contents

Physics

6.5 - Energy & Power Formulas

3 min

note 1 of 6

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 6

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 6

Potential difference is energy per charge - E = QV

When charge moves through a potential difference, energy is transferred.
The potential difference across a component is the energy transferred per coulomb of charge passing through it (1 V = 1 J/C).

Formula for energy transferred

$E = Q V$

Where:

$E$ = Energy transferred, in joules (J)
$Q$ = Charge, in coulombs (C)
$V$ = Potential difference, in volts (V)

A charge Q traced round a circuit labelled charge, battery and resistors, gaining +10 V at the battery on top and losing -5 V at each of the two resistors below. — The charge gains energy at the battery and transfers it in the two resistors.

note 4 of 6

Worked example - Energy transferred by a motor

The motor in a battery-powered toy car is connected to a 6 V battery. 90 C of charge passes through the motor while the car is running. Calculate the energy transferred.

Step 1 - Identify the values

Charge: $Q = 90$ C
Potential difference: $V = 6$ V

Step 2 - Substitute into E = QV

$E = Q V = 90 \times 6 = 540 J$

note 5 of 6

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 6 of 6

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

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

key terms

TermDefinition

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

Potential differenceThe energy transferred per coulomb of charge that moves between two points, such as across a component.

end of notes

Key points to remember

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6.6 - Fuses & EarthingNext up

contents

Physics

6.5 - Energy & Power Formulas

3 min

note 1 of 6

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 6

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 6

Potential difference is energy per charge - E = QV

When charge moves through a potential difference, energy is transferred.
The potential difference across a component is the energy transferred per coulomb of charge passing through it (1 V = 1 J/C).

Formula for energy transferred

$E = Q V$

Where:

$E$ = Energy transferred, in joules (J)
$Q$ = Charge, in coulombs (C)
$V$ = Potential difference, in volts (V)

note 4 of 6

Worked example - Energy transferred by a motor

The motor in a battery-powered toy car is connected to a 6 V battery. 90 C of charge passes through the motor while the car is running. Calculate the energy transferred.

Step 1 - Identify the values

Charge: $Q = 90$ C
Potential difference: $V = 6$ V

Step 2 - Substitute into E = QV

$E = Q V = 90 \times 6 = 540 J$

note 5 of 6

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 6 of 6

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

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

key terms

TermDefinition

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

Potential differenceThe energy transferred per coulomb of charge that moves between two points, such as across a component.

end of notes

Key points to remember

Take a quiz View flashcards

6.6 - Fuses & EarthingNext up