Electricity

Electricity

:::info Board Coverage AQA Paper 1 & 2 | Edexcel Paper 1 & 2 | OCR A Gateway P4 | WJEC P4 :::

Explore the simulation above to develop intuition for this topic.

1. Electrical Circuits

1.1 Circuit Symbols

Symbol	Component
Long thin line, short thick line	Cell
Two pairs of lines	Battery
Single horizontal line	Wire
Vertical line across wire	Switch (open/closed)
Zigzag	Resistor
Rectangle	Variable resistor
Circle with X	Lamp
Circle with A	Ammeter
Circle with V	Voltmeter
Two parallel lines (different lengths)	Capacitor
Circle with M	Motor
Three horizontal lines (decreasing)	Thermistor
Circle with arrow (diode)	Diode
Circle with arrow and bar	LED

1.2 Current, Potential Difference, and Resistance

Current ($I$) is the rate of flow of charge. It is measured in amperes (amps, A).

$I = \frac{Q}{t}$

Where $Q$ is charge in coulombs (C) and $t$ is time in seconds (s).

Potential difference ($V$) is the energy transferred per unit charge. It is measured in volts (V).

$V = \frac{W}{Q}$

Where $W$ is energy transferred in joules (J).

Resistance ($R$) is a measure of how much a component opposes the flow of current. It is Measured in ohms ($\Omega$).

1.3 What Current Actually Is

In a metal wire, current is the flow of free (delocalised) electrons. The number of electrons Passing a point per second is enormous: even a current of 1 A corresponds to about $6.25 \times 10^{18}$ electrons per second. The drift velocity of these electrons is surprisingly Slow — about 0.1 mm/s in a copper wire carrying 1 A. The signal (the electric field that Drives the electrons) propagates at nearly the speed of light, which is why a light turns on Essentially instantly when you flip the switch, even though the individual electrons move slowly.

1.4 Ohm’s Law

Ohm’s law: For a conductor at constant temperature, the current is directly proportional to the Potential difference.

$V = IR$

Worked Example. A 12 V battery is connected to a resistor of 4 $\Omega$. Calculate the current.

I = \frac{V}{R} = \frac{12}{4} = 3 \mathrm{ A

Worked Example. A lamp has a current of 0.5 A and a resistance of 60 $\Omega$. Find the Potential difference across it.

V = IR = 0.5 \times 60 = 30 \mathrm{ V

1.5 Why Ohm’s Law Is Not a Universal Law

Ohm’s law applies only to ohmic conductors ( metals at constant temperature). Many Components do not obey Ohm’s law: a filament lamp has increasing resistance with current (because The filament heats up), a diode only conducts in one direction, and a thermistor changes resistance With temperature. The term “law” is historical; it is better thought of as a property of certain Materials .

2. Series and Parallel Circuits

2.1 Series Circuits

Components are connected one after another in a single loop.

Rules:

• Current is the same through all components: $I_1 = I_2 = I_3 = \ldots$
• Total potential difference is the sum of the p.d. Across each component: V_{\mathrm{total} = V_1 + V_2 + V_3 + \ldots
• Total resistance is the sum of individual resistances: R_{\mathrm{total} = R_1 + R_2 + R_3 + \ldots
• If one component fails, the whole circuit breaks

2.2 Parallel Circuits

Components are connected in separate branches.

Rules:

• Total current is the sum of the currents in each branch: I_{\mathrm{total} = I_1 + I_2 + I_3 + \ldots
• Potential difference is the same across each branch: $V_1 = V_2 = V_3 = \ldots$
• \frac{1}{R_{\mathrm{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots
• If one branch fails, other branches continue to work

Worked Example. Two resistors of 6 $\Omega$ and 12 $\Omega$ are connected in parallel. Find the Total resistance.

\frac{1}{R_{\mathrm{total}} = \frac{1}{6} + \frac{1}{12} = \frac{2 + 1}{12} = \frac{3}{12} = \frac{1}{4} R_{\mathrm{total} = 4 \mathrm{ \Omega

2.3 Why Parallel Resistance Is Always Less Than the Smallest Individual Resistance

Adding a parallel branch provides an additional path for current to flow. More paths means less Total opposition to current, so the total resistance decreases. In the extreme case, adding a Parallel branch with zero resistance (a short circuit) makes the total resistance zero, and all Current flows through the short.

2.4 Combined Series and Parallel

Worked Example. A 3 $\Omega$ resistor is in series with a parallel combination of 6 $\Omega$ and 12 $\Omega$ resistors. Find the total resistance.

Parallel combination: R_p = 4 \mathrm{ \Omega (from above).

Total: R_{\mathrm{total} = 3 + 4 = 7 \mathrm{ \Omega.

2.5 Why Adding Resistors in Series Increases Total Resistance

Each additional series resistor adds more opposition to the flow of current. The electrons must pass Through every resistor in turn, so the total resistance is the sum of all individual resistances. This is analogous to adding length to a pipe: a longer pipe offers more resistance to fluid flow.

3. Circuit Components

3.1 I-V Characteristics

Ohmic conductor (e.g. A wire at constant temperature): The I-V graph is a straight line through The origin. Current is directly proportional to p.d.

Filament lamp: The I-V graph is a curve. As current increases, the filament heats up, increasing Resistance. This is a non-ohmic component.

Diode: Current flows in one direction only. In the forward direction, current is zero until the Threshold p.d. (about 0.6 V for a silicon diode) is reached.

3.2 Why the Filament Lamp Curve Bends

As current flows through the filament, the power dissipated ($P = I^2R$) heats the filament. The Resistance of a metal increases with temperature (because the ions vibrate more vigorously and Scatter the conduction electrons more frequently). So as the current increases, the temperature Rises, the resistance rises, and the graph of $I$ against $V$ curves away from the straight line. At Higher voltages, the same increase in voltage produces a smaller increase in current than at lower Voltages.

3.3 Thermistors and LDRs

Thermistor (NTC): Resistance decreases as temperature increases.

$R \propto \frac{1}{T}$

Applications: Thermostats, temperature sensors, engine temperature monitors.

Light Dependent Resistor (LDR): Resistance decreases as light intensity increases.

Applications: Automatic night lights, burglar alarms, light meters.

3.4 Why LDRs and Thermistors Are Useful

These components convert a non-electrical quantity (light intensity, temperature) into an electrical Quantity (resistance). This allows a circuit to respond to changes in the environment. A potential Divider containing an LDR, for example, produces an output voltage that varies with light level, Which can be used to switch a light on automatically at dusk.

3.5 Required Practical: I-V Characteristics

Method:

1. Set up the circuit with the component, ammeter (in series), and voltmeter (in parallel).
2. Use a variable resistor to change the p.d.
3. Record pairs of p.d. And current values.
4. Plot a graph of current (y-axis) against p.d. (x-axis).
5. Repeat for different components.

4. Electrical Energy and Power

4.1 Energy Transferred

$W = VIt = QV = I^2Rt = \frac{V^2}{R}t$

4.2 Power

$P = IV = I^2R = \frac{V^2}{R}$

Worked Example. A 240 V kettle has a power rating of 2.4 kW. Calculate the current it draws.

P = IV \implies I = \frac{P}{V} = \frac{2400}{240} = 10 \mathrm{ A

Worked Example. A 12 $\Omega$ resistor has a current of 2 A flowing through it. Calculate the Power dissipated.

P = I^2R = 4 \times 12 = 48 \mathrm{ W

4.3 Why $P = I^2R$ and $P = V^2/R$ Give the Same Answer

Starting from $P = IV$ and substituting $V = IR$ gives $P = I(IR) = I^2R$. Substituting $I = V/R$ Gives $P = (V/R)V = V^2/R$. These are not different formulas; they are the same formula expressed in Terms of different pairs of variables. Use $P = I^2R$ when you know the current and resistance; use $P = V^2/R$ when you know the voltage and resistance; use $P = IV$ when you know the current and Voltage.

4.4 Domestic Electricity

In the UK, mains electricity is supplied at approximately 230 V (AC, 50 Hz).

The national grid is a network of cables and transformers that distributes electricity from Power stations to consumers.

• Step-up transformers increase the voltage at power stations (reducing current, which reduces energy losses in cables)
• Step-down transformers decrease the voltage for safe domestic use

Power rating: Every electrical appliance has a power rating (in watts or kilowatts) that tells You the rate at which it transfers energy.

Worked Example. A 2 kW heater is used for 3 hours. Calculate the energy transferred and the cost At 15 p per kWh.

\mathrm{Energy = 2 \times 3 = 6 \mathrm{ kWh \mathrm{Cost = 6 \times 15 = 90 \mathrm{ pence

4.5 Why the National Grid Uses High Voltages

Power dissipated in transmission cables is $P = I^2R$. For a given cable resistance $R$The power Loss is proportional to $I^2$. Since P_{\mathrm{transmitted} = VIIncreasing $V$ allows $I$ to Decrease for the same transmitted power. If the voltage is doubled, the current is halved, and the Power loss is reduced to one-quarter. This is the entire reason for the national grid’s Step-up/step-down transformer system.

Worked Example. A power station transmits 500 kW through cables of resistance 2 $\Omega$.

At 10,000 V: $I = 500000/10000 = 50$ A. Power loss = $50^2 \times 2 = 5000$ W.

At 250,000 V: $I = 500000/250000 = 2$ A. Power loss = $2^2 \times 2 = 8$ W.

The higher voltage reduces losses by a factor of 625.

5. Mains Electricity and Safety

5.1 AC and DC

Direct current (DC): Current flows in one direction only. Supplied by batteries and cells.

Alternating current (AC): Current changes direction periodically. The UK mains supply is AC with A frequency of 50 Hz and a peak voltage of about 325 V (giving an RMS voltage of approximately 230 V).

5.2 RMS Voltage and Peak Voltage

The root-mean-square (RMS) voltage of an AC supply is the DC voltage that would deliver the same Average power to a resistor.

V_{\mathrm{rms} = \frac{V_{\mathrm{peak}}{\sqrt{2}}

For the UK mains: V_{\mathrm{peak} = 230 \times \sqrt{2} \approx 325 V. When we say the mains Voltage is 230 V, we mean the RMS voltage. The peak voltage (the maximum instantaneous voltage) is 325 V. This matters for insulation ratings: insulation must withstand the peak voltage, not the RMS Voltage.

5.3 Cables and Plugs

A standard UK three-pin plug has:

• Live wire (brown): Carries the alternating potential difference from the supply
• Neutral wire (blue): Completes the circuit at 0 V
• Earth wire (green and yellow): Safety wire that carries current to earth if a fault occurs

5.4 Why the Earth Wire Is a Safety Feature

If a metal-cased appliance develops a fault (for example, a live wire touches the metal casing), the Earth wire provides a low-resistance path to earth. A large current flows, which blows the fuse or Trips the circuit breaker, disconnecting the appliance before anyone touches the live casing. Without the earth wire, the casing would remain live, and touching it could be fatal.

5.5 Fuses and Circuit Breakers

Fuse: A thin wire that melts and breaks the circuit if the current exceeds a certain value. Always choose a fuse with a rating slightly higher than the normal operating current.

Circuit breaker: An electromagnetic switch that trips when the current is too high. Can be Reset, unlike a fuse.

Residual Current Device (RCD): Compares the current in the live and neutral wires. If they Differ (current is leaking to earth, perhaps through a person), the RCD trips in milliseconds, far Faster than a fuse. Required in modern installations for circuits supplying sockets that may be used Outdoors.

5.6 Electrical Hazards

Hazard	Cause	Prevention
Electric shock	Contact with live wire	Insulation, earthing, double insulation
Fire	Overheating due to too much current	Fuses, circuit breakers, correct wire thickness
Electrocution	Wet conditions + electricity	Keep appliances away from water

6. Static Electricity

6.1 Electric Charge

When certain materials are rubbed together, electrons are transferred from one material to the Other:

• The material that gains electrons becomes negatively charged
• The material that loses electrons becomes positively charged

Only electrons move; protons remain fixed in the nucleus.

6.2 Electric Fields

A charged object creates an electric field around it. This field exerts a force on other Charged objects:

• Like charges repel
• Unlike charges attract

Electric field lines:

• Point away from positive charges
• Point towards negative charges
• Never cross
• Closer lines = stronger field

6.3 Sparks and Lightning

When the electric field between two charged objects is strong enough, electrons jump across the gap, Creating a spark. Lightning is a large-scale example of this phenomenon.

Worked Example. Explain how a photocopier uses static electricity.

The drum inside the photocopier is given a charge. Light is shone onto the document, and the Reflected light discharges specific areas of the drum. Toner (charged powder) sticks only to the Discharged areas. The toner is then transferred to paper and heated to fix it in place.

6.4 Electrostatic Precipitators

Industrial electrostatic precipitators remove particulate matter from exhaust gases. The gas passes Through a grid of wires maintained at a high voltage. The wires ionise the air, charging the dust Particles. The charged particles are then attracted to oppositely charged plates, where they collect And can be removed. This is how power stations reduce particulate pollution.

7. Internal Resistance (Higher Tier)

7.1 Concept

A real battery has internal resistance $r$Which causes energy to be dissipated inside the Battery itself when current flows. The terminal p.d. Is less than the EMF:

$V = \varepsilon - Ir$

Where $\varepsilon$ is the EMF (the total energy per unit charge supplied by the battery).

7.2 Why Internal Resistance Matters

As the current drawn from a battery increases, the voltage lost across the internal resistance ($Ir$) increases, and the terminal voltage decreases. A car battery has very low internal resistance (about 0.01 $\Omega$) so it can deliver the large current needed to start the engine without the Voltage dropping too much. A school power supply has higher internal resistance to limit the maximum Current and protect the components.

Worked Example. A battery of EMF 9 V and internal resistance 0.5 $\Omega$ is connected to a 4 $\Omega$ resistor. Find the current and terminal voltage.

I = \frac{\varepsilon}{R + r} = \frac{9}{4 + 0.5} = \frac{9}{4.5} = 2 \mathrm{ A

V = \varepsilon - Ir = 9 - 2 \times 0.5 = 8 \mathrm{ V

Worked Examples

Example 1:

A typical exam question on Electricity requires you to apply your knowledge to an unfamiliar context. Read the question carefully, identify the key concept being tested, and structure your answer using the appropriate terminology.

Example 2:

Multi-step problems in Electricity often combine two or more concepts. Break the problem down: identify what you need to find, recall the relevant formula or principle, substitute values, and state your answer with correct units or formatting.

Common Pitfalls

• Confusing series and parallel circuit rules. Current adds in parallel; resistance adds in series.
• Placing ammeters in parallel or voltmeters in series. Ammeters must be in series (low resistance); voltmeters must be in parallel (high resistance).
• Forgetting that $P = I^2R$ and $P = V^2/R$ only apply to resistive components.
• Using the wrong fuse rating. The fuse should be slightly above the normal operating current, not much higher.
• Confusing the live and neutral wires. The live wire is at the supply voltage; the neutral wire is at approximately 0 V.
• Forgetting to convert units (kW to W, hours to seconds) in energy calculations.
• Assuming the total resistance in parallel is the sum of individual resistances. It is the reciprocal of the sum of reciprocals.
• Confusing EMF and terminal p.d. EMF is the total energy per unit charge supplied; terminal p.d. Is what is available to the external circuit (EMF minus internal losses).
• Measuring current with a voltmeter or voltage with an ammeter. Ammeters measure current and are placed in series; voltmeters measure potential difference and are placed in parallel.

Practice Questions

1. A circuit has a 9 V battery connected to two resistors of 3 $\Omega$ and 6 $\Omega$ in series. Calculate the total resistance, the current, and the p.d. Across each resistor.

2. Three resistors of 4 $\Omega$6 $\Omega$And 12 $\Omega$ are connected in parallel to a 12 V supply. Calculate the current through each resistor and the total current.

3. A 60 W bulb is connected to the 230 V mains. Calculate the current it draws and suggest an appropriate fuse rating.

4. Explain why a filament lamp is a non-ohmic conductor, referring to its I-V characteristic.

5. An electric heater has a power rating of 1.5 kW. Calculate the energy transferred when it is used for 45 minutes, giving your answer in kWh and in joules.

6. Describe the function of each of the three wires in a UK mains plug.

7. A thermistor has a resistance of 200 $\Omega$ at 20^{\circ}\mathrm{C and 50 $\Omega$ at 80^{\circ}\mathrm{C. It is connected in series with a 100 $\Omega$ fixed resistor to a 12 V supply. Calculate the current at each temperature.

8. Explain why the national grid uses high voltages for transmission.

9. Calculate the cost of running a 200 W television for 6 hours per day for 30 days, at a rate of 20 p per kWh.

10. Describe how static electricity is used in a spray paint gun, explaining the role of electric fields.

11. A battery of EMF 12 V and internal resistance 1.5 $\Omega$ is connected to an external circuit. The terminal p.d. Is 10.5 V. Calculate the current and the external resistance.

12. A 3 $\Omega$ resistor and a 6 $\Omega$ resistor are connected in parallel, and this combination is connected in series with a 2 $\Omega$ resistor to a 12 V supply. Calculate the current through each resistor and the power dissipated in each.

13. Explain the difference between a fuse and a circuit breaker. State one advantage of each.

14. A hairdryer rated at 2000 W is connected to the 230 V mains. Calculate the current it draws. Explain why it is important not to use an extension lead rated for 5 A with this appliance.

15. Two identical resistors of resistance $R$ are connected first in series and then in parallel to the same battery. Show that the ratio of the power dissipated in the parallel arrangement to the power dissipated in the series arrangement is 4:1.

8. Worked Example: Combined Series-Parallel Circuit with Power

A 12 \mathrm{ V battery (internal resistance 0.5 \mathrm{ \Omega) is connected to a 3 \mathrm{ \Omega resistor in series with a parallel combination of 6 \mathrm{ \Omega and 12 \mathrm{ \Omega resistors.

Step 1: Find the parallel resistance.

\frac{1}{R_p} = \frac{1}{6} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \implies R_p = 4 \mathrm{ \Omega

Step 2: Total resistance (including internal resistance).

R_{\mathrm{total} = R_{\mathrm{internal} + R_{\mathrm{series} + R_p = 0.5 + 3 + 4 = 7.5 \mathrm{ \Omega

Step 3: Total current.

I = \frac{\varepsilon}{R_{\mathrm{total}} = \frac{12}{7.5} = 1.6 \mathrm{ A

Step 4: Terminal PD.

V = \varepsilon - Ir = 12 - 1.6 \times 0.5 = 11.2 \mathrm{ V

Step 5: Current through each parallel resistor.

I_6 = \frac{V_p}{6} = \frac{11.2 - 1.6 \times 3}{6} = \frac{6.4}{6} = 1.067 \mathrm{ A

I_{12} = \frac{V_p}{12} = \frac{6.4}{12} = 0.533 \mathrm{ A

Check: 1.067 + 0.533 = 1.6 \mathrm{ A. Correct.

Step 6: Power dissipated in each resistor.

P_3 = I^2 R_3 = 1.6^2 \times 3 = 7.68 \mathrm{ W

P_6 = I_6^2 \times 6 = 1.067^2 \times 6 = 6.83 \mathrm{ W

P_{12} = I_{12}^2 \times 12 = 0.533^2 \times 12 = 3.41 \mathrm{ W

9. Worked Example: Potential Divider with a Thermistor

A potential divider circuit consists of a 10 \mathrm{ k\Omega fixed resistor and an NTC Thermistor in series with a 12 \mathrm{ V supply. At 20^{\circ}\mathrm{CThe thermistor Resistance is 10 \mathrm{ k\Omega. At 80^{\circ}\mathrm{CThe thermistor resistance is 1 \mathrm{ k\Omega. The output voltage is taken across the thermistor.

At 20^{\circ}\mathrm{C:

V_{\mathrm{out} = 12 \times \frac{10000}{10000 + 10000} = 12 \times \frac{1}{2} = 6 \mathrm{ V

At 80^{\circ}\mathrm{C:

V_{\mathrm{out} = 12 \times \frac{1000}{10000 + 1000} = 12 \times \frac{1}{11} = 1.09 \mathrm{ V

As temperature increases, the thermistor resistance decreases, and the output voltage decreases. This circuit could be used to trigger a heater when the temperature drops (the output voltage rises When it is cold).

10. Why Current Is the Same Everywhere in a Series Circuit

Charge is conserved: charge cannot accumulate at any point in a circuit. If the current were Different at different points in a series circuit, charge would build up at the junction where the Current changes, which is impossible in a steady-state DC circuit. This is analogous to water Flowing through a single pipe: the same amount of water passes through every cross-section per Second. The same argument, applied to junctions in parallel circuits, gives Kirchhoff’s first law: The total current entering a junction equals the total current leaving it.

11. Derivation: Power Formulas from Ohm’s Law

Starting from $P = IV$:

• Substitute $V = IR$: $P = I \times IR = I^2 R$
• Substitute $I = V/R$: $P = \frac{V}{R} \times V = \frac{V^2}{R}$

These are all equivalent expressions. Use $P = IV$ when you know both current and voltage. Use $P = I^2R$ when you know current and resistance (useful for finding power dissipated in a specific Resistor in a circuit). Use $P = V^2/R$ when you know voltage and resistance.

12. Why the National Grid Uses High Voltages: Quantitative Analysis

Consider transmitting power $P$ through cables of total resistance $R$.

At voltage $V$The current is $I = P/V$And the power lost in the cables is:

P_{\mathrm{loss} = I^2 R = \frac{P^2 R}{V^2}

This shows that power loss is inversely proportional to $V^2$. If the transmission voltage is Increased by a factor of 10, the power loss decreases by a factor of 100. If a power station Generates 500 \mathrm{ MW and the cable resistance is 5 \mathrm{ \Omega:

• At 25 \mathrm{ kV: I = 20000 \mathrm{ A P_{\mathrm{loss} = 20000^2 \times 5 = 2 \times 10^9 \mathrm{ W = 2000 MW (four times the generated power — absurd)
• At 400 \mathrm{ kV: I = 1250 \mathrm{ A P_{\mathrm{loss} = 1250^2 \times 5 = 7.8 \times 10^6 \mathrm{ W = 7.8 MW (1.6% of generated power — feasible)

13. Worked Example: Cost of Electrical Appliances

A household has the following daily usage:

• Fridge: 150 W for 24 hours
• TV: 120 W for 5 hours
• Washing machine: 2000 W for 1 hour
• Lights (5 bulbs): 60 W each for 6 hours
• Kettle: 2000 W for 0.3 hours

Calculate the daily energy consumption and the monthly cost at 20 p/kWh.

E_{\mathrm{fridge} = 0.15 \times 24 = 3.6 \mathrm{ kWh E_{\mathrm{TV} = 0.12 \times 5 = 0.6 \mathrm{ kWh E_{\mathrm{washing} = 2.0 \times 1 = 2.0 \mathrm{ kWh E_{\mathrm{lights} = 5 \times 0.06 \times 6 = 1.8 \mathrm{ kWh E_{\mathrm{kettle} = 2.0 \times 0.3 = 0.6 \mathrm{ kWh

E_{\mathrm{total} = 3.6 + 0.6 + 2.0 + 1.8 + 0.6 = 8.6 \mathrm{ kWh per day

Monthly cost (30 days): 8.6 \times 30 \times 20 = 5160 \mathrm{ pence = \pounds 51.60

The fridge is the largest consumer despite its low power rating, because it runs continuously. This Illustrates the importance of considering both power and usage time.

14. Worked Example: I-V Characteristic of a Filament Lamp

A filament lamp is connected to a variable power supply. The following measurements are recorded:

PD (V)	Current (A)	Resistance ($\Omega$)
2.0	0.20	10.0
4.0	0.35	11.4
6.0	0.45	13.3
8.0	0.52	15.4
10.0	0.57	17.5
12.0	0.60	20.0

The resistance increases from 10 $\Omega$ at 2 V to 20 $\Omega$ at 12 V. This confirms that the Filament lamp is non-ohmic: the resistance is not constant. The I-V graph curves away from the Straight line because the filament heats up as the current increases.

15. Summary Table: Series vs Parallel Circuits

Property	Series	Parallel
Current	Same through all components	Splits at junctions; adds up
Voltage	Splits across components; adds up	Same across all branches
Resistance	R_{\mathrm{total} = R_1 + R_2 + \cdots	1/R_{\mathrm{total} = 1/R_1 + 1/R_2 + \cdots
If one component fails	Whole circuit breaks	Other branches continue working
Adding more resistors	Increases R_{\mathrm{total}	Decreases R_{\mathrm{total}
Example	Old Christmas tree lights	Household wiring

16. Safety Features Summary

Feature	How It Works	What It Protects
Fuse	Thin wire melts at a set current	Prevents fire from overheating
Circuit breaker	Electromagnetic switch trips	Prevents fire; reusable
Earth wire	Low-resistance path to ground	Prevents electric shock from faulty casing
Double insulation	Plastic casing, no earth needed	Prevents shock from internal faults
RCD	Compares live and neutral current	Detects current leakage to earth; very fast
Insulation	Non-conductive coating on wires	Prevents contact with live conductors

17. Practice Questions (Additional)

16. A circuit contains a 9 \mathrm{ V battery, a 100 \mathrm{ \Omega resistor in series with a parallel combination of 200 \mathrm{ \Omega and 300 \mathrm{ \Omega resistors. Calculate the current through each resistor and the power dissipated in the 200 \mathrm{ \Omega resistor.

17. A student connects an ammeter in parallel with a resistor. Explain why this is dangerous and what will happen.

18. A 6 \mathrm{ V battery with internal resistance 0.3 \mathrm{ \Omega is connected to an external circuit. The terminal PD is 5.4 \mathrm{ V. Calculate the current and the external resistance.

19. Design a potential divider circuit using a 9 \mathrm{ V battery and two resistors that produces an output voltage of 3 \mathrm{ V. Specify the values of both resistors.

20. Explain the difference between a thermistor and an LDR. Describe a sensing circuit for each and explain how the output voltage changes as the sensed quantity changes.

21. A hairdryer rated at 1800 \mathrm{ W is connected to the 230 \mathrm{ V mains. Calculate the current it draws. If the mains cable has a resistance of 0.5 \mathrm{ \OmegaCalculate the power lost in the cable and the voltage reaching the hairdryer.

22. Three identical resistors, each of resistance $R$Are connected to a battery. Calculate the total resistance when they are connected (a) all in series, (b) all in parallel, and (c) two in parallel with the third in series.

23. Explain why the resistance of a filament lamp increases with temperature, while the resistance of a thermistor (NTC) decreases with temperature.

24. A mobile phone charger has an output of 5 \mathrm{ V and 2 \mathrm{ A. If the phone battery has a capacity of 3000 \mathrm{ mAhHow long does it take to charge from empty? Calculate the energy transferred to the battery during a full charge.

25. A circuit contains a variable resistor, a fixed resistor of 470 \mathrm{ \OmegaAnd an LED in series with a 9 \mathrm{ V battery. The LED requires a minimum current of 10 \mathrm{ mA to light and has a forward voltage drop of 2 \mathrm{ V. Calculate the range of resistance values for the variable resistor that keeps the LED lit without exceeding a maximum current of 20 \mathrm{ mA.

Extended Worked Examples

Example 26: Energy and Cost of Household Appliances

A household uses the following appliances daily: a 2 \mathrm{ kW heater for 4 hours, six 10 \mathrm{ W LED lights for 6 hours, a 200 \mathrm{ W TV for 3 hours, and a 3 \mathrm{ kW Kettle for 0.5 \mathrm{ hours. Electricity costs 28 \mathrm{p per kWh. Calculate the daily and Monthly (30-day) cost.

Step 1: Energy used by each appliance

• Heater: E = Pt = 2 \times 4 = 8 \mathrm{ kWh
• Lights: E = 0.01 \times 6 \times 6 = 0.36 \mathrm{ kWh
• TV: E = 0.2 \times 3 = 0.6 \mathrm{ kWh
• Kettle: E = 3 \times 0.5 = 1.5 \mathrm{ kWh

Step 2: Total daily energy

E_{\mathrm{total} = 8 + 0.36 + 0.6 + 1.5 = 10.46 \mathrm{ kWh

Step 3: Daily and monthly cost

\mathrm{Daily cost = 10.46 \times 28 = 292.9 \mathrm{ p = \pounds 2.93

\mathrm{Monthly cost = 292.9 \times 30 = 8787 \mathrm{ p = \pounds 87.87

:::info Info Heating appliances dominate household electricity bills. Switching to a more efficient heater or Improving insulation can significantly reduce costs. :::

Example 27: Parallel Circuit with Multiple Branches

Three resistors of $100 \Omega$, $200 \Omega$And $300 \Omega$ are connected in parallel across a 12 \mathrm{ V battery. Calculate the current through each resistor, the total current, and the Total resistance.

Step 1: Current through each resistor (Ohm’s law)

I_1 = \frac{V}{R_1} = \frac{12}{100} = 0.120 \mathrm{ A = 120 \mathrm{ mA

I_2 = \frac{V}{R_2} = \frac{12}{200} = 0.060 \mathrm{ A = 60 \mathrm{ mA

I_3 = \frac{V}{R_3} = \frac{12}{300} = 0.040 \mathrm{ A = 40 \mathrm{ mA

Step 2: Total current

I_{\mathrm{total} = I_1 + I_2 + I_3 = 120 + 60 + 40 = 220 \mathrm{ mA

Step 3: Total resistance

R_{\mathrm{total} = \frac{V}{I_{\mathrm{total}} = \frac{12}{0.220} = 54.5 \Omega

Check using the reciprocal formula:

$\frac{1}{R} = \frac{1}{100} + \frac{1}{200} + \frac{1}{300} = 0.01 + 0.005 + 0.00333 = 0.01833$

$R = \frac{1}{0.01833} = 54.5 \Omega$

Confirmed.

Example 28: Internal Resistance and Maximum Current

A battery has EMF 9 \mathrm{ V and internal resistance $1.5 \Omega$. What is the maximum current That can be drawn, and what is the terminal PD at this current?

Step 1: Maximum current (short circuit)

I_{\max} = \frac{\mathcal{E}}{r} = \frac{9}{1.5} = 6 \mathrm{ A

This occurs when the external resistance is zero (short circuit).

Step 2: Terminal PD at maximum current

V = \mathcal{E} - Ir = 9 - 6 \times 1.5 = 9 - 9 = 0 \mathrm{ V

All the EMF is “lost” across the internal resistance. The battery delivers maximum current but zero Useful voltage to the external circuit.

Step 3: Power dissipated in the battery

P_{\mathrm{internal} = I^2 r = 36 \times 1.5 = 54 \mathrm{ W

This energy is converted to heat inside the battery, which can cause it to overheat or even explode.

:::caution Warning The energy is dissipated as heat inside the battery. Never short-circuit a battery. :::

Common Pitfalls Extended

Pitfall 6: Adding Currents in Series Circuits

In a series circuit, the current is the same through every component. Do not add currents in Series. Only add currents when they meet at a junction (Kirchhoff’s first law).

Pitfall 7: Forgetting to Subtract the LED/Component Voltage

When calculating the required series resistor for an LED, the resistor voltage is the supply voltage minus the LED forward voltage: V_R = V_{\mathrm{supply} - V_{\mathrm{LED}. Using just the Supply voltage gives an incorrectly high resistance and the LED will not light.

Pitfall 8: Confusing AC and DC in Power Calculations

For DC: $P = IV = I^2R = V^2/R$ using the steady values. For AC: the same formulas apply but using RMS values. The peak power is twice the average power for a sinusoidal AC supply: P_{\mathrm{peak} = 2P_{\mathrm{avg}.

Additional Practice Problems

26. A 12 \mathrm{ V car battery has internal resistance $0.05 \Omega$. The starter motor draws 200 \mathrm{ A. Calculate (a) the terminal PD, (b) the power delivered to the starter, and (c) the power wasted in the battery.

27. Three identical resistors each of $R$ are connected to a battery. Calculate the total resistance for all possible arrangements and rank them from largest to smallest.

28. An electric shower has a power rating of 8.5 \mathrm{ kW and operates at 230 \mathrm{ V. Calculate the current it draws and explain why it requires a dedicated high-current circuit.

29. A student sets up a circuit with a thermistor and a fixed resistor in series with a 6 \mathrm{ V battery. The thermistor has resistance $2000 \Omega$ at 20°\mathrm{C and $500 \Omega$ at 60°\mathrm{C. Calculate the voltage across the thermistor at each temperature.

30. Explain the difference between conventional current and electron flow. Why do we still use conventional current in circuit diagrams?

Practice Problems

Question 1: Ohm’s Law calculation

A resistor has a resistance of $50 \Omega$. When a 12 \mathrm{ V battery is connected across it, Calculate (a) the current through the resistor and (b) the power dissipated.

Answer

(a) $V = IR$So I = V/R = 12/50 = 0.24 \mathrm{ A.

(b) P = VI = 12 \times 0.24 = 2.88 \mathrm{ W.

Question 2: Series and parallel circuits

Two $6 \Omega$ resistors are connected in parallel, and this combination is connected in series with A $4 \Omega$ resistor to a 12 \mathrm{ V supply. Calculate the total resistance, the total Current, and the potential difference across each component.

Answer

Parallel combination: $1/R_p = 1/6 + 1/6 = 2/6 = 1/3$So $R_p = 3 \Omega$.

Total resistance: R_{\mathrm{total} = 4 + 3 = 7 \Omega.

Total current: I = V/R_{\mathrm{total} = 12/7 = 1.71 \mathrm{ A.

PD across the $4 \Omega$ resistor: V_4 = IR = 1.71 \times 4 = 6.86 \mathrm{ V.

PD across the parallel combination: V_p = 12 - 6.86 = 5.14 \mathrm{ V.

Question 3: Cost of electricity

A 2 \mathrm{ kW heater is used for 5 hours per day for 30 days. If electricity costs 15 \mathrm{ p/kWhCalculate the total cost.

Answer

Energy used = P \times t = 2 \times 5 \times 30 = 300 \mathrm{ kWh.

Cost = 300 \times 15 = 4500 \mathrm{ p = \pounds 45.00.

Question 4: Current, charge, and time

A torch bulb passes a current of 0.3 \mathrm{ A for 2 minutes. Calculate the charge that flows Through the bulb.

Answer

Q = It = 0.3 \times 120 = 36 \mathrm{ C.

Question 5: Potential difference and energy transfer

A $10 \Omega$ resistor in a circuit has a current of 0.5 \mathrm{ A flowing through it. Calculate The energy transferred by the resistor in 30 seconds.

Answer

E = VIt = IR \times I \times t = I^2 Rt = 0.5^2 \times 10 \times 30 = 0.25 \times 10 \times 30 = 75 \mathrm{ J.

Summary

This topic covers the key concepts of Electricity for GCSE Physics. Focus on understanding the fundamental principles, practising with exam-style questions, and applying your knowledge to unfamiliar contexts.