Energy Transferred in Circuits

This lesson covers the calculation of energy transferred in electrical circuits, including the use of kilowatt-hours and electricity bill calculations, as required by the Edexcel GCSE Combined Science specification (1SC0).

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Energy Transferred — Two Key Equations

Equation 1: E = Pt

$E = Pt$E=Pt

where:

• E = energy transferred in joules (J)
• P = power in watts (W)
• t = time in seconds (s)

This equation is a rearrangement of the power equation $P = E/t$P=E/t.

Equation 2: E = QV

$E = QV$E=QV

where:

• E = energy transferred in joules (J)
• Q = charge in coulombs (C)
• V = potential difference in volts (V)

This equation comes from the definition of potential difference: $V = E/Q$V=E/Q.

Equation	Best used when you know...
$E = Pt$E=Pt	Power and time
$E = QV$E=QV	Charge and voltage

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Worked Examples Using E = Pt

Worked Example 1

A 2000 W kettle is switched on for 3 minutes. How much energy is transferred?

Convert time: $t = 3 \times 60 = 180 \text{ s}$t=3×60=180 s

$E = Pt = 2000 \times 180 = 360\,000 \text{ J} = 360 \text{ kJ}$E=Pt=2000×180=360000 J=360 kJ

Worked Example 2

A 40 W lamp is on for 2 hours. How much energy does it transfer?

Convert time: $t = 2 \times 3600 = 7200 \text{ s}$t=2×3600=7200 s

$E = Pt = 40 \times 7200 = 288\,000 \text{ J} = 288 \text{ kJ}$E=Pt=40×7200=288000 J=288 kJ

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Worked Examples Using E = QV

Worked Example 3

A charge of 50 C passes through a 12 V battery. How much energy is transferred?

$E = QV = 50 \times 12 = 600 \text{ J}$E=QV=50×12=600 J

Worked Example 4

A current of 3 A flows through a 230 V heater for 10 seconds. How much energy is transferred?

First find the charge: $Q = It = 3 \times 10 = 30 \text{ C}$Q=It=3×10=30 C

Then: $E = QV = 30 \times 230 = 6900 \text{ J}$E=QV=30×230=6900 J

Exam Tip: You could also solve Worked Example 4 using $E = Pt$E=Pt and $P = IV$P=IV: $P = 3 \times 230 = 690 \text{ W}$P=3×230=690 W, then $E = 690 \times 10 = 6900 \text{ J}$E=690×10=6900 J. Both methods give the same answer.

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Kilowatt-Hours (kWh)

For domestic electricity, energy is measured in kilowatt-hours (kWh) rather than joules, because joules are too small a unit for household energy use.

Definition

1 kilowatt-hour is the energy transferred by a 1 kW device operating for 1 hour.

$E \text{ (kWh)} = P \text{ (kW)} \times t \text{ (hours)}$E (kWh)=P (kW)×t (hours)

Converting Between kWh and Joules

$1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3\,600\,000 \text{ J} = 3.6 \text{ MJ}$1 kWh=1000 W×3600 s=3600000 J=3.6 MJ

Exam Tip: When using the kWh equation, make sure power is in kilowatts and time is in hours. This is different from the standard SI units used in $E = Pt$E=Pt.

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Worked Examples Using kWh

Worked Example 5

A 3 kW oven is used for 2 hours. How many kWh of energy does it use?

$E = P \times t = 3 \times 2 = 6 \text{ kWh}$E=P×t=3×2=6 kWh

Worked Example 6

A 100 W lamp is on for 10 hours. How many kWh of energy does it use?

Convert power: $P = 100 \text{ W} = 0.1 \text{ kW}$P=100 W=0.1 kW

$E = 0.1 \times 10 = 1 \text{ kWh}$E=0.1×10=1 kWh

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Calculating Electricity Costs

The cost of electricity is calculated using:

$\text{Cost} = \text{Energy (kWh)} \times \text{Price per kWh}$Cost=Energy (kWh)×Price per kWh

Worked Example 7

A 2.5 kW kettle is used for a total of 0.5 hours per day. Electricity costs 34p per kWh. What is the daily cost?

Step 1 — Energy used:

$E = 2.5 \times 0.5 = 1.25 \text{ kWh}$E=2.5×0.5=1.25 kWh

Step 2 — Cost:

$\text{Cost} = 1.25 \times 34 = 42.5 \text{p}$Cost=1.25×34=42.5p

Worked Example 8

A household uses 12 kWh of electricity per day. The tariff is 28p per kWh with a daily standing charge of 46p. What is the total daily cost?

$\text{Energy cost} = 12 \times 28 = 336 \text{p}$Energy cost=12×28=336p

$\text{Total cost} = 336 + 46 = 382 \text{p} = £3.82$Total cost=336+46=382p=£3.82

Exam Tip: Exam questions often give the price per kWh in pence. Make sure your final answer is in the correct unit — either pence or pounds. Show your conversion if needed.

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Reducing Electricity Bills

Method	How It Reduces Cost
Use energy-efficient appliances (e.g. LED lamps)	Lower power rating means fewer kWh used
Turn off devices when not in use	Reduces the time (t) in $E = Pt$E=Pt
Insulate the home	Less energy needed for heating
Use appliances at off-peak times	Lower tariff rates (if on an economy tariff)
Switch to a cheaper energy provider	Lower price per kWh

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Energy and Charge in Components

When charge flows through a component with a potential difference across it, energy is transferred:

$E = QV$E=QV

This energy can be transferred to different stores:

• In a resistor or filament lamp — to the thermal store (heating).
• In a motor — to the kinetic store (movement).
• In an LED — to the thermal store and emitted as light (radiation).
• In a buzzer — to the thermal store and emitted as sound (mechanical waves).

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Summary

• Energy transferred in a circuit can be calculated using $E = Pt$E=Pt or $E = QV$E=QV.
• For domestic electricity, energy is measured in kilowatt-hours (kWh): $E = P \text{(kW)} \times t \text{(hours)}$E=P(kW)×t(hours).
• $1 \text{ kWh} = 3.6 \text{ MJ} = 3\,600\,000 \text{ J}$1 kWh=3.6 MJ=3600000 J.
• Electricity cost: $\text{Cost} = \text{kWh} \times \text{price per kWh}$Cost=kWh×price per kWh.
• To reduce bills: use efficient appliances, reduce usage time and insulate the home.
• Energy is transferred to different stores depending on the component (thermal, kinetic, light, sound).

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Extended Worked Examples

Worked Example 9 — Four Routes to the Same Answer

A 12 V battery drives 0.5 A through a circuit for 5 minutes. Calculate the energy transferred using four different routes.

Preliminary: $t = 5 \times 60 = 300 \text{ s}$t=5×60=300 s, $Q = It = 0.5 \times 300 = 150 \text{ C}$Q=It=0.5×300=150 C, $P = IV = 0.5 \times 12 = 6 \text{ W}$P=IV=0.5×12=6 W.