contents

Physics

3.14 - Energy & Power Formulas

4 min

note 1 of 7

Energy transferred from power and time

The total energy an appliance transfers depends on its power and how long it's switched on for.
Power is the electrical energy an appliance transfers each second, measured in watts (W), where 1 W = 1 J/s.

Formula for energy transferred

$E = P t$

Where:

$E$ = Energy transferred, in joules (J)
$P$ = Power, in watts (W)
$t$ = Time, in seconds (s)

note 2 of 7

Worked example - Comparing two kettles

A 2,000 W kettle is switched on for 3 minutes. How long, in minutes, would a 3,000 W kettle take to transfer the same amount of energy?

Step 1 - Convert the time to seconds

$t = 3 \times 60 = 180 s$

Step 2 - Find the energy transferred by the 2,000 W kettle

$E = P t = 2,000 \times 180 = 360,000 J$

Step 3 - Rearrange E = Pt for the time taken by the 3,000 W kettle

$t = \frac{E}{P} = \frac{360,000}{3,000} = 120 s$

Step 4 - Convert the time back to minutes

$t = 120 \div 60 = 2 minutes$

note 3 of 7

Power ratings of appliances

Appliances are labelled with a power rating - the maximum power the appliance transfers when in use.
A higher power rating transfers energy faster, which means an appliance needs less time to do the same job.
A higher power rating doesn't always mean more useful energy is transferred - an inefficient appliance transfers more energy to wasted thermal stores.

note 4 of 7

Formula for electrical power

$P = I V$

Where:

$P$ = Power, in watts (W)
$I$ = Current, in amperes (A)
$V$ = Potential difference, in volts (V)

Electrical power can be found from the potential difference across an appliance and the current through it.

note 5 of 7

Worked example - Finding the current with P = IV

A 1.8 kW electric iron is connected to a 230 V supply. Calculate the current through the iron. Give your answer to two significant figures.

Step 1 - Convert the power to watts

1.8 kW = 1,800 W

Step 2 - Rearrange P = IV for the current

$I = \frac{P}{V}$

Step 3 - Substitute the values

$I = \frac{1,800}{230} = 7.82... = 7.8 A (to 2 s.f.)$

note 6 of 7

Power from current or p.d. - P = I²R and P = V²/R

The power of a component can be calculated from its resistance using either current or potential difference:

$P = I^{2} R$

$P = \frac{V^{2}}{R}$

Where:

$P$ = Power, in watts (W)
$I$ = Current, in amperes (A)
$V$ = Potential difference, in volts (V)
$R$ = Resistance, in ohms (Ω)

note 7 of 7

Worked example - Power from current and resistance

A fan heater connected to a 230 V supply has a current of 6.0 A through it. Calculate the power of the heater and its resistance. Give each answer to two significant figures.

Step 1 - Find the power with P = IV

$\begin{matrix} P & = I V \\ = 6.0 \times 230 = 1,380 \\ = 1,400 W (to 2 s.f.) \end{matrix}$

Step 2 - Rearrange P = I²R for the resistance

$\begin{matrix} R & = \frac{P}{I^{2}} \\ = \frac{1,380}{{6.0}^{2}} = \frac{1,380}{36} \\ = 38.33... = 38 Ω (to 2 s.f.) \end{matrix}$

key terms

TermDefinition

PowerThe rate at which energy is transferred or work is done.

Power ratingThe maximum power an appliance works at, which is the most energy it can transfer each second.

end of notes

Key points to remember

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3.15 - Fuses & EarthingNext up

contents

Physics

3.14 - Energy & Power Formulas

4 min

note 1 of 7

Energy transferred from power and time

The total energy an appliance transfers depends on its power and how long it's switched on for.
Power is the electrical energy an appliance transfers each second, measured in watts (W), where 1 W = 1 J/s.

Formula for energy transferred

$E = P t$

Where:

$E$ = Energy transferred, in joules (J)
$P$ = Power, in watts (W)
$t$ = Time, in seconds (s)

note 2 of 7

Worked example - Comparing two kettles

A 2,000 W kettle is switched on for 3 minutes. How long, in minutes, would a 3,000 W kettle take to transfer the same amount of energy?

Step 1 - Convert the time to seconds

$t = 3 \times 60 = 180 s$

Step 2 - Find the energy transferred by the 2,000 W kettle

$E = P t = 2,000 \times 180 = 360,000 J$

Step 3 - Rearrange E = Pt for the time taken by the 3,000 W kettle

$t = \frac{E}{P} = \frac{360,000}{3,000} = 120 s$

Step 4 - Convert the time back to minutes

$t = 120 \div 60 = 2 minutes$

note 3 of 7

Power ratings of appliances

Appliances are labelled with a power rating - the maximum power the appliance transfers when in use.
A higher power rating transfers energy faster, which means an appliance needs less time to do the same job.
A higher power rating doesn't always mean more useful energy is transferred - an inefficient appliance transfers more energy to wasted thermal stores.

note 4 of 7

Formula for electrical power

$P = I V$

Where:

$P$ = Power, in watts (W)
$I$ = Current, in amperes (A)
$V$ = Potential difference, in volts (V)

Electrical power can be found from the potential difference across an appliance and the current through it.

note 5 of 7

Worked example - Finding the current with P = IV

A 1.8 kW electric iron is connected to a 230 V supply. Calculate the current through the iron. Give your answer to two significant figures.

Step 1 - Convert the power to watts

1.8 kW = 1,800 W

Step 2 - Rearrange P = IV for the current

$I = \frac{P}{V}$

Step 3 - Substitute the values

$I = \frac{1,800}{230} = 7.82... = 7.8 A (to 2 s.f.)$

note 6 of 7

Power from current or p.d. - P = I²R and P = V²/R

The power of a component can be calculated from its resistance using either current or potential difference:

$P = I^{2} R$

$P = \frac{V^{2}}{R}$

Where:

$P$ = Power, in watts (W)
$I$ = Current, in amperes (A)
$V$ = Potential difference, in volts (V)
$R$ = Resistance, in ohms (Ω)

note 7 of 7

Worked example - Power from current and resistance

A fan heater connected to a 230 V supply has a current of 6.0 A through it. Calculate the power of the heater and its resistance. Give each answer to two significant figures.

Step 1 - Find the power with P = IV

$\begin{matrix} P & = I V \\ = 6.0 \times 230 = 1,380 \\ = 1,400 W (to 2 s.f.) \end{matrix}$

Step 2 - Rearrange P = I²R for the resistance

$\begin{matrix} R & = \frac{P}{I^{2}} \\ = \frac{1,380}{{6.0}^{2}} = \frac{1,380}{36} \\ = 38.33... = 38 Ω (to 2 s.f.) \end{matrix}$

key terms

TermDefinition

PowerThe rate at which energy is transferred or work is done.

Power ratingThe maximum power an appliance works at, which is the most energy it can transfer each second.

end of notes

Key points to remember

Take a quiz View flashcards

3.15 - Fuses & EarthingNext up

3.14 - Energy & Power Formulas

Energy transferred from power and time

Formula for energy transferred

Where:

Worked example - Comparing two kettles

Step 1 - Convert the time to seconds

Step 2 - Find the energy transferred by the 2,000 W kettle

Step 3 - Rearrange E = Pt for the time taken by the 3,000 W kettle

Step 4 - Convert the time back to minutes

Power ratings of appliances

Formula for electrical power

Where:

Worked example - Finding the current with P = IV

Step 1 - Convert the power to watts

Step 2 - Rearrange P = IV for the current

Step 3 - Substitute the values

Power from current or p.d. - P = I2R and P = V2/R

Where:

Worked example - Power from current and resistance

Step 1 - Find the power with P = IV

Step 2 - Rearrange P = I2R for the resistance

Key points to remember

3.14 - Energy & Power Formulas

Energy transferred from power and time

Formula for energy transferred

Where:

Worked example - Comparing two kettles

Step 1 - Convert the time to seconds

Step 2 - Find the energy transferred by the 2,000 W kettle

Step 3 - Rearrange E = Pt for the time taken by the 3,000 W kettle

Step 4 - Convert the time back to minutes

Power ratings of appliances

Formula for electrical power

Where:

Worked example - Finding the current with P = IV

Step 1 - Convert the power to watts

Step 2 - Rearrange P = IV for the current

Step 3 - Substitute the values

Power from current or p.d. - P = I2R and P = V2/R

Where:

Worked example - Power from current and resistance

Step 1 - Find the power with P = IV

Step 2 - Rearrange P = I2R for the resistance

Key points to remember

Power from current or p.d. - P = I²R and P = V²/R

Step 2 - Rearrange P = I²R for the resistance

Power from current or p.d. - P = I²R and P = V²/R

Step 2 - Rearrange P = I²R for the resistance