Electrical energy and electrical power

1. Overview

This topic explores how electrical circuits serve as a system for transferring energy from a power source (like a battery or the mains) to various components where it is converted into useful forms, such as light or heat. Understanding these calculations allows us to determine the efficiency of appliances and calculate the financial cost of the electricity we consume at home.

Key Definitions

• Electrical Power ($P$): The rate at which electrical energy is transferred by a circuit or a component.
• Electrical Energy ($E$): The total amount of work done by an electric current over a period of time.
• Watt (W): The SI unit of power, equivalent to one Joule per second ($1\text{ J/s}$).
• Kilowatt-hour (kWh): A unit of energy representing the energy used by a $1\text{ kW}$ appliance running for $1\text{ hour}$.
• Mains Supply: The general-purpose alternating current (AC) electric power supply provided to homes and businesses.

Core Content

Energy Transfer in Circuits

Electric circuits are energy transfer systems.

1. The Source: An electrical cell, battery, or the mains supply provides the energy.
2. The Transfer: Electrons flow through the circuit, carrying this energy.
3. The Components: Components like lamps (light), heaters (thermal), or motors (kinetic) convert the electrical energy into other forms.
4. The Surroundings: Eventually, most of this energy is dissipated into the surroundings, usually as heat.

Electrical Power

Power is calculated using the current and potential difference (voltage) across a component.

• Formula: $P = I \times V$
• 📊A simple circuit containing a battery, an ammeter in series, and a voltmeter in parallel across a bulb to show how $I$ and $V$ are measured.

Worked Example: A toaster is connected to a $230\text{V}$ mains supply. If the current flowing through it is $4\text{A}$, what is the power?

• $P = I \times V$
• $P = 4\text{A} \times 230\text{V} = 920\text{W}$

Electrical Energy

Energy is power multiplied by time. In standard SI units, time must be in seconds.

• Formula: $E = P \times t$ (or $E = I \times V \times t$)

Worked Example: How much energy is transferred by a $60\text{W}$ bulb left on for $5$ minutes?

• Convert time to seconds: $5 \times 60 = 300\text{s}$
• $E = P \times t = 60\text{W} \times 300\text{s} = 18,000\text{J}$ (or $18\text{kJ}$)

The Kilowatt-hour (kWh) and Cost

Electricity companies measure energy in kilowatt-hours because Joules are too small for household bills.

• $1\text{ kWh} = \text{Power (in kW)} \times \text{Time (in hours)}$
• $\text{Total Cost} = \text{Energy used (in kWh)} \times \text{Price per unit}$

Worked Example: An air conditioner rated at $2\text{kW}$ is used for $10$ hours. If $1\text{kWh}$ costs $$0.15$, what is the total cost?

1. $\text{Energy} = 2\text{kW} \times 10\text{h} = 20\text{kWh}$
2. $\text{Cost} = 20\text{kWh} \times $0.15 = $3.00$

Extended Content (Extended Only)

There are no specific supplemental requirements for section 4.2.5; all students should focus on the core objectives listed above.

Key Equations

Word Equation	Symbol Equation	Units
$\text{Power} = \text{Current} \times \text{Voltage}$	$P = I \times V$	Watts (W), Amps (A), Volts (V)
$\text{Energy} = \text{Power} \times \text{Time}$	$E = P \times t$	Joules (J), Watts (W), Seconds (s)
$\text{Energy (kWh)} = \text{Power (kW)} \times \text{Time (h)}$	$E = P \times t$	kWh, Kilowatts (kW), Hours (h)

Common Mistakes to Avoid

• ❌ Wrong: Using minutes or hours when calculating energy in Joules. ✓ Right: Always convert time to seconds ($1\text{ minute} = 60\text{s}$) unless calculating kWh.
• ❌ Wrong: Calculating cost using Power in Watts. ✓ Right: Divide Watts by $1000$ to get Kilowatts (kW) before calculating kWh (e.g., $500\text{W} = 0.5\text{kW}$).
• ❌ Wrong: Stating that a fuse provides a path to earth. ✓ Right: The Earth wire provides a path to earth; the fuse is a safety device that melts and breaks the circuit if current is too high.
• ❌ Wrong: Confusing logic gate symbols in electricity diagrams. ✓ Right: Remember that an AND gate has a flat input side and a D-shaped output; do not confuse it with OR gates.
• ❌ Wrong: Forgetting that standard mains voltage is $230\text{V}$ (in many exam contexts) and using incorrect values.

Exam Tips

1. Check your units first: Before starting a calculation, look at the units provided. If the question asks for Energy in Joules but gives time in minutes, convert to seconds immediately.
2. Show your working: In kWh cost questions, examiners often give marks for the intermediate step of converting Power from Watts to Kilowatts.
3. Identify the relationship: If a question gives you Current and Voltage but asks for Energy, remember you must first find Power ($P=IV$) and then multiply by time ($E=Pt$).

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A small electric heater is used to warm a room. The heater has a power rating of 800 W and is connected to a 200 V mains supply.

(a) Define power. [1]

(b) Calculate the current flowing through the heater. [3]

(c) State what happens to the temperature of the room if the heater is left on for a long time. [1]

Worked Solution:

(a)

1. Power is the rate at which energy is transferred. [Definition of power]

How to earn full marks:

• Correct definition of power, mentioning rate and energy.

(b)

1. $P = VI$ [Recall the power equation]
2. $800 = 200 \times I$ [Substitute the values]
3. $I = \frac{800}{200} = 4 \text{ A}$ [Calculate the current]

How to earn full marks:

• Correct formula stated or implied.
• Correct substitution of values.
• Correct answer with unit. $\boxed{4 \text{ A}}$

(c)

1. The temperature of the room increases. [The heater transfers energy to the room]

How to earn full marks:

• Correctly state that temperature increases.

Common Pitfall: Remember that power is the rate of energy transfer. Some students confuse power with energy itself. Also, always include the correct SI unit in your final answer, which in this case is Amperes (A) for current.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

A student is investigating the power output of a small electric motor. The motor lifts a mass vertically.

(a) State the equation linking gravitational potential energy (GPE), mass (m), gravitational field strength (g), and height (h). [1]

(b) The motor lifts a 0.25 kg mass a height of 1.2 m in 4.0 s. Calculate:

(i) The gain in GPE of the mass. Assume $g = 9.8 \text{ N/kg}$. [2]

(ii) The power output of the motor. [3]

Worked Solution:

(a)

1. $GPE = mgh$ [Recall the GPE equation]

How to earn full marks:

• Correct formula relating GPE, mass, gravity and height.

(b)(i)

1. $GPE = mgh = 0.25 \times 9.8 \times 1.2 = 2.94 \text{ J}$ [Substitute the values into the equation]

How to earn full marks:

• Correct substitution of values.
• Correct answer with unit. $\boxed{2.94 \text{ J}}$

(b)(ii)

1. $Power = \frac{Energy}{Time}$ [Recall the power equation]
2. $Power = \frac{2.94 \text{ J}}{4.0 \text{ s}} = 0.735 \text{ W}$ [Substitute the values and calculate]

How to earn full marks:

• Correct formula stated or implied.
• Correct substitution, using the calculated GPE from part (i).
• Correct answer with unit. $\boxed{0.735 \text{ W}}$

Common Pitfall: Be careful to use the GPE calculated in part (i) for the energy value in part (ii). A common mistake is to try and use voltage or current values directly without calculating the energy first. Also, remember to include the correct units (Joules for GPE and Watts for power).

Exam-Style Question 3 — Extended Response [8 marks]

Question:

A mobile phone charger contains a transformer. The input voltage to the transformer is 230 V AC and the output voltage is 5.0 V AC.

(a) Explain why a transformer is necessary in a mobile phone charger. [2]

(b) The primary coil of the transformer has 1150 turns. Calculate the number of turns in the secondary coil. [3]

(c) The current in the secondary coil is 1.2 A. Assuming the transformer is 100% efficient, calculate the current in the primary coil. [3]

Worked Solution:

(a)

1. The transformer is needed to step down the high mains voltage (230 V) to a lower voltage (5.0 V). [State the function of the transformer]
2. This lower voltage is safer and more suitable for charging the mobile phone battery. [Explain the safety reason]

How to earn full marks:

• Mention stepping down the voltage.
• Explain why the lower voltage is needed (safety/suitability for charging).

(b)

1. $\frac{V_p}{V_s} = \frac{N_p}{N_s}$ [Recall the transformer equation]
2. $\frac{230 \text{ V}}{5.0 \text{ V}} = \frac{1150}{N_s}$ [Substitute the values]
3. $N_s = \frac{5.0 \text{ V} \times 1150}{230 \text{ V}} = 25$ [Calculate the number of turns]

How to earn full marks:

• Correct formula stated or implied.
• Correct substitution of values.
• Correct answer with unit (though "turns" is acceptable). $\boxed{25}$

(c)

1. $V_pI_p = V_sI_s$ [Recall the power equation for a transformer]
2. $230 \text{ V} \times I_p = 5.0 \text{ V} \times 1.2 \text{ A}$ [Substitute the values]
3. $I_p = \frac{5.0 \text{ V} \times 1.2 \text{ A}}{230 \text{ V}} = 0.026 \text{ A}$ (or $26 \text{ mA}$) [Calculate the current]

How to earn full marks:

• Correct formula stated or implied.
• Correct substitution of values.
• Correct answer with unit. $\boxed{0.026 \text{ A}}$

Common Pitfall: When using the transformer equations, ensure you correctly identify primary and secondary voltages and number of turns. Students often mix up the subscripts, leading to incorrect calculations. Also, remember that the number of turns is a ratio and doesn't require a unit, but current always needs its unit (Amperes).

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A student wants to investigate how the brightness of a filament lamp changes with the power supplied to it. The student uses a variable resistor (rheostat) to change the voltage across the lamp.

(a)

Draw a circuit diagram that would allow the student to vary the voltage across the lamp and measure both the voltage and the current through the lamp. [3]

(b) The student records the following data:

Voltage (V)	Current (A)
1.0	0.15
2.0	0.25
3.0	0.33
4.0	0.40

Calculate the electrical power for each voltage setting, and add a column for Power (W) to the table. [3]

(c) Describe how the brightness of the lamp changes as the power increases, and suggest a reason for this change. [3]

Worked Solution:

(a)

1. 📊A circuit diagram showing a power supply, a variable resistor (rheostat), an ammeter, a voltmeter, and a filament lamp connected in series with the ammeter and rheostat, and the voltmeter connected in parallel with the lamp. The rheostat should be drawn using the symbol with an arrow.
   *[Circuit diagram showing the correct arrangement of components]*

How to earn full marks:

• Power supply, lamp, ammeter and rheostat are in series.
• Voltmeter is in parallel with the lamp.
• Correct symbols for all components, including the rheostat.

(b)

1. $P = VI$ [Recall the power equation]
2. For 1.0 V: $P = 1.0 \text{ V} \times 0.15 \text{ A} = 0.15 \text{ W}$ [Calculate power for the first voltage]
3. For 2.0 V: $P = 2.0 \text{ V} \times 0.25 \text{ A} = 0.50 \text{ W}$ [Calculate power for the second voltage]
4. For 3.0 V: $P = 3.0 \text{ V} \times 0.33 \text{ A} = 0.99 \text{ W}$ [Calculate power for the third voltage]
5. For 4.0 V: $P = 4.0 \text{ V} \times 0.40 \text{ A} = 1.60 \text{ W}$ [Calculate power for the fourth voltage]

Voltage (V)	Current (A)	Power (W)
1.0	0.15	0.15
2.0	0.25	0.50
3.0	0.33	0.99
4.0	0.40	1.60

How to earn full marks:

• Correct formula stated or implied.
• Correct calculation of power for all rows.
• Correct units (W) in the table.

(c)

1. As the power increases, the lamp gets brighter. [State the relationship between power and brightness]
2. This is because more electrical energy is being converted into light and heat energy per second. [Explain energy conversion]
3. The filament gets hotter, emitting more light due to increased thermal energy. [Explain the effect on the filament]

How to earn full marks:

• Correctly state that brightness increases with power.
• Explanation of energy conversion (electrical to light/heat).
• Relate the increased power to the filament temperature and light emission.

Common Pitfall: When drawing circuit diagrams, make sure you use the correct symbols for each component. Also, remember that the ammeter is always in series with the component you're measuring the current through, and the voltmeter is always in parallel. When calculating power, double-check your multiplication and always include the unit (Watt).