstood
  3. 3Ohm's law, readunderstood
  4. 4Different components, readunderstood
  5. 5Series circuits, readunderstood
  6. 6Parallel circuits, readunderstood
  7. 7Power in circuits, readunderstood
  8. 8Mains electricity and safety, readunderstood

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

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

  • 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 loop with a battery of two cells on the left, marked + at the top and − at the bottom, with the label 3 V. A closed switch labelled switch and a lit lamp labelled lamp are on the top wire. Arrowheads on the wires point clockwise. Step 1: the loop alone. Step 2: the words cell gives and 3 J each C are in the middle. Step 3: the words lamp takes and 3 J each C are in the middle. Labels: 3 V, switch, lamp, +, −, cell gives, lamp takes, 3 J each C.+−3 Vswitchlamp

    1 A 3 V battery and a lamp in one loop. The charge goes round it.

  2. A loop with a battery of two cells on the left, marked + at the top and − at the bottom, with the label 3 V. A closed switch labelled switch and a lit lamp labelled lamp are on the top wire. Arrowheads on the wires point clockwise. Step 1: the loop alone. Step 2: the words cell gives and 3 J each C are in the middle. Step 3: the words lamp takes and 3 J each C are in the middle. Labels: 3 V, switch, lamp, +, −, cell gives, lamp takes, 3 J each C.+−3 Vswitchlampcell gives3 J each C

    2 Each coulomb that goes through the battery gains 3 J of energy.

  3. A loop with a battery of two cells on the left, marked + at the top and − at the bottom, with the label 3 V. A closed switch labelled switch and a lit lamp labelled lamp are on the top wire. Arrowheads on the wires point clockwise. Step 1: the loop alone. Step 2: the words cell gives and 3 J each C are in the middle. Step 3: the words lamp takes and 3 J each C are in the middle. Labels: 3 V, switch, lamp, +, −, cell gives, lamp takes, 3 J each C.+−3 Vswitchlamplamp takes3 J each C

    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

A table with the columns Source and Energy each C. Row 1: 1.5 V cell, 1.5 J. Row 2: 9 V battery, 9 J. Row 3: 3 V battery, 3 J.SourceEnergy each C1.5 V cell1.5 J3 V battery3 J9 V battery9 J

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?

  1. Write the equation: E = QV.
  2. Put in the numbers: E = 8 C × 3 V.
  3. 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?

  1. Change the energy to joules: 0.27 kJ = 0.27 × 1 000 = 270 J.
  2. We want V. Turn E = QV round to V = E ÷ Q.
  3. 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.

A list of seven circuit symbols, each on a short wire with its name beside it: cell, battery, lamp, switch, resistor, ammeter, voltmeter. The ammeter is a circle with the letter A and the voltmeter is a circle with the letter V. The switch is drawn open. Labels: cell, battery, lamp, switch, resistor, ammeter, A, voltmeter, V.cellbatterylampswitchresistorAammeterVvoltmeter
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.

A loop with a battery of two cells and a lamp. An ammeter is in the loop and reads 0.5 A. A voltmeter is joined by two wires across the lamp and reads 3 V. Labels: A, 0.5 A, V, 3 V.A0.5 AV3 V
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?

  1. Change the current to amps: 500 mA = 500 ÷ 1 000 = 0.5 A.
  2. Change the time to seconds: 2 min = 2 × 60 = 120 s.
  3. Find the charge: Q = It = 0.5 A × 120 s = 60 C.
  4. 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 loop with a battery of two cells on the left, marked + at the top and − at the bottom, with the label 3 V. On the top wire is a voltmeter, a circle with the letter V, labelled about 3 V. On the right wire is an unlit lamp labelled lamp, with no rays round it. Labels: 3 V, V, about 3 V, lamp, +, −.+−3 VVabout 3 Vlamp
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.

  1. 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.
  2. 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.
  3. 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 table with the columns Cells and Bulb. Row 1: one cell, dim. Row 2: two, same way, brighter. Row 3: two, opposite, dark.CellsBulbone celldimtwo, same waybrightertwo, oppositedark

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.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A battery transfers 84 J of energy to 14 C of charge. What is the p.d. across the battery?

    Tap your answer.

    1. Not this one. You multiplied the energy by the charge. The p.d. is the energy for each coulomb, so divide: V = E ÷ Q. Try another.

    2. Not this one. You took the charge away from the energy. The p.d. is the energy divided by the charge. Try another.

    3. Yes, that is right. V = E ÷ Q = 84 J ÷ 14 C = 6 V. Each coulomb gets 6 J.

    4. Not this one. You divided the charge by the energy, the wrong way round. Use V = E ÷ Q. Try another.

    See the answer

    6 V

    V = E ÷ Q = 84 J ÷ 14 C = 6 V. Each coulomb gets 6 J.

    • If you chose A: You multiplied the energy by the charge. The p.d. is the energy for each coulomb, so divide: V = E ÷ Q.
    • If you chose B: You took the charge away from the energy. The p.d. is the energy divided by the charge.
    • If you chose D: You divided the charge by the energy, the wrong way round. Use V = E ÷ Q.

  2. A 9 V battery drives 5 C of charge round a circuit. How much energy does it transfer?

    Tap your answer.

    1. Not this one. You divided the p.d. by the charge. Energy is the charge times the p.d.: E = QV. Try another.

    2. Not this one. You added 9 and 5. Energy is a product: E = QV. Try another.

    3. Not this one. You divided the charge by the p.d. Multiply the two instead. Try another.

    4. Yes, that is right. E = QV = 5 C × 9 V = 45 J.

    See the answer

    45 J

    E = QV = 5 C × 9 V = 45 J.

    • If you chose A: You divided the p.d. by the charge. Energy is the charge times the p.d.: E = QV.
    • If you chose B: You added 9 and 5. Energy is a product: E = QV.
    • If you chose C: You divided the charge by the p.d. Multiply the two instead.

  3. Which statement is right about how to measure the p.d. across a lamp?

    Tap your answer.

    1. Yes, that is right. The p.d. is across two points, so the voltmeter is joined across the lamp, side by side with it.

    2. Not this one. In the loop, a voltmeter blocks the current, and its reading is nearly the battery p.d., not the p.d. across the lamp. Go to S3.10.2. Try another.

    3. Not this one. An ammeter has a tiny resistance. Across the lamp it would give the current a path round the lamp, and the lamp would go dark. Go to S3.10.1. Try another.

    4. Not this one. An ammeter in series reads the current through the lamp, not the p.d. across it. Go to S3.10.1. Try another.

    See the answer

    Join a voltmeter in parallel with the lamp

    The p.d. is across two points, so the voltmeter is joined across the lamp, side by side with it.

    • If you chose B: In the loop, a voltmeter blocks the current, and its reading is nearly the battery p.d., not the p.d. across the lamp. Go to S3.10.2.
    • If you chose C: An ammeter has a tiny resistance. Across the lamp it would give the current a path round the lamp, and the lamp would go dark. Go to S3.10.1.
    • If you chose D: An ammeter in series reads the current through the lamp, not the p.d. across it. Go to S3.10.1.

  4. The battery in the picture is three 1.5 V cells in a line. A voltmeter is across the lamp. How much energy does the battery give to each coulomb of charge that goes through it?

    A loop with a battery of three cells, an ammeter that reads 0.3 A and a lit lamp. A voltmeter is joined by two wires across the lamp. Labels: A, 0.3 A, V.A0.3 AV
    Tap your answer.

    1. Not this one. That is the energy from one cell. The three cells are in a line and their p.d.s add. Go to S3.10.2. Try another.

    2. Yes, that is right. The p.d.s of cells in a line add: 3 × 1.5 V = 4.5 V. Each coulomb gets 4.5 J.

    3. Not this one. That is the current in amps, not the energy for each coulomb. Add the p.d.s of the cells. Try another.

    4. Not this one. You multiplied the p.d. by the number of cells again. The cells add: 1.5 V + 1.5 V + 1.5 V. Try another.

    See the answer

    4.5 J

    The p.d.s of cells in a line add: 3 × 1.5 V = 4.5 V. Each coulomb gets 4.5 J.

    • If you chose A: That is the energy from one cell. The three cells are in a line and their p.d.s add. Go to S3.10.2.
    • If you chose C: That is the current in amps, not the energy for each coulomb. Add the p.d.s of the cells.
    • If you chose D: You multiplied the p.d. by the number of cells again. The cells add: 1.5 V + 1.5 V + 1.5 V.

  5. The p.d. across a lamp is 5 V. What does that tell you?

    Tap your answer.

    1. Not this one. Volts and amps are different. The p.d. is energy for each coulomb. The current depends on the lamp's resistance too. Try another.

    2. Not this one. The lamp does not store charge. The same charge flows in and out. Try another.

    3. Not this one. A p.d. is not used up each second. It is energy for each coulomb, whatever the time. Try another.

    4. Yes, that is right. 5 V is 5 J for each coulomb. So 1 C that passes through the lamp hands over 5 J to the surroundings.

    See the answer

    Each coulomb that goes through the lamp transfers 5 J

    5 V is 5 J for each coulomb. So 1 C that passes through the lamp hands over 5 J to the surroundings.

    • If you chose A: Volts and amps are different. The p.d. is energy for each coulomb. The current depends on the lamp's resistance too.
    • If you chose B: The lamp does not store charge. The same charge flows in and out.
    • If you chose C: A p.d. is not used up each second. It is energy for each coulomb, whatever the time.

  6. From an earlier chapter

    A current of 1.5 A flows for 8 s. How much charge passes?

    Tap your answer.

    1. Not this one. You divided the current by the time. Charge is current times time. Go to S4.12.1. Try another.

    2. Yes, that is right. Q = It = 1.5 A × 8 s = 12 C.

    3. Not this one. You added the two numbers. Charge is a product: Q = It. Go to S4.12.1. Try another.

    4. Not this one. You divided the time by the current. Charge is current times time. Go to S4.12.1. Try another.

    See the answer

    12 C

    Q = It = 1.5 A × 8 s = 12 C.

    • If you chose A: You divided the current by the time. Charge is current times time. Go to S4.12.1.
    • If you chose C: You added the two numbers. Charge is a product: Q = It. Go to S4.12.1.
    • If you chose D: You divided the time by the current. Charge is current times time. Go to S4.12.1.

  7. A 12 V battery transfers energy to 6 C of charge. Find the energy.

    See the answer

    72 J

    E = QV = 6 C × 12 V = 72 J.

  8. A lamp transfers 100 J of energy as 20 C of charge passes through it. Find the p.d. across the lamp.

    See the answer

    5 V

    V = E ÷ Q = 100 J ÷ 20 C = 5 V.

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.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

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