Chapter 7 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

S4.12.7 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

  • 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

  • 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 loop with a battery labelled 3 V made of two cells, an ammeter that reads 0.5 A and a lamp. A voltmeter is joined across the lamp and reads 3 V. Labels: 3 V, A, 0.5 A, V, 3 V.3 VA0.5 AV3 V
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?

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

Answer: 1.5 W

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. In the middle are the words each second, 0.5 C × 3 J/C and = 1.5 J. Labels: 3 V, switch, lamp, +, −, each second, 0.5 C × 3 J/C, = 1.5 J.+−3 Vswitchlampeach second0.5 C × 3 J/C= 1.5 J
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.

A graph with the axes power (W) and current (A). The current runs from 0 to 3 A in steps of 0.5 and the power from 0 to 36 W in steps of 6. A curve starts at the origin and climbs more and more steeply, labelled R = 4 Ω. Two dots on the curve have dashed lines to both axes: 1 A and 4 W, and 2 A and 16 W. Labels: power (W), current (A), R = 4 Ω.R = 4 Ω0.511.522.53612182430360current (A)power (W)
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?

  1. Write the equation: P = I²R.
  2. Square the current: 2 A × 2 A = 4.
  3. 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

A table with the columns Device, P (W) and I (A). Row 1: lamp, 40, 0.17. Row 2: fan, 60, 0.26. Row 3: iron, 1 000, 4.35. Row 4: kettle, 2 300, 10.DeviceP (W)I (A)lamp400.17fan600.26iron1 0004.35kettle2 30010

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.

Two circuits, one above the other, each with a supply on the left, a fuse labelled fuse on the live wire along the top, a wire labelled neutral along the bottom and a heater on the right. In the top circuit the wire in the fuse is whole and the heater is on, with wavy heat lines. In the bottom circuit the wire in the fuse is melted apart and the heater is off. Labels: live, neutral, fuse, supply, heater, live, neutral, fuse, supply, heater.fuseliveneutralsupplyheaterfuseliveneutralsupplyheater
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?

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

  1. 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.
  2. Write them down. If only the power in W is marked, find the current from I = P ÷ V.
  3. 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 table with the columns V (V), I (A) and P (W). Row 1: 2.5, 0.3, 0.75.V (V)I (A)P (W)2.50.30.75

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.

Check yourself

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

  1. A motor takes a current of 2.5 A from a 12 V battery. What power does it transfer?

    Tap your answer.

    1. Not this one. You divided the p.d. by the current. The power is the current times the p.d.: P = IV. Go to S4.9.4. Try another.

    2. Not this one. You added the two numbers. P = IV is a product. Try another.

    3. Not this one. You divided the current by the p.d., the wrong way round. Multiply them. Try another.

    4. Yes, that is right. P = IV = 2.5 A × 12 V = 30 W. The motor transfers 30 J each second.

    See the answer

    30 W

    P = IV = 2.5 A × 12 V = 30 W. The motor transfers 30 J each second.

    • If you chose A: You divided the p.d. by the current. The power is the current times the p.d.: P = IV. Go to S4.9.4.
    • If you chose B: You added the two numbers. P = IV is a product.
    • If you chose C: You divided the current by the p.d., the wrong way round. Multiply them.

  2. A current of 3 A flows through a resistor of 5 Ω. What power does the resistor transfer?

    Tap your answer.

    1. Not this one. You multiplied the current by the resistance. That gives a p.d., 15 V. The power is I²R: square the current first. Try another.

    2. Not this one. You squared the p.d. instead of the current. Use I²R: 3 × 3 = 9, then × 5. Try another.

    3. Yes, that is right. P = I²R = 3 A × 3 A × 5 Ω = 9 × 5 = 45 W.

    4. Not this one. You added the two numbers. P = I²R is a product. Try another.

    See the answer

    45 W

    P = I²R = 3 A × 3 A × 5 Ω = 9 × 5 = 45 W.

    • If you chose A: You multiplied the current by the resistance. That gives a p.d., 15 V. The power is I²R: square the current first.
    • If you chose B: You squared the p.d. instead of the current. Use I²R: 3 × 3 = 9, then × 5.
    • If you chose D: You added the two numbers. P = I²R is a product.

  3. An iron is marked 1 380 W and runs from 230 V. What current does it take?

    Tap your answer.

    1. Yes, that is right. I = P ÷ V = 1 380 W ÷ 230 V = 6 A.

    2. Not this one. You divided the p.d. by the power, the wrong way round. The current is P ÷ V. Try another.

    3. Not this one. You multiplied the power by the p.d. Turn P = IV round: I = P ÷ V. Try another.

    4. Not this one. You added the two numbers. Divide the power by the p.d. Try another.

    See the answer

    6 A

    I = P ÷ V = 1 380 W ÷ 230 V = 6 A.

    • If you chose B: You divided the p.d. by the power, the wrong way round. The current is P ÷ V.
    • If you chose C: You multiplied the power by the p.d. Turn P = IV round: I = P ÷ V.
    • If you chose D: You added the two numbers. Divide the power by the p.d.

  4. A toaster takes 4.5 A. The fuses available are 3 A, 5 A, 7 A and 13 A. Which fuse should it have?

    Tap your answer.

    1. Not this one. That is below the normal current of 4.5 A. It would melt every time the toaster is used. Go to S3.10.8. Try another.

    2. Yes, that is right. The fuse must be just above the normal current of 4.5 A. The smallest rating above 4.5 A is 5 A.

    3. Not this one. That is well above 4.5 A. A fault could send much more than the normal current through the wires without melting it. Try another.

    4. Not this one. That is far above 4.5 A. A fault could send nearly three times the normal current through the wires without melting it. Go to S3.10.8. Try another.

    See the answer

    The 5 A fuse

    The fuse must be just above the normal current of 4.5 A. The smallest rating above 4.5 A is 5 A.

    • If you chose A: That is below the normal current of 4.5 A. It would melt every time the toaster is used. Go to S3.10.8.
    • If you chose C: That is well above 4.5 A. A fault could send much more than the normal current through the wires without melting it.
    • If you chose D: That is far above 4.5 A. A fault could send nearly three times the normal current through the wires without melting it. Go to S3.10.8.

  5. The current through a resistor is doubled and its resistance stays the same. What happens to the power?

    Tap your answer.

    1. Not this one. Power depends on the square of the current: P = I²R. A doubled current gives a power four times as big. Try another.

    2. Not this one. The power changes with the current. P = I²R is bigger when I is bigger. Try another.

    3. Yes, that is right. P = I²R. If I doubles, I² becomes four times as big, and R stays the same. So the power is four times as big.

    4. Not this one. A bigger current gives more power, not less. Try another.

    See the answer

    It becomes four times as big

    P = I²R. If I doubles, I² becomes four times as big, and R stays the same. So the power is four times as big.

    • If you chose A: Power depends on the square of the current: P = I²R. A doubled current gives a power four times as big.
    • If you chose B: The power changes with the current. P = I²R is bigger when I is bigger.
    • If you chose D: A bigger current gives more power, not less.

  6. A 60 W lamp is on for 2 minutes. How much energy does it transfer?

    Tap your answer.

    1. Not this one. That is the energy in one minute, 60 W × 60 s. The lamp is on for two minutes. Try another.

    2. Not this one. You multiplied by 2 but did not change the minutes to seconds. 2 min is 120 s. Try another.

    3. Not this one. You divided the power by the time. Energy is power times time: E = Pt. Try another.

    4. Yes, that is right. 2 min = 120 s. E = Pt = 60 W × 120 s = 7 200 J.

    See the answer

    7 200 J

    2 min = 120 s. E = Pt = 60 W × 120 s = 7 200 J.

    • If you chose A: That is the energy in one minute, 60 W × 60 s. The lamp is on for two minutes.
    • If you chose B: You multiplied by 2 but did not change the minutes to seconds. 2 min is 120 s.
    • If you chose C: You divided the power by the time. Energy is power times time: E = Pt.

  7. From an earlier chapter

    A branch is added to a parallel circuit. What happens to the total resistance?

    Tap your answer.

    1. Not this one. That is how a resistor added in series behaves. A new branch is a new path, so the total resistance falls. Go to S4.12.6. Try another.

    2. Yes, that is right. The new branch adds current at the same p.d., so the battery gives more current in all. R = V ÷ I falls.

    3. Not this one. A new path lets more current through at the same p.d., so the total resistance changes. Go to S4.12.6. Try another.

    See the answer

    It falls

    The new branch adds current at the same p.d., so the battery gives more current in all. R = V ÷ I falls.

    • If you chose A: That is how a resistor added in series behaves. A new branch is a new path, so the total resistance falls. Go to S4.12.6.
    • If you chose C: A new path lets more current through at the same p.d., so the total resistance changes. Go to S4.12.6.

  8. A lamp takes 3.5 A from a 12 V supply. Find its power.

    See the answer

    42 W

    P = IV = 3.5 A × 12 V = 42 W.

  9. A current of 4 A flows through a resistor of 3 Ω. Find the power it transfers.

    See the answer

    48 W

    P = I²R = 4 A × 4 A × 3 Ω = 16 × 3 = 48 W.

  10. A 2.5 kW heater is switched on for 3 hours. How much energy does it transfer, in kilowatt-hours?

    See the answer

    7.5 kWh

    E = Pt = 2.5 kW × 3 h = 7.5 kWh.

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.

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

No internet tomorrow? Save the whole book as one file (977 KB). It opens on any phone.