Power and Energy Transfer in Circuits

This lesson covers electrical power and energy transfer in circuits, including the key power equations and the calculation of energy in joules and kilowatt-hours, as required by AQA GCSE Combined Science Trilogy (8464, section 6.2.1).

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What Is Electrical Power?

Power is the rate of energy transfer — how quickly energy is transferred from one form to another.

Power is measured in watts (W). One watt is one joule per second:

$1 \text{ W} = 1 \text{ J/s}$1 W=1 J/s

Higher power = faster energy transfer.

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Power Equations

AQA requires you to know and use three power equations:

Equation 1: P = IV

$P = I \times V$P=I×V

Where:

• $P$P = power in watts (W)
• $I$I = current in amperes (A)
• $V$V = potential difference in volts (V)

Equation 2: P = I²R

$P = I^2 \times R$P=I2×R

Where:

• $P$P = power in watts (W)
• $I$I = current in amperes (A)
• $R$R = resistance in ohms (Ω)

This equation is derived by substituting $V = IR$V=IR into $P = IV$P=IV:

$P = I \times (IR) = I^2R$P=I×(IR)=I2R

Equation 3: P = V²/R

$P = \frac{V^2}{R}$P=RV2​

This is derived by substituting $I = V/R$I=V/R into $P = IV$P=IV:

$P = \frac{V}{R} \times V = \frac{V^2}{R}$P=RV​×V=RV2​

Which Equation to Use?

Known Quantities	Use
Current and voltage	$P = IV$P=IV
Current and resistance	$P = I^2R$P=I2R
Voltage and resistance	$P = V^2/R$P=V2/R

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Energy Transfer Equations

Equation 1: E = Pt

$E = P \times t$E=P×t

Where:

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

Equation 2: E = QV

$E = Q \times V$E=Q×V

Where:

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

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

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The Kilowatt-Hour (kWh)

For household electricity bills, energy is measured in kilowatt-hours (kWh), not joules.

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

One kilowatt-hour is the energy transferred by a 1 kW device running for 1 hour:

$1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3{,}600{,}000 \text{ J} = 3.6 \times 10^6 \text{ J}$1 kWh=1000 W×3600 s=3,600,000 J=3.6×106 J

Calculating Electricity Costs

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

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Worked Examples

Worked Example 1 — Using P = IV

A 230 V kettle draws a current of 10 A. Calculate the power.

$P = I \times V = 10 \times 230 = 2300 \text{ W} = 2.3 \text{ kW}$P=I×V=10×230=2300 W=2.3 kW

Worked Example 2 — Using P = I²R

A current of 4 A flows through a 10 Ω resistor. Calculate the power dissipated.

$P = I^2 \times R = 4^2 \times 10 = 16 \times 10 = 160 \text{ W}$P=I2×R=42×10=16×10=160 W

Worked Example 3 — Using P = V²/R

A 12 V battery is connected to a 48 Ω resistor. Calculate the power.

$P = \frac{V^2}{R} = \frac{12^2}{48} = \frac{144}{48} = 3 \text{ W}$P=RV2​=48122​=48144​=3 W

Worked Example 4 — Using E = Pt

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

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

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

Worked Example 5 — Using E = QV

A charge of 100 C passes through a 12 V battery. How much energy is transferred?

$E = Q \times V = 100 \times 12 = 1200 \text{ J}$E=Q×V=100×12=1200 J

Worked Example 6 — Electricity Cost

A 3 kW oven is used for 2 hours. Electricity costs 34p per kWh. What is the cost?

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

$\text{Cost} = 6 \times 34 = 204 \text{p} = £2.04$Cost=6×34=204p=£2.04

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Power and Resistance — Key Relationships

Scenario	Relationship	Explanation
Fixed current, increasing resistance	Power increases	$P = I^2R$P=I2R — more resistance means more energy dissipated as heat
Fixed voltage, increasing resistance	Power decreases	$P = V^2/R$P=V2/R — more resistance means less current and less power

This is why:

• Heating elements use high-resistance wire (to dissipate lots of energy as heat).
• Power cables use thick, low-resistance copper (to minimise energy wasted as heat during transmission).

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Typical Power Ratings

Device	Typical Power
LED lamp	5–10 W
Filament lamp	40–100 W
Laptop	30–90 W
Kettle	2000–3000 W
Electric shower	7000–10 500 W

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Choosing a Fuse

$I = \frac{P}{V}$I=VP​

Calculate the normal current, then choose the fuse rating just above this value.

Worked Example 7

A 700 W vacuum cleaner on a 230 V supply. Which fuse: 3 A, 5 A or 13 A?

$I = \frac{700}{230} \approx 3.04 \text{ A}$I=230700​≈3.04 A

A 3 A fuse is too close to the operating current and may blow. A 5 A fuse is the best choice.

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Common Mistakes

Mistake	Correction
Using kW in $E = Pt$E=Pt when t is in seconds	If time is in seconds, power must be in watts to get joules. Use kW only with hours to get kWh
Confusing power and energy	Power is the rate of energy transfer; energy is the total amount transferred
Forgetting to convert minutes to seconds	Always check units before substituting into equations
Using the wrong power equation	Choose based on the quantities given in the question

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Summary

• Power is the rate of energy transfer, measured in watts (W).
• Three power equations: $P = IV$P=IV, $P = I^2R$P=I2R, $P = V^2/R$P=V2/R.
• Energy transferred: $E = Pt$E=Pt (joules) or $E = QV$E=QV (joules).
• Kilowatt-hours: $E = P \text{(kW)} \times t \text{(hours)}$E=P(kW)×t(hours); 1 kWh = 3.6 × 10⁶ J.
• Cost = energy (kWh) × price per kWh.
• Choose a fuse with a rating just above the normal operating current.
• For fixed current, more resistance = more power dissipated.
• For fixed voltage, more resistance = less power dissipated.

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Extended Worked Examples