Energy, power and time

Definition

Joule

One joule is the energy transferred when a charge of one coulomb moves through a potential difference of one volt.

Energy transferred by an electrical circuit is measured in joules, and the symbol for the unit is $J\text{J}$ .
The first of the two energy equations is $E = P t$ , where $E$ is the energy transferred in joules, $P$ is the power in watts and $t$ is the time in seconds.
Both quantities on the right must already be in those units. A power quoted in kilowatts is converted to watts, and a time quoted in minutes or hours is converted to seconds, before either is substituted.
The equation rearranges to $P=EtP=\dfrac{E}{t}$ and $t=EPt=\dfrac{E}{P}$ .
Read directly from the equation, a device running at $W1\ \text{W}$ for $s1\ \text{s}$ transfers $J1\ \text{J}$ .
The joule is a small unit on household scales. Boiling a full kettle transfers a few hundred thousand joules, so answers are often written in standard form or in kilojoules, where $J1\ \text{kJ}=1000\ \text{J}$ .
The value calculated is the total energy transferred out of the supply and into the component during that time, so it is always positive and it always grows as the time grows.

Example

Energy from power and time

A kettle of power $kW2.2\ \text{kW}$ is switched on for $3.0$ minutes.
Converting the power gives $WP=2.2\times1000=2200\ \text{W}$ .
Converting the time gives $st=3.0\times60=180\ \text{s}$ .
The equation is $E = P t$ .
Substituting gives $JE=2200\times180=396\,000\ \text{J}$ .
The energy transferred is $J3.96\times10^{5}\ \text{J}$ , which is $kJ396\ \text{kJ}$ .

Building the energy equation

The second equation is built from two results that are already available, so it does not have to be memorised on its own.
The charge that passes a point in a circuit is given by $Q = I t$ , with $Q$ in coulombs, $I$ in amperes and $t$ in seconds.
The energy that charge transfers is given by $E = Q V$ , where $V$ is the potential difference in volts across the component the charge passes through.
Substituting the first result into the second replaces $Q$ with $I t$ , which gives $E = (I t) V$ .
Written in the usual order this is $E = I V t$ , with $E$ in joules, $I$ in amperes, $V$ in volts and $t$ in seconds.
The chain reads as a sentence. The current and the time together fix how much charge passes, and the potential difference fixes how much energy each coulomb of that charge delivers.
Both equations give the same energy for the same component over the same time, so $E = P t$ and $E = I V t$ must agree.
The data decide which equation to use. A power and a time point to $E = P t$ , while a current, a potential difference and a time point to $E = I V t$ .

Using the equations

The reliable method has four stages: write the equation, convert every quantity into its base unit, substitute, then state the answer with its unit.
Time conversions cost more marks than anything else in this work. $s1\ \text{minute}=60\ \text{s}$ , $s1\ \text{hour}=3600\ \text{s}$ and $s2.5\ \text{hours}=9000\ \text{s}$ .
Power conversions follow the same pattern, since $W1\ \text{kW}=1000\ \text{W}$ , so $W2.2\ \text{kW}=2200\ \text{W}$ and $W0.80\ \text{kW}=800\ \text{W}$ .
A question that supplies a charge rather than a current is answered through $E = Q V$ instead, and any charge in millicoulombs is converted first, since $C1\ \text{mC}=1\times10^{-3}\ \text{C}$ .
Large answers are normally given in standard form to two or three significant figures, matching the precision of the least precise value in the question.
A size check catches most slips. A mains appliance running for a few minutes transfers of the order of $J10^{5}\ \text{J}$ , so an answer of a few hundred joules for a kettle points to a missed time conversion.

Example

Energy from current and voltage

A lamp on a $V230\ \text{V}$ supply carries a current of $A0.26\ \text{A}$ and is left on for $2.0$ hours.
Converting the time gives $st=2.0\times3600=7200\ \text{s}$ .
The equation is $E = I V t$ .
Substituting gives $E=0.26×230×7200E=0.26\times230\times7200$ .
Taking the first product, $0.26×230=59.80.26\times230=59.8$ , and then $56059.8\times7200=430\,560$ .
The energy transferred is $J4.3\times10^{5}\ \text{J}$ to two significant figures.

Finding a missing quantity

Either equation can be rearranged to make any one of its quantities the subject, and the energy is then the value that is given rather than the value that is wanted.
From $E = P t$ the rearrangements are $P=EtP=\dfrac{E}{t}$ and $t=EPt=\dfrac{E}{P}$ .
From $E = I V t$ the rearrangements are $I=EVtI=\dfrac{E}{Vt}$ , $V=EItV=\dfrac{E}{It}$ and $t=EIVt=\dfrac{E}{IV}$ .
The two quantities that are not wanted are multiplied together first and the energy is then divided by that product, which keeps the arithmetic to a single division.
A question often hides the time in another unit, so a running time of $2.5$ minutes must become $s150\ \text{s}$ before it is used.
The answer carries the unit of the quantity that was found: $A\text{A}$ for a current, $V\text{V}$ for a potential difference, $W\text{W}$ for a power and $s\text{s}$ for a time.

Example

Finding the current

A motor connected to a $V12\ \text{V}$ supply transfers $J9.0\times10^{3}\ \text{J}$ in $2.5$ minutes.
Converting the time gives $st=2.5\times60=150\ \text{s}$ .
The equation is $E = I V t$ , which rearranges to $I=EVtI=\dfrac{E}{Vt}$ .
The denominator is $Vt=12×150=1800Vt=12\times150=1800$ .
Substituting gives $AI=\dfrac{9000}{1800}=5.0\ \text{A}$ .

Comparing two appliances

A comparison of two devices means something only if the same time is used for both, so the first step is to fix that time and convert it to seconds once.
With $E = P t$ and the same $t$ for each device, the energy transferred is directly proportional to the power, so a device of ten times the power transfers ten times the energy.
With $E = I V t$ and two devices on the same mains supply, $V$ and $t$ are the same for both, so the energy transferred is directly proportional to the current each device draws.
A ratio settles such a question faster than two full calculations, although the data for each device must still be checked for the units they are quoted in.
A low-power device left on for a long time can transfer more energy in total than a high-power device used briefly, because the time enters the equation exactly as the power does.

Example

Two devices, same time

A hair dryer of power $kW1.5\ \text{kW}$ and a lamp of power $W60\ \text{W}$ each run for $10$ minutes.
The shared time is $st=10\times60=600\ \text{s}$ , and the dryer's power is $W1500\ \text{W}$ .
For the dryer, $JE=Pt=1500\times600=9.0\times10^{5}\ \text{J}$ .
For the lamp, $JE=Pt=60\times600=3.6\times10^{4}\ \text{J}$ .
The ratio is $9.0×1053.6×104=25\dfrac{9.0\times10^{5}}{3.6\times10^{4}}=25$ .
The dryer transfers $25$ times as much energy as the lamp in the same time.

Common Mistake

Do not substitute a time in minutes or hours into either equation. Convert it to seconds first, every time.
Do not substitute a power in kilowatts. Convert to watts, since the joule pairs with the watt and the second.
Do not pair a current from one component with the potential difference across a different one. Both values must belong to the same component.
Do not mix the two equations in one substitution by using a power and a current together. Each equation is complete as it stands.
Do not leave an energy without its unit. The unit here is always the joule.
Do not round an intermediate value heavily. Carry the extra figures through and round only the final answer.

Exam technique

Setting out the calculation

Quote the equation, $E = P t$ or $E = I V t$ , before substituting any numbers.
Show each unit conversion on its own line, such as $st=4.0\times60=240\ \text{s}$ , so a later slip still leaves the method visible.
Rearrange first and substitute second when a missing quantity is wanted, keeping the two steps apart.
Choose the equation from the data given rather than from habit, and say which one is being used.
Give the answer with its unit and to a sensible number of significant figures, using standard form above about $J10^{4}\ \text{J}$ .
Use directly proportional only where one quantity is held fixed, for example energy against time at constant power.

Self review

State the equation linking energy, power and time, and give the unit of each quantity in it.
Build $E = I V t$ from $Q = I t$ and $E = Q V$ .
Calculate the energy transferred by a $kW1.8\ \text{kW}$ heater in $5.0$ minutes.
Find the current in a $V12\ \text{V}$ lamp that transfers $J1440\ \text{J}$ in $s60\ \text{s}$ .
Explain how a $W60\ \text{W}$ lamp left on all day can transfer more energy than a $kW2\ \text{kW}$ kettle used twice.

Energy, power and time

Definition

Joule

One joule is the energy transferred when a charge of one coulomb moves through a potential difference of one volt.

Energy transferred by an electrical circuit is measured in joules, and the symbol for the unit is $J\text{J}$ .
The first of the two energy equations is $E = P t$ , where $E$ is the energy transferred in joules, $P$ is the power in watts and $t$ is the time in seconds.
Both quantities on the right must already be in those units. A power quoted in kilowatts is converted to watts, and a time quoted in minutes or hours is converted to seconds, before either is substituted.
The equation rearranges to $P=EtP=\dfrac{E}{t}$ and $t=EPt=\dfrac{E}{P}$ .
Read directly from the equation, a device running at $W1\ \text{W}$ for $s1\ \text{s}$ transfers $J1\ \text{J}$ .
The joule is a small unit on household scales. Boiling a full kettle transfers a few hundred thousand joules, so answers are often written in standard form or in kilojoules, where $J1\ \text{kJ}=1000\ \text{J}$ .
The value calculated is the total energy transferred out of the supply and into the component during that time, so it is always positive and it always grows as the time grows.

Example

Energy from power and time

A kettle of power $kW2.2\ \text{kW}$ is switched on for $3.0$ minutes.
Converting the power gives $WP=2.2\times1000=2200\ \text{W}$ .
Converting the time gives $st=3.0\times60=180\ \text{s}$ .
The equation is $E = P t$ .
Substituting gives $JE=2200\times180=396\,000\ \text{J}$ .
The energy transferred is $J3.96\times10^{5}\ \text{J}$ , which is $kJ396\ \text{kJ}$ .

Building the energy equation

The second equation is built from two results that are already available, so it does not have to be memorised on its own.
The charge that passes a point in a circuit is given by $Q = I t$ , with $Q$ in coulombs, $I$ in amperes and $t$ in seconds.
The energy that charge transfers is given by $E = Q V$ , where $V$ is the potential difference in volts across the component the charge passes through.
Substituting the first result into the second replaces $Q$ with $I t$ , which gives $E = (I t) V$ .
Written in the usual order this is $E = I V t$ , with $E$ in joules, $I$ in amperes, $V$ in volts and $t$ in seconds.
The chain reads as a sentence. The current and the time together fix how much charge passes, and the potential difference fixes how much energy each coulomb of that charge delivers.
Both equations give the same energy for the same component over the same time, so $E = P t$ and $E = I V t$ must agree.
The data decide which equation to use. A power and a time point to $E = P t$ , while a current, a potential difference and a time point to $E = I V t$ .

Using the equations

The reliable method has four stages: write the equation, convert every quantity into its base unit, substitute, then state the answer with its unit.
Time conversions cost more marks than anything else in this work. $s1\ \text{minute}=60\ \text{s}$ , $s1\ \text{hour}=3600\ \text{s}$ and $s2.5\ \text{hours}=9000\ \text{s}$ .
Power conversions follow the same pattern, since $W1\ \text{kW}=1000\ \text{W}$ , so $W2.2\ \text{kW}=2200\ \text{W}$ and $W0.80\ \text{kW}=800\ \text{W}$ .
A question that supplies a charge rather than a current is answered through $E = Q V$ instead, and any charge in millicoulombs is converted first, since $C1\ \text{mC}=1\times10^{-3}\ \text{C}$ .
Large answers are normally given in standard form to two or three significant figures, matching the precision of the least precise value in the question.
A size check catches most slips. A mains appliance running for a few minutes transfers of the order of $J10^{5}\ \text{J}$ , so an answer of a few hundred joules for a kettle points to a missed time conversion.

Example

Energy from current and voltage

A lamp on a $V230\ \text{V}$ supply carries a current of $A0.26\ \text{A}$ and is left on for $2.0$ hours.
Converting the time gives $st=2.0\times3600=7200\ \text{s}$ .
The equation is $E = I V t$ .
Substituting gives $E=0.26×230×7200E=0.26\times230\times7200$ .
Taking the first product, $0.26×230=59.80.26\times230=59.8$ , and then $56059.8\times7200=430\,560$ .
The energy transferred is $J4.3\times10^{5}\ \text{J}$ to two significant figures.

Finding a missing quantity

Either equation can be rearranged to make any one of its quantities the subject, and the energy is then the value that is given rather than the value that is wanted.
From $E = P t$ the rearrangements are $P=EtP=\dfrac{E}{t}$ and $t=EPt=\dfrac{E}{P}$ .
From $E = I V t$ the rearrangements are $I=EVtI=\dfrac{E}{Vt}$ , $V=EItV=\dfrac{E}{It}$ and $t=EIVt=\dfrac{E}{IV}$ .
The two quantities that are not wanted are multiplied together first and the energy is then divided by that product, which keeps the arithmetic to a single division.
A question often hides the time in another unit, so a running time of $2.5$ minutes must become $s150\ \text{s}$ before it is used.
The answer carries the unit of the quantity that was found: $A\text{A}$ for a current, $V\text{V}$ for a potential difference, $W\text{W}$ for a power and $s\text{s}$ for a time.

Example

Finding the current

A motor connected to a $V12\ \text{V}$ supply transfers $J9.0\times10^{3}\ \text{J}$ in $2.5$ minutes.
Converting the time gives $st=2.5\times60=150\ \text{s}$ .
The equation is $E = I V t$ , which rearranges to $I=EVtI=\dfrac{E}{Vt}$ .
The denominator is $Vt=12×150=1800Vt=12\times150=1800$ .
Substituting gives $AI=\dfrac{9000}{1800}=5.0\ \text{A}$ .

Comparing two appliances

A comparison of two devices means something only if the same time is used for both, so the first step is to fix that time and convert it to seconds once.
With $E = P t$ and the same $t$ for each device, the energy transferred is directly proportional to the power, so a device of ten times the power transfers ten times the energy.
With $E = I V t$ and two devices on the same mains supply, $V$ and $t$ are the same for both, so the energy transferred is directly proportional to the current each device draws.
A ratio settles such a question faster than two full calculations, although the data for each device must still be checked for the units they are quoted in.
A low-power device left on for a long time can transfer more energy in total than a high-power device used briefly, because the time enters the equation exactly as the power does.

Example

Two devices, same time

A hair dryer of power $kW1.5\ \text{kW}$ and a lamp of power $W60\ \text{W}$ each run for $10$ minutes.
The shared time is $st=10\times60=600\ \text{s}$ , and the dryer's power is $W1500\ \text{W}$ .
For the dryer, $JE=Pt=1500\times600=9.0\times10^{5}\ \text{J}$ .
For the lamp, $JE=Pt=60\times600=3.6\times10^{4}\ \text{J}$ .
The ratio is $9.0×1053.6×104=25\dfrac{9.0\times10^{5}}{3.6\times10^{4}}=25$ .
The dryer transfers $25$ times as much energy as the lamp in the same time.

Common Mistake

Do not substitute a time in minutes or hours into either equation. Convert it to seconds first, every time.
Do not substitute a power in kilowatts. Convert to watts, since the joule pairs with the watt and the second.
Do not pair a current from one component with the potential difference across a different one. Both values must belong to the same component.
Do not mix the two equations in one substitution by using a power and a current together. Each equation is complete as it stands.
Do not leave an energy without its unit. The unit here is always the joule.
Do not round an intermediate value heavily. Carry the extra figures through and round only the final answer.

Exam technique

Setting out the calculation

Quote the equation, $E = P t$ or $E = I V t$ , before substituting any numbers.
Show each unit conversion on its own line, such as $st=4.0\times60=240\ \text{s}$ , so a later slip still leaves the method visible.
Rearrange first and substitute second when a missing quantity is wanted, keeping the two steps apart.
Choose the equation from the data given rather than from habit, and say which one is being used.
Give the answer with its unit and to a sensible number of significant figures, using standard form above about $J10^{4}\ \text{J}$ .
Use directly proportional only where one quantity is held fixed, for example energy against time at constant power.

Self review

State the equation linking energy, power and time, and give the unit of each quantity in it.
Build $E = I V t$ from $Q = I t$ and $E = Q V$ .
Calculate the energy transferred by a $kW1.8\ \text{kW}$ heater in $5.0$ minutes.
Find the current in a $V12\ \text{V}$ lamp that transfers $J1440\ \text{J}$ in $s60\ \text{s}$ .
Explain how a $W60\ \text{W}$ lamp left on all day can transfer more energy than a $kW2\ \text{kW}$ kettle used twice.

10.3.4 Energy transferred by an electrical circuit

Energy, power and time

Energy from power and time

Building the energy equation

Using the equations

Energy from current and voltage

Finding a missing quantity

Finding the current

Comparing two appliances

Two devices, same time

Setting out the calculation

Review 10.3.4 Energy transferred by an electrical circuit by teaching Genie

10.3.4 Energy transferred by an electrical circuit

Energy, power and time

Energy from power and time

Building the energy equation

Using the equations

Energy from current and voltage

Finding a missing quantity

Finding the current

Comparing two appliances

Two devices, same time

Setting out the calculation

Review 10.3.4 Energy transferred by an electrical circuit by teaching Genie

10.3.4 Energy transferred by an electrical circuit

Energy, power and time

Energy from power and time

Building the energy equation

Using the equations

Energy from current and voltage

Finding a missing quantity

Finding the current

Comparing two appliances

Two devices, same time

Setting out the calculation

Review 10.3.4 Energy transferred by an electrical circuit by teaching Genie

10.3.4 Energy transferred by an electrical circuit Revision Guide

10.3.4 Energy transferred by an electrical circuit

Energy, power and time

Energy from power and time

Building the energy equation

Using the equations

Energy from current and voltage

Finding a missing quantity

Finding the current

Comparing two appliances

Two devices, same time

Setting out the calculation

Review 10.3.4 Energy transferred by an electrical circuit by teaching Genie

10.3.4 Energy transferred by an electrical circuit Revision Guide