Power

What you'll learn

What electrical power means as a rate of energy transfer.
How to use $P = V I$ , $P = I^2R$ and $P = V^2/R$ .
How to calculate electrical energy transfer using $W = V I t$ .
How the kilowatt-hour links physics to real electricity bills.

Power starts with energy transfer

You have already met energy as something transferred between stores. In circuits, electrical energy may be transferred to thermal energy in a heater, light in a lamp, or kinetic energy in a motor.

Power tells you how quickly that transfer happens. A high-power device transfers energy faster than a low-power device.

Definition

Power

Power, $P$ , is the rate of energy transfer or the rate of doing work. It is measured in watts, W, where one watt means one joule per second.

The general relationship is:

\frac{W}{t}

where $W$ is the energy transferred or work done in joules, J, and $t$ is the time in seconds, s.

Example

Finding power from energy and time

A lamp transfers $J600\ \text{J}$ of energy in $s30\ \text{s}$ , so use the rate relationship $P = W / t$ .
Substitute the values with units:
$\frac{600\ \text{J}}{30\ \text{s}} = 20\ \text{J s}^{-1}$ .
Since $W1\ \text{J s}^{-1} = 1\ \text{W}$ , the lamp has power $W20\ \text{W}$ .

Electrical prerequisites

Before using the power equations, you need three circuit quantities.

Current, $I$ , is the rate of flow of charge. It is measured in amperes, A.

Potential difference, $V$ , is the energy transferred per unit charge between two points. It is measured in volts, V.

Resistance, $R$ , describes how much a component opposes current. It is measured in ohms, $Ω\Omega$ , and is linked to potential difference and current by $V = I R$ .

In this topic, a load means the component receiving electrical energy from the circuit, such as a resistor, lamp, heater or motor. To measure power in a load, the ammeter goes in series to measure $I$ , and the voltmeter goes in parallel across the load to measure $V$ .

Circuit showing a load with ammeter in series, voltmeter in parallel, and labels for power and energy transfer

Tip

Ammeter and voltmeter placement

An ammeter measures the current through a component, so it must be in series with it. A voltmeter measures the potential difference across a component, so it must be in parallel with it.

Electrical power: $P = V I$

For a component, electrical power is given by:

P = V I

where $P$ is power in watts, W, $V$ is potential difference in volts, V, and $I$ is current in amperes, A.

This equation makes sense from the units:

volts mean joules per coulomb
amperes mean coulombs per second
multiplying gives joules per second, which is watts

Key Idea

The meaning of P = VI

A component has a high power if each coulomb transfers a lot of energy, high $V$ , and/or lots of charge flows each second, high $I$ .

Example

Calculating power from voltage and current

A motor operates at $V12.0\ \text{V}$ with a current of $A3.5\ \text{A}$ , so use $P = V I$ .
Substitute the values:
$(12.0\ \text{V})(3.5\ \text{A}) = 42\ \text{W}$ .
Interpret the result: the motor transfers energy at a rate of $s−142\ \text{J s}^{-1}$ .

Power in a resistor

For a resistor or a component with a known resistance at that operating point, you can combine $P = V I$ with $V = I R$ .

Starting with $P = V I$ :

If you know $I$ and $R$ , replace $V$ with $I R$ :

P = I^2R

If you know $V$ and $R$ , replace $I$ with $V / R$ :

P = V^2/R

So the three OCR equations are:

P = V I

P = I^2R

P = V^2/R

Tip

Choosing the quickest power equation

Use the equation that matches the quantities you are given: $V$ and $I$ gives $P = V I$ ; $I$ and $R$ gives $P = I^2R$ ; $V$ and $R$ gives $P = V^2/R$ .

Example

Finding the power dissipated in a resistor

A resistor has resistance $Ω18.0\ \Omega$ and potential difference $V12.0\ \text{V}$ across it, so choose $P = V^2/R$ because $V$ and $R$ are known.
Substitute carefully, including the square on $V$ :
$\frac{(12.0\ \text{V})^2}{18.0\ \Omega} = 8.00\ \text{W}$ .
Check using current:
$\frac{V}{R} = \frac{12.0\ \text{V}}{18.0\ \Omega} = 0.667\ \text{A}$ , and $(12.0\ \text{V})(0.667\ \text{A}) \approx 8.00\ \text{W}$ .

Common Mistake

Squaring the wrong quantity

In $P = I^2R$ , only the current is squared. In $P = V^2/R$ , only the potential difference is squared. The resistance is not squared in either equation.

Common Mistake

When resistance changes

For components such as filament lamps, resistance can change as temperature changes. Use the resistance value for the actual operating conditions given in the question.

Energy transferred: $W = V I t$

Power is energy transferred per second, so energy transferred is:

W = P t

Using $P = V I$ gives the electrical energy equation:

W = V I t

where $W$ is the energy transferred in joules, J, $V$ is potential difference in volts, V, $I$ is current in amperes, A, and $t$ is time in seconds, s.

Key Idea

Energy from electrical power

$W = V I t$ is just “energy equals power times time”, with electrical power written as $V I$ .

Example

Calculating electrical energy transferred

A kettle operates at $V230\ \text{V}$ with a current of $A9.0\ \text{A}$ for $s180\ \text{s}$ , so use $W = V I t$ .
Substitute the values:
$(230\ \text{V})(9.0\ \text{A})(180\ \text{s}) = 372600\ \text{J}$ .
Quote sensibly to two significant figures:
$\approx 3.7 \times 10^5\ \text{J}$ .

The kilowatt-hour

Domestic electrical energy use is often much larger than a few joules, so electricity companies use the kilowatt-hour, usually written as kW h or kWh on bills.

Definition

Kilowatt-hour

One kilowatt-hour is the energy transferred by a power of one kilowatt operating for one hour.

A kilowatt is $W1000\ \text{W}$ and one hour is $s3600\ \text{s}$ , so:

J1\ \text{kW h} = 1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^6\ \text{J}

So the kilowatt-hour is a unit of energy, not power.

Common Mistake

kW h is not kW per hour

A kilowatt-hour means kilowatts multiplied by hours. It is an energy unit. Do not treat it as “kilowatts per hour”.

Calculating the cost of energy

For electricity bills, use:

h\text{energy used in kW h} = \text{power in kW} \times \text{time in h}

Then:

h\text{cost} = \text{energy used in kW h} \times \text{price per kW h}

This links the circuit physics to real decisions about energy use: the same device costs more to run if it has a larger power, is used for longer, or electricity costs more per kW h.

Example

Calculating the cost of running an appliance

A microwave has power $W850\ \text{W}$ and is used for $min12\ \text{min}$ each day for 30 days. Convert to billing units:
$kW850\ \text{W} = 0.850\ \text{kW}$ , and total time is $h12\ \text{min} \times 30 = 360\ \text{min} = 6.00\ \text{h}$ .
Calculate the energy in kW h:
$h\text{energy} = (0.850\ \text{kW})(6.00\ \text{h}) = 5.10\ \text{kW h}$ .
If electricity costs $h−129\ \text{p}\ \text{kW h}^{-1}$ , calculate the cost:
$p≈£1.48\text{cost} = 5.10 \times 29\ \text{p} = 148\ \text{p} \approx \text{£}1.48$ .

Tip

Which time unit should you use?

For $W = V I t$ , use seconds so the energy comes out in joules. For electricity cost, use hours with power in kilowatts so the energy comes out in kW h.

Exam technique

In the exam

Choose your equation from the quantities given: $V$ and $I$ for $P = V I$ , $I$ and $R$ for $P = I^2R$ , or $V$ and $R$ for $P = V^2/R$ .
Keep unit systems consistent: volts, amperes and seconds give joules; kilowatts and hours give kilowatt-hours.
Check the answer size: small electronics may be a few watts, kettles and heaters are often a few kilowatts, and kW h is energy, not power.

Self review

Check yourself

A $Ω6.0\ \Omega$ resistor carries a current of $A2.0\ \text{A}$ . Which power equation is quickest?
How many joules are in $h0.50\ \text{kW h}$ ?
A device is rated at $W750\ \text{W}$ and runs for $min20\ \text{min}$ . What conversions are needed before calculating cost in p per kW h?

What you'll learn

What electrical power means as a rate of energy transfer.
How to use $P = V I$ , $P = I^2R$ and $P = V^2/R$ .
How to calculate electrical energy transfer using $W = V I t$ .
How the kilowatt-hour links physics to real electricity bills.

Power starts with energy transfer

Power tells you how quickly that transfer happens. A high-power device transfers energy faster than a low-power device.

Definition

Power

Power, $P$ , is the rate of energy transfer or the rate of doing work. It is measured in watts, W, where one watt means one joule per second.

The general relationship is:

\frac{W}{t}

where $W$ is the energy transferred or work done in joules, J, and $t$ is the time in seconds, s.

Example

Finding power from energy and time

A lamp transfers $J600\ \text{J}$ of energy in $s30\ \text{s}$ , so use the rate relationship $P = W / t$ .
Substitute the values with units:
$\frac{600\ \text{J}}{30\ \text{s}} = 20\ \text{J s}^{-1}$ .
Since $W1\ \text{J s}^{-1} = 1\ \text{W}$ , the lamp has power $W20\ \text{W}$ .

Electrical prerequisites

Before using the power equations, you need three circuit quantities.

Current, $I$ , is the rate of flow of charge. It is measured in amperes, A.

Potential difference, $V$ , is the energy transferred per unit charge between two points. It is measured in volts, V.

Resistance, $R$ , describes how much a component opposes current. It is measured in ohms, $Ω\Omega$ , and is linked to potential difference and current by $V = I R$ .

Circuit showing a load with ammeter in series, voltmeter in parallel, and labels for power and energy transfer

Tip

Ammeter and voltmeter placement

An ammeter measures the current through a component, so it must be in series with it. A voltmeter measures the potential difference across a component, so it must be in parallel with it.

Electrical power: $P = V I$

For a component, electrical power is given by:

P = V I

where $P$ is power in watts, W, $V$ is potential difference in volts, V, and $I$ is current in amperes, A.

This equation makes sense from the units:

volts mean joules per coulomb
amperes mean coulombs per second
multiplying gives joules per second, which is watts

Key Idea

The meaning of P = VI

A component has a high power if each coulomb transfers a lot of energy, high $V$ , and/or lots of charge flows each second, high $I$ .

Example

Calculating power from voltage and current

A motor operates at $V12.0\ \text{V}$ with a current of $A3.5\ \text{A}$ , so use $P = V I$ .
Substitute the values:
$(12.0\ \text{V})(3.5\ \text{A}) = 42\ \text{W}$ .
Interpret the result: the motor transfers energy at a rate of $s−142\ \text{J s}^{-1}$ .

Power in a resistor

For a resistor or a component with a known resistance at that operating point, you can combine $P = V I$ with $V = I R$ .

Starting with $P = V I$ :

If you know $I$ and $R$ , replace $V$ with $I R$ :

P = I^2R

If you know $V$ and $R$ , replace $I$ with $V / R$ :

P = V^2/R

So the three OCR equations are:

P = V I

P = I^2R

P = V^2/R

Tip

Choosing the quickest power equation

Use the equation that matches the quantities you are given: $V$ and $I$ gives $P = V I$ ; $I$ and $R$ gives $P = I^2R$ ; $V$ and $R$ gives $P = V^2/R$ .

Example

Finding the power dissipated in a resistor

A resistor has resistance $Ω18.0\ \Omega$ and potential difference $V12.0\ \text{V}$ across it, so choose $P = V^2/R$ because $V$ and $R$ are known.
Substitute carefully, including the square on $V$ :
$\frac{(12.0\ \text{V})^2}{18.0\ \Omega} = 8.00\ \text{W}$ .
Check using current:
$\frac{V}{R} = \frac{12.0\ \text{V}}{18.0\ \Omega} = 0.667\ \text{A}$ , and $(12.0\ \text{V})(0.667\ \text{A}) \approx 8.00\ \text{W}$ .

Common Mistake

Squaring the wrong quantity

In $P = I^2R$ , only the current is squared. In $P = V^2/R$ , only the potential difference is squared. The resistance is not squared in either equation.

Common Mistake

When resistance changes

For components such as filament lamps, resistance can change as temperature changes. Use the resistance value for the actual operating conditions given in the question.

Energy transferred: $W = V I t$

Power is energy transferred per second, so energy transferred is:

W = P t

Using $P = V I$ gives the electrical energy equation:

W = V I t

where $W$ is the energy transferred in joules, J, $V$ is potential difference in volts, V, $I$ is current in amperes, A, and $t$ is time in seconds, s.

Key Idea

Energy from electrical power

$W = V I t$ is just “energy equals power times time”, with electrical power written as $V I$ .

Example

Calculating electrical energy transferred

A kettle operates at $V230\ \text{V}$ with a current of $A9.0\ \text{A}$ for $s180\ \text{s}$ , so use $W = V I t$ .
Substitute the values:
$(230\ \text{V})(9.0\ \text{A})(180\ \text{s}) = 372600\ \text{J}$ .
Quote sensibly to two significant figures:
$\approx 3.7 \times 10^5\ \text{J}$ .

The kilowatt-hour

Domestic electrical energy use is often much larger than a few joules, so electricity companies use the kilowatt-hour, usually written as kW h or kWh on bills.

Definition

Kilowatt-hour

One kilowatt-hour is the energy transferred by a power of one kilowatt operating for one hour.

A kilowatt is $W1000\ \text{W}$ and one hour is $s3600\ \text{s}$ , so:

J1\ \text{kW h} = 1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^6\ \text{J}

So the kilowatt-hour is a unit of energy, not power.

Common Mistake

kW h is not kW per hour

A kilowatt-hour means kilowatts multiplied by hours. It is an energy unit. Do not treat it as “kilowatts per hour”.

Calculating the cost of energy

For electricity bills, use:

h\text{energy used in kW h} = \text{power in kW} \times \text{time in h}

Then:

h\text{cost} = \text{energy used in kW h} \times \text{price per kW h}

This links the circuit physics to real decisions about energy use: the same device costs more to run if it has a larger power, is used for longer, or electricity costs more per kW h.

Example

Calculating the cost of running an appliance

A microwave has power $W850\ \text{W}$ and is used for $min12\ \text{min}$ each day for 30 days. Convert to billing units:
$kW850\ \text{W} = 0.850\ \text{kW}$ , and total time is $h12\ \text{min} \times 30 = 360\ \text{min} = 6.00\ \text{h}$ .
Calculate the energy in kW h:
$h\text{energy} = (0.850\ \text{kW})(6.00\ \text{h}) = 5.10\ \text{kW h}$ .
If electricity costs $h−129\ \text{p}\ \text{kW h}^{-1}$ , calculate the cost:
$p≈£1.48\text{cost} = 5.10 \times 29\ \text{p} = 148\ \text{p} \approx \text{£}1.48$ .

Tip

Which time unit should you use?

For $W = V I t$ , use seconds so the energy comes out in joules. For electricity cost, use hours with power in kilowatts so the energy comes out in kW h.

Exam technique

In the exam

Choose your equation from the quantities given: $V$ and $I$ for $P = V I$ , $I$ and $R$ for $P = I^2R$ , or $V$ and $R$ for $P = V^2/R$ .
Keep unit systems consistent: volts, amperes and seconds give joules; kilowatts and hours give kilowatt-hours.
Check the answer size: small electronics may be a few watts, kettles and heaters are often a few kilowatts, and kW h is energy, not power.

Self review

Check yourself

A $Ω6.0\ \Omega$ resistor carries a current of $A2.0\ \text{A}$ . Which power equation is quickest?
How many joules are in $h0.50\ \text{kW h}$ ?
A device is rated at $W750\ \text{W}$ and runs for $min20\ \text{min}$ . What conversions are needed before calculating cost in p per kW h?

What you'll learn

Power starts with energy transfer

Electrical prerequisites

Electrical power: $P = V I$

Power in a resistor

Energy transferred: $W = V I t$

The kilowatt-hour

Calculating the cost of energy

Review Power by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Power

What you'll learn

Power starts with energy transfer

Electrical prerequisites

Electrical power: $P = V I$

Power in a resistor

Energy transferred: $W = V I t$

The kilowatt-hour

Calculating the cost of energy

Review Power by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Power

What you'll learn

Power starts with energy transfer

Electrical prerequisites

Electrical power: P=VIP = VIP=VI

Power in a resistor

Energy transferred: W=VItW = VItW=VIt

The kilowatt-hour

Calculating the cost of energy

Review Power by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Power Revision Guide

Power

What you'll learn

Power starts with energy transfer

Electrical prerequisites

Electrical power: P=VIP = VIP=VI

Power in a resistor

Energy transferred: W=VItW = VItW=VIt

The kilowatt-hour

Calculating the cost of energy

Review Power by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Power Revision Guide

Electrical power: $P = V I$

Energy transferred: $W = V I t$

Electrical power: $P = V I$

Energy transferred: $W = V I t$