Energy transfer in a circuit

A circuit is a way of transferring energy from the supply to the components, with the current acting as the pathway.
The supply transfers energy to the charge, and each component transfers that energy electrically to another store or away from the circuit.
1. In a resistor or a heater, the energy is transferred to the thermal store of the component and then by heating to the surroundings.
2. In a filament lamp, the energy is transferred to the thermal store of the filament and away by light and by heating.
3. In a motor, most of the energy is transferred to the kinetic store of the spinning parts.
The energy transferred when charge $Q$ passes through a potential difference $V$ is given by $E = Q V$ .
1. $E$ is the energy transferred in joules, $J\text{J}$ , $Q$ is the charge in coulombs, $C\text{C}$ , and $V$ is the potential difference in volts, $V\text{V}$ .
The current is not used up in a component, because it is the energy carried by the charge that is transferred.

Power

Definition

Power

The rate at which energy is transferred.

Definition

Watt

The unit of power, equal to an energy transfer of one joule every second.

Power tells you how quickly a device transfers energy, not how much energy it transfers in total.
Power is the energy transferred divided by the time taken, $P=EtP = \dfrac{E}{t}$ $P = \frac{E}{t}$ , which you can rearrange to $E = P t$ .
1. $P$ is the power in watts, $W\text{W}$ , $E$ is the energy in joules, $J\text{J}$ , and $t$ is the time in seconds, $s\text{s}$ .
One watt is one joule per second, so $J/s1\ \text{W} = 1\ \text{J/s}$ , and a $W60\ \text{W}$ lamp transfers $J60\ \text{J}$ of energy every second.
Large powers are given in kilowatts, where $W1\ \text{kW} = 1000\ \text{W}$ , so a $kW2\ \text{kW}$ kettle transfers $J2000\ \text{J}$ every second.

Power, potential difference and current

The power of any circuit device depends on the potential difference across it and the current through it.
1. The potential difference is the energy transferred by each coulomb of charge.
2. The current is the number of coulombs passing each second.
3. Multiplying energy per coulomb by coulombs per second gives energy per second, which is the power.
This gives the equation $P = V I$ , where $P$ is in watts, $V$ is in volts and $I$ is in amperes.
Doubling either the potential difference or the current, with the other kept the same, doubles the rate at which the device transfers energy.
Substituting $V = I R$ into $P = V I$ gives a second equation, $P = I^{2} R$ , which is useful when the current and the resistance are known.
1. Because the current is squared, doubling the current through a fixed resistor makes it transfer energy four times as quickly, which is why large currents heat wires so strongly.

Energy over time

Definition

Kilowatt-hour

The energy transferred by a device with a power of one kilowatt when it runs for one hour.

The total energy transferred by a device depends on its power and on how long it is used, $E = P t$ .
With power in watts and time in seconds, the energy comes out in joules.
With power in kilowatts and time in hours, the energy comes out in kilowatt-hours, $kW h\text{kW h}$ $kW h$ , which is the unit used on electricity bills.
1. One kilowatt-hour is equal to $J1000\ \text{W} \times 3600\ \text{s} = 3\,600\,000\ \text{J}$ .
A device with a higher power transfers energy faster, but a lower-power device left on for longer can still transfer more energy in total.
The power rating on an appliance, such as $V230\ \text{V}$ , $kW2.3\ \text{kW}$ , gives the power it transfers when connected to the stated potential difference.

Example

Power and energy for a heater

A heater has a potential difference of $V12\ \text{V}$ across it and a current of $A3.0\ \text{A}$ through it, and it is switched on for $5.0$ minutes.

12\ \text{V} \times 3.0\ \text{A} = 36\ \text{W}

\times 60 = 300\ \text{s}

36\ \text{W} \times 300\ \text{s} = 10\,800\ \text{J}

The heater transfers $J36\ \text{J}$ every second, and $J10\,800\ \text{J}$ in the five minutes.

Example

Heating effect in a resistor

A current of $A0.50\ \text{A}$ flows through a $Ω20\ \Omega$ resistor.

I^2R = (0.50\ \text{A})^2 \times 20\ \Omega = 5.0\ \text{W}

The current is then doubled to $A1.0\ \text{A}$ .

(1.0\ \text{A})^2 \times 20\ \Omega = 20\ \text{W}

Doubling the current has made the power four times larger.

Example

Energy used by a kettle

A kettle is rated at $V230\ \text{V}$ , $kW2.3\ \text{kW}$ , and it is used for $3.0$ minutes.

\dfrac{P}{V} = \dfrac{2300\ \text{W}}{230\ \text{V}} = 10\ \text{A}

\dfrac{3.0}{60} = 0.050\ \text{h}

2.3\ \text{kW} \times 0.050\ \text{h} = 0.115\ \text{kW h}

The same energy in joules is $J2300\ \text{W} \times 180\ \text{s} = 414\,000\ \text{J}$ .

Common Mistake

Do not confuse power and energy, because power in watts is the rate of transfer, while energy in joules or kilowatt-hours is the amount transferred.
Do not write that the kilowatt-hour is a unit of power, because it is a unit of energy.
Do not write about an electrical energy store, because no such store exists and electricity is a way of transferring energy between stores.
Do not leave time in minutes when the power is in watts, because $E = P t$ needs seconds to give joules.

Exam technique

Power calculations

Choose the power equation that matches the values given, using $P = V I$ for potential difference and current, $P = I^2R$ for current and resistance, and $\dfrac{E}{t}$ for energy and time.
In a two-step question, such as finding energy from potential difference, current and time, find the power first and then use $E = P t$ .
When you explain why power depends on potential difference and current, link energy per coulomb with coulombs per second.

Self review

Define power and state its unit.
Write down the equation linking power, potential difference and current.
Explain why the power of a device equals the potential difference multiplied by the current.
Calculate the energy transferred by a $kW1.5\ \text{kW}$ heater in $h2.0\ \text{h}$ , in kilowatt-hours.
Describe what happens to the power transferred in a resistor when the current through it is doubled.

Energy transfer in a circuit

A circuit is a way of transferring energy from the supply to the components, with the current acting as the pathway.
The supply transfers energy to the charge, and each component transfers that energy electrically to another store or away from the circuit.
1. In a resistor or a heater, the energy is transferred to the thermal store of the component and then by heating to the surroundings.
2. In a filament lamp, the energy is transferred to the thermal store of the filament and away by light and by heating.
3. In a motor, most of the energy is transferred to the kinetic store of the spinning parts.
The energy transferred when charge $Q$ passes through a potential difference $V$ is given by $E = Q V$ .
1. $E$ is the energy transferred in joules, $J\text{J}$ , $Q$ is the charge in coulombs, $C\text{C}$ , and $V$ is the potential difference in volts, $V\text{V}$ .
The current is not used up in a component, because it is the energy carried by the charge that is transferred.

Power

Definition

Power

The rate at which energy is transferred.

Definition

Watt

The unit of power, equal to an energy transfer of one joule every second.

Power tells you how quickly a device transfers energy, not how much energy it transfers in total.
Power is the energy transferred divided by the time taken, $P=EtP = \dfrac{E}{t}$ $P = \frac{E}{t}$ , which you can rearrange to $E = P t$ .
1. $P$ is the power in watts, $W\text{W}$ , $E$ is the energy in joules, $J\text{J}$ , and $t$ is the time in seconds, $s\text{s}$ .
One watt is one joule per second, so $J/s1\ \text{W} = 1\ \text{J/s}$ , and a $W60\ \text{W}$ lamp transfers $J60\ \text{J}$ of energy every second.
Large powers are given in kilowatts, where $W1\ \text{kW} = 1000\ \text{W}$ , so a $kW2\ \text{kW}$ kettle transfers $J2000\ \text{J}$ every second.

Power, potential difference and current

The power of any circuit device depends on the potential difference across it and the current through it.
1. The potential difference is the energy transferred by each coulomb of charge.
2. The current is the number of coulombs passing each second.
3. Multiplying energy per coulomb by coulombs per second gives energy per second, which is the power.
This gives the equation $P = V I$ , where $P$ is in watts, $V$ is in volts and $I$ is in amperes.
Doubling either the potential difference or the current, with the other kept the same, doubles the rate at which the device transfers energy.
Substituting $V = I R$ into $P = V I$ gives a second equation, $P = I^{2} R$ , which is useful when the current and the resistance are known.
1. Because the current is squared, doubling the current through a fixed resistor makes it transfer energy four times as quickly, which is why large currents heat wires so strongly.

Energy over time

Definition

Kilowatt-hour

The energy transferred by a device with a power of one kilowatt when it runs for one hour.

The total energy transferred by a device depends on its power and on how long it is used, $E = P t$ .
With power in watts and time in seconds, the energy comes out in joules.
With power in kilowatts and time in hours, the energy comes out in kilowatt-hours, $kW h\text{kW h}$ $kW h$ , which is the unit used on electricity bills.
1. One kilowatt-hour is equal to $J1000\ \text{W} \times 3600\ \text{s} = 3\,600\,000\ \text{J}$ .
A device with a higher power transfers energy faster, but a lower-power device left on for longer can still transfer more energy in total.
The power rating on an appliance, such as $V230\ \text{V}$ , $kW2.3\ \text{kW}$ , gives the power it transfers when connected to the stated potential difference.

Example

Power and energy for a heater

A heater has a potential difference of $V12\ \text{V}$ across it and a current of $A3.0\ \text{A}$ through it, and it is switched on for $5.0$ minutes.

12\ \text{V} \times 3.0\ \text{A} = 36\ \text{W}

\times 60 = 300\ \text{s}

36\ \text{W} \times 300\ \text{s} = 10\,800\ \text{J}

The heater transfers $J36\ \text{J}$ every second, and $J10\,800\ \text{J}$ in the five minutes.

Example

Heating effect in a resistor

A current of $A0.50\ \text{A}$ flows through a $Ω20\ \Omega$ resistor.

I^2R = (0.50\ \text{A})^2 \times 20\ \Omega = 5.0\ \text{W}

The current is then doubled to $A1.0\ \text{A}$ .

(1.0\ \text{A})^2 \times 20\ \Omega = 20\ \text{W}

Doubling the current has made the power four times larger.

Example

Energy used by a kettle

A kettle is rated at $V230\ \text{V}$ , $kW2.3\ \text{kW}$ , and it is used for $3.0$ minutes.

\dfrac{P}{V} = \dfrac{2300\ \text{W}}{230\ \text{V}} = 10\ \text{A}

\dfrac{3.0}{60} = 0.050\ \text{h}

2.3\ \text{kW} \times 0.050\ \text{h} = 0.115\ \text{kW h}

The same energy in joules is $J2300\ \text{W} \times 180\ \text{s} = 414\,000\ \text{J}$ .

Common Mistake

Do not confuse power and energy, because power in watts is the rate of transfer, while energy in joules or kilowatt-hours is the amount transferred.
Do not write that the kilowatt-hour is a unit of power, because it is a unit of energy.
Do not write about an electrical energy store, because no such store exists and electricity is a way of transferring energy between stores.
Do not leave time in minutes when the power is in watts, because $E = P t$ needs seconds to give joules.

Exam technique

Power calculations

Choose the power equation that matches the values given, using $P = V I$ for potential difference and current, $P = I^2R$ for current and resistance, and $\dfrac{E}{t}$ for energy and time.
In a two-step question, such as finding energy from potential difference, current and time, find the power first and then use $E = P t$ .
When you explain why power depends on potential difference and current, link energy per coulomb with coulombs per second.

Self review

Define power and state its unit.
Write down the equation linking power, potential difference and current.
Explain why the power of a device equals the potential difference multiplied by the current.
Calculate the energy transferred by a $kW1.5\ \text{kW}$ heater in $h2.0\ \text{h}$ , in kilowatt-hours.
Describe what happens to the power transferred in a resistor when the current through it is doubled.

3.2.12 Power transfer in circuit devices

Energy transfer in a circuit

Power

Power, potential difference and current

Energy over time

Power and energy for a heater

Heating effect in a resistor

Energy used by a kettle

Power calculations

Review 3.2.12 Power transfer in circuit devices by teaching Genie

3.2.12 Power transfer in circuit devices

Energy transfer in a circuit

Power

Power, potential difference and current

Energy over time

Power and energy for a heater

Heating effect in a resistor

Energy used by a kettle

Power calculations

Review 3.2.12 Power transfer in circuit devices by teaching Genie

3.2.12 Power transfer in circuit devices

Energy transfer in a circuit

Power

Power, potential difference and current

Energy over time

Power and energy for a heater

Heating effect in a resistor

Energy used by a kettle

Power calculations

Review 3.2.12 Power transfer in circuit devices by teaching Genie

3.2.12 Power transfer in circuit devices Revision Guide

3.2.12 Power transfer in circuit devices

Energy transfer in a circuit

Power

Power, potential difference and current

Energy over time

Power and energy for a heater

Heating effect in a resistor

Energy used by a kettle

Power calculations

Review 3.2.12 Power transfer in circuit devices by teaching Genie

3.2.12 Power transfer in circuit devices Revision Guide