P8: Electrical Power and Energy
Electrical power, energy transfers, the National Grid, transformers, AC/DC and domestic energy calculations.
Electrical Power
Power is the rate at which energy is transferred. In electrical circuits, power depends on both the current and the potential difference.
Power Equation (P = VI)
P = VI
P = power in watts (W)
V = potential difference in volts (V)
I = current in amperes (A)
Worked Example
An electric kettle runs on the mains (230 V) and draws a current of 10 A. Calculate its power.
P = VI = 230 × 10 = 2300 W = 2.3 kW
Alternative Power Equations
P = I²R
P = V² / R
These are derived by substituting V = IR or I = V/R into P = VI.
Worked Example
A heater has a resistance of 20 Ω and carries a current of 5 A. Calculate the power using P = I²R.
P = I²R = 5² × 20 = 25 × 20 = 500 W
Worked Example
A 12 V lamp has a resistance of 8 Ω. Calculate its power using P = V²/R.
P = V² / R = 12² / 8 = 144 / 8 = 18 W
Energy Transferred
The energy transferred by an electrical device depends on its power and how long it is used for.
Energy and Power
E = Pt
E = energy in joules (J)
P = power in watts (W)
t = time in seconds (s)
Worked Example
A 2 kW heater is left on for 3 hours. Calculate the energy transferred in joules.
P = 2000 W, t = 3 × 3600 = 10800 s
E = Pt = 2000 × 10800 = 21 600 000 J = 21.6 MJ
Energy and Charge
E = QV
E = energy in joules (J)
Q = charge in coulombs (C)
V = potential difference in volts (V)
Worked Example
A charge of 150 C flows through a motor with a p.d. of 12 V across it. Calculate the energy transferred.
E = QV = 150 × 12 = 1800 J
AC and DC
There are two types of electrical supply: alternating current (AC) and direct current (DC).
Alternating Current (AC)
- The current continuously changes direction
- The p.d. alternates between positive and negative values
- Mains electricity in the UK is AC at 50 Hz with a peak voltage of about 325 V
- The quoted mains voltage of 230 V is the effective (RMS) value
Direct Current (DC)
- The current flows in one direction only
- The p.d. is always positive
- Batteries and solar cells produce DC
- Many electronic devices require DC and contain rectifiers to convert AC to DC
| Property | AC | DC |
|---|---|---|
| Direction of current | Changes continuously | One direction only |
| Source | Mains supply, generators | Batteries, solar cells |
| Voltage | Can be changed easily with transformers | Cannot be changed with transformers |
| Use | Power transmission, household supply | Portable devices, electronics |
The National Grid
The National Grid is the network of cables and transformers that connects power stations to consumers across the UK.
How the National Grid Works
- Power stations generate electricity at about 25 000 V (25 kV)
- Step-up transformers increase the voltage to 132 000 V to 400 000 V (132 kV to 400 kV) for transmission
- High voltage is used for transmission to reduce energy losses in the cables
- Step-down transformers reduce the voltage to 230 V for domestic use
Why High Voltage is Used
- For a given power, a higher voltage means a lower current (P = VI)
- Lower current means less energy is wasted as heat in the cables (P = I²R)
- This makes transmission more efficient and reduces costs
- Thick cables are used to reduce resistance, but high voltage is the main method of reducing losses
You must be able to explain why the National Grid uses high voltage for transmission. The key chain is: high voltage → low current → less energy wasted as heat (P = I²R).
Transformers
Transformers change the voltage of an AC supply. They only work with AC because they require a changing magnetic field.
Types of Transformer
- Step-up transformer — increases voltage; more turns on the secondary coil than the primary
- Step-down transformer — decreases voltage; fewer turns on the secondary coil than the primary
Transformer Equation
Vp / Vs = np / ns
Vp = primary voltage
Vs = secondary voltage
np = number of turns on primary coil
ns = number of turns on secondary coil
Worked Example
A step-down transformer has 200 turns on the primary coil and 20 turns on the secondary coil. The primary voltage is 230 V. Calculate the secondary voltage.
Vs = Vp × ns / np = 230 × 20 / 200 = 23 V
Transformer Efficiency (Higher Tier)
- In an ideal transformer: Vp × Ip = Vs × Is
Real transformers are not 100% efficient — some energy is lost as heat in the coils and core
Worked Example (Higher)
A transformer has a primary voltage of 230 V and primary current of 0.5 A. The secondary voltage is 23 V. Assuming the transformer is 100% efficient, calculate the secondary current.
VpIp = VsIs
230 × 0.5 = 23 × Is
Is = 115 / 23 = 5 A
Domestic Energy and kilowatt-hours
Energy companies charge for electricity in kilowatt-hours (kWh) rather than joules because joules are very small units.
Energy in kilowatt-hours
E = Pt
E = energy in kilowatt-hours (kWh)
P = power in kilowatts (kW)
t = time in hours (h)
Worked Example
A 2.5 kW electric fire is used for 4 hours. Calculate the energy transferred in kWh. If electricity costs 28p per kWh, calculate the cost.
E = Pt = 2.5 × 4 = 10 kWh
Cost = 10 × 28 = 280p = £2.80
Worked Example
A 60 W lamp is left on for 10 hours. Calculate the energy in kWh and the cost at 28p per kWh.
P = 60 W = 0.06 kW
E = 0.06 × 10 = 0.6 kWh
Cost = 0.6 × 28 = 16.8p
Fuses and Circuit Breakers
Fuses and circuit breakers protect circuits from dangerously high currents that could cause fires or damage.
How Fuses Work
- A fuse contains a thin wire that melts if the current exceeds its rating
- Once melted, the circuit is broken and current stops flowing
- Fuses must be replaced after they have blown
- Common fuse ratings are 3 A, 5 A and 13 A
- The fuse rating should be slightly above the normal operating current of the device
Circuit Breakers
- Circuit breakers trip (switch off) when the current exceeds the safe value
- They can be reset by pressing a button or flipping a switch
- They are more convenient than fuses because they do not need to be replaced
- Residual current circuit breakers (RCCBs) detect differences between live and neutral currents, switching off in milliseconds
Practice Questions
1. A hair dryer runs on 230 V and draws a current of 6.5 A. Calculate its power.
P = VI = 230 × 6.5 = 1495 W = 1.495 kW
2. A heater has a resistance of 15 Ω and is connected to a 230 V supply. Calculate the power using P = V²/R.
P = V²/R = 230²/15 = 52900/15 = 3527 W = 3.5 kW
3. A 3 kW kettle is used for 20 minutes. Calculate the energy transferred in joules and in kWh.
In joules: E = Pt = 3000 × 1200 = 3 600 000 J = 3.6 MJ
In kWh: E = 3 × (20/60) = 3 × 0.333 = 1 kWh
4. Explain why the National Grid transmits electricity at very high voltages.
Transmitting at high voltage means a low current (P = VI). A low current reduces energy wasted as heat in the transmission cables (P = I²R), making the system more efficient.
5. A step-up transformer has 100 turns on the primary coil and 2000 turns on the secondary coil. The primary voltage is 25 000 V. Calculate the secondary voltage.
Vs = Vp × ns/np = 25 000 × 2000/100 = 25 000 × 20 = 500 000 V = 500 kV
🔢 Maths Skills
Mathematical Skills for this Topic
Rearranging P = VI: To find voltage: V = P/I. To find current: I = P/V. These are straightforward rearrangements but must be used with consistent units. Power in watts, voltage in volts, current in amperes. If power is given in kW, convert to watts first (multiply by 1000).
Using P = I²R and P = V²/R: These alternative forms are useful when you know two of the three variables but not all three. P = I²R is used when you know current and resistance. P = V²/R is used when you know voltage and resistance. To rearrange P = I²R to find I: I = √(P/R). To rearrange P = V²/R to find V: V = √(P × R).
Calculating cost using kWh: Energy in kWh = power in kW × time in hours. Cost = energy in kWh × price per kWh. Always convert watts to kilowatts (divide by 1000) and minutes to hours (divide by 60). For example, a 1500 W heater for 30 minutes: E = 1.5 × 0.5 = 0.75 kWh. At 28p/kWh, cost = 0.75 × 28 = 21p.
Unit conversions: 1 kW = 1000 W, 1 MW = 1 000 000 W. 1 kWh = 3 600 000 J (since 1 W = 1 J/s and there are 3600 s in an hour). 1 J = 1 Ws. To convert from kWh to J: multiply by 3 600 000. To convert from J to kWh: divide by 3 600 000. These conversions appear frequently in exam questions.
Transformer calculations: The equation Vp/Vs = np/ns involves ratios. Cross-multiply to solve: Vs = Vp × ns/np. For 100% efficient transformers: VpIp = VsIs. Rearrange to find the unknown current.
⚠️ Common Misconceptions
Watch Out!
Students often think that a higher power rating always means more energy is used. Wrong: A device with a higher power rating always uses more energy than a lower-power device. Correct: Energy depends on both power AND time (E = Pt). A 3 kW kettle used for 2 minutes transfers 0.1 kWh, while a 60 W lamp left on for 10 hours transfers 0.6 kWh. The lower-power lamp used more energy because it was on for much longer. Power tells you the rate of energy transfer, not the total energy.
Students often think that kWh is a unit of power. Wrong: A kilowatt-hour (kWh) is a unit of power, like the kilowatt. Correct: A kilowatt-hour is a unit of ENERGY, not power. It is the energy transferred by a 1 kW device running for 1 hour. 1 kWh = 1000 W × 3600 s = 3 600 000 J. Kilowatts (kW) measure power; kilowatt-hours (kWh) measure energy. Electricity bills charge for energy in kWh, not power in kW.
✍️ 6-Mark Question
Extended Answer Question
6 marks: Explain why the National Grid uses step-up transformers to increase voltage before transmission and step-down transformers to decrease voltage for domestic use.
Power stations generate electricity at around 25 000 V. Before transmitting electricity across the country through the National Grid, step-up transformers increase the voltage to between 132 000 V and 400 000 V. This is done to reduce the current in the transmission cables. Since P = VI, for a given power being transmitted, a higher voltage means a lower current.
A lower current is important because energy is wasted as heat in the transmission cables. The power wasted as heat is given by P = I²R, where R is the resistance of the cables. Because the current is squared in this equation, even a small reduction in current leads to a large reduction in wasted energy. For example, doubling the voltage halves the current, which reduces the power wasted to one quarter of its previous value. This makes transmission much more efficient.
However, such high voltages are far too dangerous for domestic use. Before the electricity reaches homes, step-down transformers reduce the voltage to 230 V for safe use in household appliances. The step-down transformer has fewer turns on the secondary coil than the primary coil. The combination of step-up and step-down transformers ensures that electricity is transmitted efficiently over long distances while remaining safe for consumers.
Mark scheme: 2 marks for explaining step-up transformers reduce current (P = VI), 2 marks for explaining why low current reduces waste (P = I²R), 1 mark for explaining step-down transformers for safety, 1 mark for linking the complete chain from power station to home
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
The table below shows the power ratings and typical daily usage times for three household appliances.
| Appliance | Power | Daily Use |
|---|---|---|
| Kettle | 2.2 kW | 12 min |
| Fridge | 0.15 kW | 24 h |
| Tumble dryer | 2.5 kW | 45 min |
Electricity costs 28p per kWh. Calculate the annual electricity cost for each appliance and evaluate which appliance offers the greatest potential for energy savings. Justify your answer.
Answer: Kettle: E = 2.2 × (12/60) = 0.44 kWh/day. Annual = 0.44 × 365 = 160.6 kWh. Cost = 160.6 × 28 = 4497p = £44.97. Fridge: E = 0.15 × 24 = 3.6 kWh/day. Annual = 3.6 × 365 = 1314 kWh. Cost = 1314 × 28 = 36 792p = £367.92. Tumble dryer: E = 2.5 × (45/60) = 1.875 kWh/day. Annual = 1.875 × 365 = 684.4 kWh. Cost = 684.4 × 28 = 19 163p = £191.63.
The fridge has by far the highest annual cost at £367.92 because it runs 24 hours a day, even though its power is relatively low. This demonstrates that E = Pt — time matters as much as power. The greatest potential for energy savings lies in replacing the fridge with a more efficient model (A+++ rated), as even a small improvement in efficiency over 24 hours of daily use would save significant energy over a year. Reducing tumble dryer use (e.g. air-drying clothes) would also save significant amounts.
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