Electrical power.pptx

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Electrical power

Electricity

Lesson 6 of 10

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Warm-up

Electricity

Lesson 6 of 10

Answer each one, then check.

1

What is the unit of power?

2

Write the equation linking energy and time for power.

3

What is V = IR used for?

4

Convert 2.3 kW to watts.

5

Work out 3².

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Warm-up - Answers

Electricity

Lesson 6 of 10

1

What is the unit of power?

Watt (W)

2

Write the equation linking energy and time for power.

P = E ÷ t

3

What is V = IR used for?

Linking pd, current and resistance

4

Convert 2.3 kW to watts.

2300 W

5

Work out 3².

9

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Learning Objectives

Electricity

Lesson 6 of 10

1

Explain how the power transfer in a device is related to the pd across it and the current through it.

2

Recall and apply P = VI.

3

Recall and apply P = I²R.

4

Rearrange to find current, pd or resistance.

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Electricity

Lesson 6 of 10

ELECTRICAL POWER

power = potential difference × current P = VI

Also P = I²R (power = current squared × resistance). Power is in watts (W), pd in volts (V), current in amperes (A) and resistance in ohms (Ω).

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Power in a Circuit

Electricity

Lesson 6 of 10

A device with a bigger pd or current has a greater power.

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Using P = VI

Electricity

Lesson 6 of 10

A kettle has a current of 10 A when connected to the 230 V mains. Calculate its power.

1

Write the equation

P = VI

2

Substitute

P = 230 × 10

3

Answer

P = 2300 W

ANSWER

2300 W (2.3 kW)

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Using P = I²R

Electricity

Lesson 6 of 10

A current of 6.0 A flows through a 5.0 Ω resistor. Calculate the power.

1

Write the equation

P = I²R

2

Square the current first

6.0² = 36

3

Substitute

P = 36 × 5.0

4

Answer

P = 180 W

ANSWER

180 W

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Finding the Current

Electricity

Lesson 6 of 10

A 60 W lamp is connected to the 230 V mains. Calculate the current.

1

Rearrange

I = P ÷ V

2

Substitute

I = 60 ÷ 230

3

Answer

I = 0.26 A

ANSWER

0.26 A

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Which Equation?

Electricity

Lesson 6 of 10

Look at what you are given.

You have

Use

V and I

P = VI

I and R

P = I²R

V and R

Find I first with I = V ÷ R, then P = VI

E and t

P = E ÷ t

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Common Mistakes

Electricity

Lesson 6 of 10

Take care with squares.

Squaring

Square only the current, then multiply by the resistance.

Units

Convert mA to A and kW to W first.

Rating labels

The power rating of an appliance is at its normal pd.

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Key Terms

Electricity

Lesson 6 of 10

Power

The rate at which energy is transferred.

Watt

One joule per second.

Power rating

The power an appliance transfers in normal use.

Kilowatt

1000 watts.

Potential difference

Energy transferred per coulomb of charge.

Current

The rate of flow of charge.

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Electricity

Lesson 6 of 10

YOUR TASK

Rating Plate

10 minutes

A hair dryer is rated 1500 W at 230 V. Calculate the current. A toaster takes 4.0 A at 230 V; find its power.

1

Rearrange P = VI for current.

2

Then calculate the toaster power.

WHAT A GOOD ANSWER SHOWS

Hair dryer: 1500 ÷ 230 = 6.5 A. Toaster: 230 × 4.0 = 920 W.

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Can I...?

Electricity

Lesson 6 of 10

Recall P = VI.

Recall P = I²R.

Choose the right equation.

Rearrange for current.

Square the current correctly.

Convert units to W and A.

Explain what power means.

Give the unit.

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Summary & Exam Focus

P = VI.

P = I²R.

Power is in watts.

Higher power means faster energy transfer.

EXAM FOCUS

A kettle draws 10 A from the 230 V mains. Calculate its power. (2 marks)

Choose P = VI, substitute and give W.

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