Cambridge Lower Secondary CheckpointStage 6

Physics: Electricity and magnetism

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Physics: Electricity and magnetism
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1. Electric Charge and Current

Electricity is all about the movement of tiny particles called electrons. These electrons carry a property called electric charge, measured in coulombs (C). When these charges flow through a conductor, like a copper wire, we have an electric current. Think of it like water flowing through a pipe: the water molecules are like the charge, and the flow of water is the current. We define electric current (I) as the rate of flow of electric charge (Q). A current of 1 ampere (A) means that 1 coulomb of charge is flowing past a point every second. Current is measured using an ammeter, which must be connected in series in the circuit.

I = Q / t

Key term

Electric Current: The rate of flow of electric charge, measured in amperes (A).

Common pitfall

A common mistake is forgetting to convert time into seconds when using the formula Q = I × t. Always use standard SI units in calculations.

Fun fact

A typical lightning bolt carries a current of about 30,000 amperes and transfers 15 coulombs of charge.

Worked example 13 marks

A current of 0.4 A flows through a lamp for 2 minutes. Calculate the total charge that flows through the lamp in this time.

  1. 1

    Step 1: State the formula relating charge, current, and time. Q = I × t

  2. 2

    Step 2: Convert the time from minutes to seconds. Time t = 2 minutes = 2 × 60 = 120 s.

  3. 3

    Step 3: Substitute the known values into the formula. Q = 0.4 A × 120 s.

  4. 4

    Step 4: Calculate the final answer with units. Q = 48 C.

Recap

  • Electric charge is a fundamental property of matter, carried by particles like electrons.
  • Electric current is the rate of flow of electric charge.
  • The unit of current is the ampere (A) and the unit of charge is the coulomb (C).
  • The formula linking current, charge, and time is I = Q / t.
  • Ammeters measure current and are always connected in series.

Quick check

  1. What is a current of 2 A equivalent to in coulombs per second?1 mark
  2. An ammeter measures a current of 500 mA. Convert this to amperes.1 mark

2. Voltage, Resistance, and Ohm's Law

For a current to flow, it needs a 'push'. This electrical push is called voltage, or potential difference. It's the energy given to each unit of charge by the battery or power supply. Voltage (V) is measured in volts (V) using a voltmeter, connected in parallel across a component. As charges flow through a component, they encounter opposition. This opposition to the flow of current is called resistance (R), measured in ohms (Ω). The relationship between voltage, current, and resistance for many components is described by Ohm's Law, which states that the current through a conductor is directly proportional to the voltage across it, provided the temperature remains constant. This gives us the most important equation in basic electricity: V = IR.

V = I × R

R = V / I

I = V / R

Key term

Resistance: A measure of the opposition to current flow in an electrical circuit, measured in ohms (Ω).

Examiner insight

Examiners require you to state the formula (e.g., V = IR) before substituting numbers. Full marks are often not awarded if you just write down the final answer, even if it's correct.

Fun fact

The human body's resistance can vary from 1,000 ohms (if wet) to over 100,000 ohms (if dry). This is why wet hands and electricity are a very dangerous combination.

Worked example 14 marks

A resistor is connected to a 12 V power supply and a current of 3.0 A flows through it.(a) Calculate the resistance of the resistor.(b) What current would flow if the resistor was connected to a 6.0 V supply?

  1. 1

    (a) Step 1: State Ohm's Law rearranged for resistance. R = V / I.

  2. 2

    (a) Step 2: Substitute the given values. R = 12 V / 3.0 A.

  3. 3

    (a) Step 3: Calculate the resistance. R = 4.0 Ω.

  4. 4

    (b) Step 1: State Ohm's Law rearranged for current. I = V / R.

  5. 5

    (b) Step 2: Substitute the new voltage and the calculated resistance. I = 6.0 V / 4.0 Ω.

  6. 6

    (b) Step 3: Calculate the new current. I = 1.5 A.

Recap

  • Voltage (potential difference) is the energy per unit charge, providing the 'push' for the current.
  • Resistance is the opposition to the flow of current.
  • Ohm's Law (V = IR) links voltage, current, and resistance for ohmic conductors.
  • Voltmeters measure voltage and are connected in parallel.
  • The resistance of a component is constant only if its temperature is constant.

Quick check

  1. A component has a resistance of 10 Ω. If a voltage of 5 V is applied across it, what is the current?2 marks

3. Series and Parallel Circuits

Components in a circuit can be connected in two main ways: series or parallel. In a series circuit, components are connected end-to-end, forming a single path for the current. The current is the same through all components (I_total = I1 = I2). The total voltage is shared between the components (V_total = V1 + V2). The total resistance is the sum of individual resistances (R_total = R1 + R2). If one bulb breaks in a series circuit, the entire circuit stops working. In a parallel circuit, components are connected on separate branches. The current splits to go down each branch and then recombines (I_total = I1 + I2). The voltage across each branch is the same (V_total = V1 = V2). The total resistance is found using the formula 1/R_total = 1/R1 + 1/R2. If one bulb breaks in a parallel circuit, the others can remain lit.

Series: R_total = R1 + R2 + ...

Series: V_total = V1 + V2 + ...

Parallel: 1/R_total = 1/R1 + 1/R2 + ...

Parallel: I_total = I1 + I2 + ...

Key term

Parallel Circuit: A circuit in which the current divides into two or more paths before recombining to complete the circuit.

Common pitfall

The most common error is calculating 1/R_total in a parallel circuit problem and forgetting to do the final step of inverting the answer to find R_total.

Worked example 15 marks

Two resistors, R1 = 6 Ω and R2 = 3 Ω, are connected in parallel to a 12 V battery. Calculate:(a) the total resistance of the circuit, and(b) the total current flowing from the battery.

  1. 1

    (a) Step 1: State the formula for total resistance in parallel. 1/R_total = 1/R1 + 1/R2.

  2. 2

    (a) Step 2: Substitute the values. 1/R_total = 1/6 + 1/3.

  3. 3

    (a) Step 3: Find a common denominator to add the fractions. 1/R_total = 1/6 + 2/6 = 3/6.

  4. 4

    (a) Step 4: Invert the result to find R_total. R_total = 6/3 = 2 Ω.

  5. 5

    (b) Step 1: Use Ohm's Law with the total resistance and total voltage. I_total = V_total / R_total.

  6. 6

    (b) Step 2: Substitute the values. I_total = 12 V / 2 Ω.

  7. 7

    (b) Step 3: Calculate the total current. I_total = 6 A.

Recap

  • In series circuits, current is constant and voltage is shared.
  • In parallel circuits, voltage is constant and current is shared.
  • Total resistance in series is the sum of the individual resistances (R_total = R1 + R2).
  • Total resistance in parallel is always less than the smallest individual resistance.
  • Household wiring uses parallel circuits so that appliances can be operated independently.

Quick check

  1. If you add another resistor in series to a circuit, does the total resistance increase or decrease?1 mark
  2. If you add another resistor in parallel to a circuit, does the total resistance increase or decrease?1 mark

4. Electrical Power and Energy

Electrical power is the rate at which electrical energy is converted into other forms, such as light, heat, or sound. A more powerful appliance converts energy more quickly. Power (P) is measured in watts (W), where 1 watt is 1 joule per second. The power dissipated by a component is calculated by multiplying the voltage across it by the current through it (P = VI). Using Ohm's Law (V = IR), we can also derive two other useful forms of the power equation: P = I²R and P = V²/R. The total electrical energy (E) converted by an appliance is its power multiplied by the time it is on for (E = Pt). Energy is measured in joules (J).

P = V × I

P = I² × R

P = V² / R

E = P × t

E = V × I × t

Key term

Power: The rate at which energy is transferred or converted, measured in watts (W).

Common pitfall

When calculating energy using E = Pt, students often forget to convert the time from minutes or hours into seconds, leading to an incorrect answer in joules.

Worked example 15 marks

A hair dryer is rated at 1800 W and is used on a 230 V mains supply.(a) Calculate the current drawn by the hair dryer.(b) Calculate the energy it converts in 5 minutes.

  1. 1

    (a) Step 1: State the formula for power. P = VI. Rearrange for current: I = P / V.

  2. 2

    (a) Step 2: Substitute the values. I = 1800 W / 230 V.

  3. 3

    (a) Step 3: Calculate the current. I ≈ 7.83 A.

  4. 4

    (b) Step 1: State the formula for energy. E = P × t.

  5. 5

    (b) Step 2: Convert the time to seconds. t = 5 minutes × 60 s/min = 300 s.

  6. 6

    (b) Step 3: Substitute the power and time. E = 1800 W × 300 s.

  7. 7

    (b) Step 4: Calculate the energy in joules. E = 540,000 J or 540 kJ.

Recap

  • Power is the rate of energy transfer, measured in watts (W).
  • The main formula for electrical power is P = VI.
  • Energy transferred is calculated using E = Pt, where time must be in seconds for the energy to be in joules.
  • Fuses are chosen with a rating slightly higher than the normal operating current of an appliance.
  • Higher power ratings mean an appliance converts energy faster.

Quick check

  1. A 60 W bulb is left on for 10 seconds. How much energy does it use?2 marks
  2. An appliance has a power of 2 kW. What is this in watts?1 mark

5. Magnetism and Electromagnetism

Magnetism is a force exerted by magnets. Magnets have two poles, north and south. Like poles repel, and opposite poles attract. A magnet creates a magnetic field around it, which is a region where other magnetic materials or magnets experience a force. We can represent these fields using magnetic field lines, which point from the north pole to the south pole. A surprising link exists between electricity and magnetism: an electric current produces a magnetic field. This is called electromagnetism. For a long, straight wire, the field lines are concentric circles around the wire. For a coil of wire, called a solenoid, the field inside is strong and uniform, similar to that of a bar magnet. You can find the direction of the field using the Right-Hand Grip Rule: point your thumb in the direction of the conventional current, and your fingers curl in the direction of the magnetic field.

Key term

Electromagnet: A temporary magnet made from a coil of wire (a solenoid) with an iron core, which acts as a magnet only when an electric current flows through the wire.

Examiner insight

For questions about electromagnets, examiners look for clear statements linking a change (e.g., more coils) to the effect (a stronger magnetic field). Be specific.

Fun fact

The world's most powerful electromagnets, used in research labs, can create magnetic fields over a million times stronger than the Earth's magnetic field.

Worked example 12 marks

A student creates an electromagnet by wrapping insulated wire around an iron nail and connecting it to a power supply. Describe two ways they could increase the strength of the electromagnet.

  1. 1

    Method 1: Increase the current flowing through the coil of wire. A larger current produces a stronger magnetic field.

  2. 2

    Method 2: Increase the number of turns (coils) of wire around the iron nail. A greater number of turns per unit length concentrates the magnetic field, making it stronger.

Recap

  • Magnets have north and south poles; like poles repel, opposites attract.
  • A magnetic field is a region where a magnetic force can be felt.
  • An electric current flowing through a wire creates a magnetic field around it.
  • A solenoid is a coil of wire that produces a strong, uniform magnetic field inside it.
  • The strength of an electromagnet can be increased by increasing the current, increasing the number of turns, or adding an iron core.

Quick check

  1. Draw the magnetic field pattern around a straight, current-carrying wire.2 marks

6. The Motor Effect

When a wire carrying an electric current is placed in a magnetic field, the two fields interact, and the wire experiences a force. This is known as the motor effect. The force is greatest when the wire is at 90 degrees to the magnetic field lines. The direction of the force can be found using Fleming's Left-Hand Rule. Hold your left hand with your thumb, first finger, and second finger all at right angles to each other. Your First finger points in the direction of the magnetic Field (North to South). Your seCond finger points in the direction of the Current (conventional, + to -). Your ThuMb then points in the direction of the Motion (the force). This effect is the principle behind electric motors, where a coil of wire in a magnetic field is made to rotate continuously.

F ∝ B (Magnetic field strength)

F ∝ I (Current)

F ∝ L (Length of wire in field)

Key term

Motor Effect: The force experienced by a current-carrying conductor when it is placed in a magnetic field.

Examiner insight

Examiners will often test your ability to apply Fleming's Left-Hand Rule in unfamiliar situations. Practice with diagrams where the field or current are in various directions (e.g., into/out of the page).

Common pitfall

A very common mistake is confusing Fleming's Left-Hand Rule (for motors) with his Right-Hand Rule (for generators). Remember: your 'left' motor car gets you moving.

Worked example 13 marks

A horizontal wire is placed in a uniform magnetic field that is directed vertically downwards. The current in the wire flows from left to right. Determine the direction of the force on the wire.

  1. 1

    Step 1: Identify the directions of the field and current. Field (First finger) is downwards. Current (seCond finger) is from left to right.

  2. 2

    Step 2: Apply Fleming's Left-Hand Rule. Point your first finger down and your second finger to the right.

  3. 3

    Step 3: Observe the direction of your thumb. Your thumb points out of the page.

  4. 4

    Step 4: State the direction of the force. The force on the wire is directed out of the page.

Recap

  • A current-carrying wire in a magnetic field experiences a force (the motor effect).
  • The size of the force depends on the current, magnetic field strength, and length of the wire.
  • Fleming's Left-Hand Rule is used to find the direction of the force.
  • Remember: First finger = Field, seCond finger = Current, ThuMb = Motion.
  • The motor effect is used to make electric motors rotate.

Quick check

  1. What happens to the force on the wire if the direction of the current is reversed?1 mark

7. Electromagnetic Induction

Just as electricity can create magnetism, magnetism can be used to create electricity. This process is called electromagnetic induction. If you move a wire through a magnetic field, or move a magnet near a wire, a voltage (also called an electromotive force or e.m.f.) is induced across the ends of the wire. If the wire is part of a complete circuit, a current will flow. This happens because the wire is 'cutting' through the magnetic field lines. To get a bigger induced voltage, you can move the wire faster, use a stronger magnet, or use a coil with more turns of wire. The direction of the induced current can be found using Fleming's Right-Hand Rule, which is similar to the left-hand rule but used for generators. This principle is fundamental to how we generate electricity in power stations.

Key term

Electromagnetic Induction: The process of inducing a voltage (e.m.f.) in a conductor by moving it through a magnetic field or by changing the magnetic field around it.

Examiner insight

To get full marks for explaining induction, you must use the phrase 'cutting magnetic field lines'. Simply saying 'the wire is in a magnetic field' is not enough.

Fun fact

Contactless payment cards and wireless phone chargers work using electromagnetic induction to transfer energy over a short distance without any physical connection.

Worked example 14 marks

A student pushes the north pole of a bar magnet into a coil of wire connected to a sensitive ammeter.(a) Explain why the ammeter shows a reading.(b) What happens to the reading if the magnet is held stationary inside the coil?(c) What could the student do to induce a larger current?

  1. 1

    (a) As the magnet moves, its magnetic field lines are cut by the coil of wire. This induces a voltage (e.m.f.) across the coil, which causes a current to flow, detected by the ammeter.

  2. 2

    (b) If the magnet is stationary, no field lines are being cut. Therefore, no voltage is induced, and the ammeter reading returns to zero.

  3. 3

    (c) To induce a larger current, the student could: (1) move the magnet faster, (2) use a stronger magnet, or (3) use a coil with more turns of wire.

Recap

  • A voltage is induced when a conductor cuts magnetic field lines.
  • This process is called electromagnetic induction.
  • No voltage is induced if the conductor and magnet are stationary relative to each other.
  • The size of the induced voltage can be increased by moving faster, using a stronger magnet, or using more turns in a coil.
  • Electromagnetic induction is the principle behind electrical generators.

Quick check

  1. A wire is moved parallel to magnetic field lines. Is a voltage induced? Explain why.2 marks

8. Transformers

A transformer is a device that changes the size of an alternating voltage. It consists of two coils of wire, a primary coil and a secondary coil, wrapped around a soft iron core. When an alternating voltage is applied to the primary coil, it creates a continuously changing magnetic field in the iron core. This changing magnetic field passes through the secondary coil and induces an alternating voltage across it. A step-up transformer has more turns on its secondary coil than its primary (Ns > Np) and it increases the voltage. A step-down transformer has fewer turns on its secondary coil (Ns < Np) and it decreases the voltage. The ratio of the voltages is equal to the ratio of the number of turns. For a 100% efficient transformer, the power input equals the power output (VpIp = VsIs). Transformers are crucial for the National Grid, stepping up voltage for efficient transmission over long distances and then stepping it down for safe use in homes.

Vp / Vs = Np / Ns

Vp × Ip = Vs × Is (for a 100% efficient transformer)

Key term

Transformer: A device that uses electromagnetic induction to change the size of an alternating voltage, consisting of two coils wrapped around a common iron core.

Examiner insight

Examiners expect you to know why transformers are used in the national grid: stepping up voltage reduces the current (P=VI), which in turn reduces energy loss as heat in the transmission cables (since heat loss is proportional to I²R).

Common pitfall

Students often invert the turns ratio formula, writing Ns/Np instead of Np/Ns. Always place the primary coil values (Vp, Np) on one side of the fraction and the secondary values (Vs, Ns) on the other.

Worked example 15 marks

A transformer is used to step down the 230 V mains voltage to 11.5 V for a low-voltage lamp. The primary coil has 800 turns.(a) Calculate the number of turns on the secondary coil.(b) If the lamp draws a current of 2.0 A, calculate the current in the primary coil, assuming the transformer is 100% efficient.

  1. 1

    (a) Step 1: State the transformer turns-voltage relationship. Vp / Vs = Np / Ns.

  2. 2

    (a) Step 2: Rearrange the formula to find Ns. Ns = Np × (Vs / Vp).

  3. 3

    (a) Step 3: Substitute the known values. Ns = 800 × (11.5 V / 230 V).

  4. 4

    (a) Step 4: Calculate the number of secondary turns. Ns = 800 × 0.05 = 40 turns.

  5. 5

    (b) Step 1: State the power relationship for a 100% efficient transformer. Vp × Ip = Vs × Is.

  6. 6

    (b) Step 2: Rearrange to find the primary current, Ip. Ip = (Vs × Is) / Vp.

  7. 7

    (b) Step 3: Substitute the values. Ip = (11.5 V × 2.0 A) / 230 V.

  8. 8

    (b) Step 4: Calculate the primary current. Ip = 23 / 230 = 0.1 A.

Recap

  • Transformers only work with alternating current (AC).
  • They consist of a primary coil, a secondary coil, and a soft iron core.
  • A step-up transformer increases voltage (Ns > Np).
  • A step-down transformer decreases voltage (Ns < Np).
  • The formula Vp / Vs = Np / Ns relates voltages and turns.
  • In an ideal transformer, input power equals output power.

Quick check

  1. Why does a transformer not work with a DC supply from a battery?2 marks

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A 12 Ω resistor and a 4 Ω resistor are connected in series with a 24 V battery. Calculate the total current flowing from the battery.3 marks
  2. An electric motor has a power rating of 1.5 kW and is connected to a 230 V mains supply. Calculate the most suitable fuse rating for the motor from the following options: 1 A, 3 A, 5 A, 13 A. You must show your calculation.4 marks
  3. Draw a diagram of a simple DC motor. Label the magnet, the coil, the commutator, and the brushes. Use an arrow to show the direction of force on one side of the coil.5 marks
  4. A student has a 6 Ω resistor and a 12 Ω resistor. Calculate the total resistance if they are connected (a) in series, and (b) in parallel.4 marks
  5. Explain, using the principle of electromagnetic induction, how a simple AC generator (alternator) works.4 marks
  6. A charge of 180 C flows through a component in 1.5 minutes. Calculate the current in amperes.3 marks
  7. A step-up transformer has a primary coil with 50 turns and a secondary coil with 1000 turns. If the input voltage is 12 V AC, what is the output voltage?3 marks
  8. A lamp has a resistance of 50 Ω when a current of 0.4 A flows through it. Calculate the power dissipated by the lamp.3 marks
  9. Describe an experiment to plot the magnetic field pattern of a bar magnet using a plotting compass. Include a diagram of the expected result.4 marks
  10. A circuit contains a 10 Ω resistor in series with a parallel combination of two 12 Ω resistors. The entire circuit is connected to a 6 V supply. Calculate the total current drawn from the supply.6 marks

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