contents

Physics

2.7 - Energy & Power Formulas

5 min

note 1 of 8

The heating effect of a current in a resistor

When there's a current in a resistor, energy is transferred that heats the resistor up.
This energy is dissipated to the thermal energy stores of the resistor and its surroundings.

Why the resistor heats up

As free electrons flow through the resistor, they collide with the ions in the metal lattice.
Each collision transfers energy to the ions, making them vibrate more.
The more the ions vibrate, the hotter the metal becomes.

Inside a resistor, orange electrons move past rows of vibrating blue ions, under an arrow labelled free electrons flow through resistor, with labels electron, ion and electrons collide with ions, making them vibrate more.

Advantages and disadvantages

When heat is the useful output (advantage)	When heat is not the useful output (disadvantage)
Heaters and toasters use high-resistance wire to transfer energy to the surroundings	Energy transferred to thermal stores is wasted, reducing efficiency
A fuse melts and breaks the circuit if the current gets too high	Components that get too hot can be damaged, and heating can change how the circuit behaves

note 2 of 8

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 3 of 8

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 8

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 8

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 8

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 7 of 8

Formula for energy transferred from current, p.d. and time

Combining $E = P t$ and $P = I V$ gives a single equation for the energy transferred to a component.

$E = I V t$

Where:

$E$ = Energy transferred, in joules (J)
$I$ = Current, in amperes (A)
$V$ = Potential difference, in volts (V)
$t$ = Time, in seconds (s)

note 8 of 8

Worked example - Energy transferred by a motor

The motor in a battery-powered toy car is connected to a 4.5 V battery. A current of 0.60 A flows through the motor for 2.5 minutes. Calculate the energy transferred by the motor.

Step 1 - Convert the time to seconds

$t = 2.5 \times 60 = 150 s$

Step 2 - Substitute into E = IVt

$E = I V t = 0.60 \times 4.5 \times 150 = 405 J$

key terms

TermDefinition

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

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

2.8 - AC & DC CurrentNext up

contents

Physics

2.7 - Energy & Power Formulas

5 min

note 1 of 8

The heating effect of a current in a resistor

When there's a current in a resistor, energy is transferred that heats the resistor up.
This energy is dissipated to the thermal energy stores of the resistor and its surroundings.

Why the resistor heats up

As free electrons flow through the resistor, they collide with the ions in the metal lattice.
Each collision transfers energy to the ions, making them vibrate more.
The more the ions vibrate, the hotter the metal becomes.

Advantages and disadvantages

When heat is the useful output (advantage)	When heat is not the useful output (disadvantage)
Heaters and toasters use high-resistance wire to transfer energy to the surroundings	Energy transferred to thermal stores is wasted, reducing efficiency
A fuse melts and breaks the circuit if the current gets too high	Components that get too hot can be damaged, and heating can change how the circuit behaves

note 2 of 8

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 3 of 8

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 8

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 8

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 8

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 7 of 8

Formula for energy transferred from current, p.d. and time

Combining $E = P t$ and $P = I V$ gives a single equation for the energy transferred to a component.

$E = I V t$

Where:

$E$ = Energy transferred, in joules (J)
$I$ = Current, in amperes (A)
$V$ = Potential difference, in volts (V)
$t$ = Time, in seconds (s)

note 8 of 8

Worked example - Energy transferred by a motor

The motor in a battery-powered toy car is connected to a 4.5 V battery. A current of 0.60 A flows through the motor for 2.5 minutes. Calculate the energy transferred by the motor.

Step 1 - Convert the time to seconds

$t = 2.5 \times 60 = 150 s$

Step 2 - Substitute into E = IVt

$E = I V t = 0.60 \times 4.5 \times 150 = 405 J$

key terms

TermDefinition

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

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

2.8 - AC & DC CurrentNext up