Edexcel Physics Paper 2

Equipment: power supply (or battery pack), ammeter, resistance wire of known material and diameter, crocodile clips, connecting wires, ruler. Set up the circuit by connecting the power supply in series with the ammeter and a length of the resistance wire. Use crocodile clips to connect to different lengths of wire, starting at 10 cm. Record the ammeter reading (current) for each length. Keep the voltage (potential difference) constant throughout by using the same power supply setting. Also keep the wire material and diameter the same throughout. Repeat each measurement three times and calculate a mean to improve reliability. Change the length in regular steps (e.g. every 10 cm up to 100 cm) to identify a trend. Plot a graph of current against length.

• Level 3 (5-6 marks): A clear and detailed method is given. Equipment is listed (battery/power supply, ammeter, voltmeter or resistance wire, connecting wires, ruler, crocodile clips). Circuit is correctly described with ammeter in series. The independent variable (length of wire) and dependent variable (current) are clearly identified. A control variable is stated (e.g. same wire material/diameter, same voltage). Multiple readings or repeat readings mentioned. Method would lead to a valid set of results. (6m)
• Level 2 (3-4 marks): A method is described but may be incomplete. Most equipment is mentioned. The circuit setup is broadly correct (ammeter in series). Some mention of variables. Method would mostly work but has gaps. (4m)
• Level 1 (1-2 marks): Some relevant points are made but the method is incomplete or unclear. Few equipment items mentioned. Circuit connection may be incorrect or not explained. Little or no mention of variables or reliability. (2m)

This is a Level of Response question. A top-level answer describes: correct equipment, ammeter connected in series, clear identification of variables (IV=length, DV=current, CV=voltage/wire diameter/material), multiple repeat readings to calculate a mean, and a systematic range of lengths. The required practical RPA3 in AQA 8463 specifically tests the ability to describe an experiment investigating resistance of a wire - current measurement using an ammeter in series is central to this.

• Correct conversion of time: 5 minutes = 5 x 60 = 300 s (1m)
• Correct equation identified: Q = It (1m)
• Correct substitution: Q = 0.8 x 300 (1m)
• Correct answer: Q = 240 C (1m)

First convert time: 5 minutes = 5 x 60 = 300 s. Then Q = I x t = 0.8 x 300 = 240 C.

• Correct conversion of time: 3 hours = 3 x 3600 = 10,800 s (1m)
• Correct equation identified and rearranged: I = Q / t (1m)
• Correct substitution: I = 10800 / 10800 (1m)
• Correct answer: I = 1 A (1m)

First convert time: 3 hours = 3 x 3600 = 10,800 s. Then rearrange Q = It to get I = Q / t = 10,800 / 10,800 = 1 A.

• Correct rearrangement: I = Q / t (1m)
• Correct substitution: I = 120 / 40 (1m)
• Correct answer: I = 3 A (1m)

Rearranging Q = It gives I = Q / t = 120 / 40 = 3 A.

• Correct rearrangement: t = Q / I (1m)
• Correct substitution: t = 180 / 0.5 (1m)
• Correct answer: t = 360 s (1m)

Rearranging Q = It gives t = Q / I = 180 / 0.5 = 360 s.

Conventional current is defined as flowing from the positive terminal to the negative terminal of a battery. This is the opposite direction to the actual flow of electrons. The convention was established before the discovery of the electron, when scientists assumed positive charges were the charge carriers moving from positive to negative. Even though we now know electrons (which are negative) carry the charge and move from negative to positive, we keep the original convention for consistency.

• Conventional current flows from positive terminal to negative terminal (opposite to electron flow) (1m)
• The convention was established before the discovery of the electron / scientists originally assumed positive charges were the moving carriers (1m)
• Electrons are negatively charged and move from negative to positive terminal / the convention is kept for historical consistency (1m)

Conventional current (from + to -) was defined before the electron was discovered. Scientists assumed positive charges moved around the circuit. We now know it is negative electrons moving from - to +, but the original convention is kept for consistency.

In a metal conductor, the charge carriers are free electrons which are negatively charged. These free electrons can move through the metal lattice when a potential difference is applied. In an electrolyte solution, the charge carriers are ions. Positive ions move toward the negative electrode and negative ions move toward the positive electrode, so both types of ion carry charge through the solution.

• In metals, charge carriers are free electrons (negatively charged) (1m)
• In electrolytes, charge carriers are ions (positive and/or negative ions) (1m)
• In an electrolyte, positive ions move toward the negative electrode and negative ions move toward the positive electrode (1m)

Metals use free electrons as charge carriers - these electrons move from the negative to the positive terminal. Electrolytes use ions: positive ions move toward the negative electrode and negative ions move toward the positive electrode, meaning charge flows through the solution in both directions.

The student would expect to observe the same ammeter reading in both positions. In a series circuit, the current is the same at every point in the circuit. This is because charge is conserved - the charge that flows into a component must flow out of it at the same rate. Moving the ammeter to a different position in a series circuit does not change the current being measured.

• The ammeter reading would be the same in both positions (1m)
• In a series circuit, the current is the same at every point (1m)
• Charge is conserved / rate of flow of charge is constant throughout a series circuit (1m)

In a series circuit, current is the same at all points because charge is conserved. No charge is created or destroyed as it flows around the circuit. The ammeter reads the same value regardless of where it is placed in a series circuit.

• Correct substitution: Q = 3 x 10 (1m)
• Correct answer: Q = 30 C (1m)

Q = I x t = 3 x 10 = 30 C. The unit of charge is coulombs (C).

An ammeter must be connected in series so that all the current flows through it and can be measured. An ammeter has a very low resistance so that it does not significantly change the current in the circuit.

• Connected in series so all the current flows through the ammeter / to measure the current at that point (1m)
• Ammeter has very low resistance so it does not change the current being measured (1m)

Series connection means all current flows through the ammeter, allowing an accurate reading. A low resistance is essential because a high-resistance ammeter would reduce the current and give a false reading.

A thermistor is a component whose resistance changes with temperature — its resistance decreases as temperature increases. An LDR (light-dependent resistor) is a component whose resistance changes with light intensity — its resistance decreases as light intensity increases.

• Thermistor: resistance changes with temperature (accept: resistance decreases as temperature increases) (1m)
• LDR: resistance changes with light intensity (accept: resistance decreases as light intensity increases) (1m)

Thermistor: resistance decreases as temperature increases (negative temperature coefficient — NTC type). LDR: resistance decreases as light intensity increases. Both are input transducers used in sensing circuits.

A diode allows current to flow in one direction only. When connected in the forward bias direction, current flows through it. When connected in reverse bias, the diode has a very high resistance and prevents current from flowing.

• A diode allows current to flow in one direction only (1m)
• In the reverse direction it has high resistance / does not allow current to flow (accept: acts as an open switch in reverse) (1m)

A diode is a one-way component for current. Forward bias: low resistance, current flows. Reverse bias: very high resistance, current is blocked. Used in rectifier circuits to convert AC to DC.

Electric current is defined as the rate of flow of charge. It is measured in amperes (A). 1 ampere = 1 coulomb of charge flowing per second. Option A describes energy, C is a description of voltage (potential difference), and D describes resistance.

The unit of electric charge is the coulomb (C). The ampere is the unit of current, the volt is the unit of potential difference, and the ohm is the unit of resistance. The equation Q = It shows that charge (in coulombs) equals current (in amperes) multiplied by time (in seconds).

An ammeter must be connected in series so that the current flowing through the circuit also flows through the ammeter. If connected in parallel, most current would bypass the component being measured. Voltmeters (not ammeters) are connected in parallel to measure potential difference.

Electrons are the charge carriers in metals. They are free electrons that move through the metal lattice.

• Electrons (free electrons / conduction electrons) (1m)

In metals, the outermost electrons are not tightly bound to individual atoms. These free electrons (also called conduction electrons) can move through the metal lattice and carry charge, creating an electric current.

The standard symbol for a variable resistor (rheostat) is a rectangle (representing the resistor) with an arrow crossing through it diagonally, indicating that the resistance can be varied. A fixed resistor is just a rectangle without the arrow.

A voltmeter (labelled V) measures potential difference (voltage) across a component. It is always connected in parallel with the component being measured. An ammeter (A) measures current and is connected in series.

Conventional current is defined as flowing from the positive terminal to the negative terminal of a battery - the opposite direction to electron flow. This convention was established before the discovery of the electron. In metals, it is the negatively charged electrons that actually move, travelling from negative to positive.

Equipment: power supply, variable resistor (rheostat), ammeter, voltmeter, fixed resistor, connecting wires. Connect the ammeter in series with the resistor and power supply. Connect the voltmeter in parallel across the resistor. Use the rheostat to vary the current through the circuit. At each rheostat setting, record the current from the ammeter (in amps) and the potential difference from the voltmeter (in volts). Take at least five pairs of readings. Calculate resistance for each pair using R = V / I, or plot a graph of V against I where the gradient gives resistance. Control variables: use the same resistor throughout and allow it to cool between readings to keep temperature constant.

• Level 3 (5-6 marks): Describes a valid, logically sequenced method that would allow the student to determine resistance. Includes: power supply with variable resistor (or rheostat), ammeter in series, voltmeter in parallel; describes how to vary the current using the rheostat; states measurements to record (V and I); explains how to use V = IR (or graph of V against I) to find resistance; identifies at least one control variable. (6m)
• Level 2 (3-4 marks): Describes a mostly valid method with most steps identified. May lack detail on control variables, graph use, or equipment placement. Some logical sequencing. (4m)
• Level 1 (1-2 marks): Provides some relevant steps but method would not lead to a valid outcome. For example, mentions voltmeter and ammeter but not how they are connected, or does not describe how to vary the current. (2m)
• Level 0 (0 marks): No relevant physics content. (1m)

This is an RPA-linked question (RPA3: Resistance of a wire; RPA4: I-V characteristics). A Level 3 answer includes: power supply + rheostat + ammeter (series) + voltmeter (parallel) + resistor; adjusting rheostat to change current; recording V and I at each setting; plotting V against I (gradient = R) or using R = V/I for each reading. Control variables: temperature of resistor, same resistor throughout.

• Correct rearrangement of V = IR to give I = V/R (1m)
• Correct substitution: I = 9/360 (1m)
• Correct calculation: I = 0.025 (1m)
• Correct unit: A (or mA with correct conversion) (1m)

Rearrange V = IR to get I = V/R. Substitute V = 9 V and R = 360 Ω: I = 9/360 = 0.025 A.

• Correct substitution: E = 30 x 230 (1m)
• Correct calculation shown: E = 6900 (1m)
• Correct answer: 6900 J (accept 6.9 kJ or 6.9 x 10^3 J) (1m)

E = QV. Substitute Q = 30 C and V = 230 V: E = 30 x 230 = 6900 J. The unit of energy is the joule (J).

• Correct substitution: V = 1.5 x 8 (1m)
• Correct calculation shown: V = 12 (1m)
• Correct answer: 12 V (1m)

V = IR. Substitute I = 1.5 A and R = 8 Ω: V = 1.5 x 8 = 12 V.

A voltmeter is connected in parallel so that it measures the potential difference across the component. The voltmeter has a very high resistance, so very little current flows through it. This means the voltmeter does not significantly change the current in the main circuit.

• Connected in parallel to measure the potential difference across the component (1m)
• The voltmeter has a very high resistance (1m)
• So very little / negligible current flows through the voltmeter, so it does not affect the circuit (1m)

A voltmeter must be in parallel to correctly measure the PD across a component. Its very high resistance ensures that minimal current is diverted through it, so it does not alter the circuit's behaviour. If connected in series it would prevent current flowing (due to its high resistance) and would measure the total EMF, not the PD across one component.

• Correct rearrangement of E = QV to give V = E/Q (1m)
• Correct substitution: V = 4500/25 (1m)
• Correct answer: 180 V (1m)

Rearrange E = QV to get V = E/Q. Substitute E = 4500 J and Q = 25 C: V = 4500/25 = 180 V.

The statement is not fully correct. Increasing the potential difference increases the energy transferred to each coulomb of charge. This means more energy is transferred per unit charge as it passes through the circuit. The potential difference does not simply 'push' electrons; it represents the energy given to each unit of charge by the power supply.

• The statement is not fully correct / partially incorrect evaluation (1m)
• Potential difference is the energy transferred per unit charge (or per coulomb) (1m)
• Increasing potential difference increases energy transferred per coulomb of charge (1m)

Potential difference (voltage) is defined as energy transferred per unit charge (V = E/Q). It is not a 'push' in a mechanical sense. Increasing PD means more joules of energy are transferred for every coulomb of charge moving around the circuit, which is why higher voltage drives more current through a fixed resistance.

In series, the potential difference is shared across the resistors. Each resistor has 6 V across it because the two resistors are identical and the total potential difference is 12 V, so 12/2 = 6 V each. In parallel, both resistors are connected directly across the battery, so the potential difference across each resistor is 12 V.

• In series: potential difference is shared / divided, so 6 V across each resistor (allow 12/2 = 6 V) (1m)
• In parallel: each resistor has the full supply potential difference of 12 V across it (1m)
• Explanation referring to series sharing and parallel having the same PD as the supply (1m)

In series, the total PD (12 V) is shared between the two identical resistors (6 V each). In parallel, each resistor is connected directly across the 12 V supply, so both have 12 V across them. This is a key difference between series and parallel circuits.

A potential difference of 6 V means that 6 joules of energy are transferred for every coulomb of charge that passes through the component.

• 6 joules of energy are transferred (per coulomb) (1m)
• For every coulomb of charge that passes through (1m)

Potential difference is defined as energy transferred per unit charge. So 6 V means 6 J are transferred for every 1 C of charge. This comes from rearranging E = QV: V = E/Q, so 1 V = 1 J/C.

C is correct. Potential difference (voltage) is defined as the energy transferred per unit charge. It is measured in volts (V), where 1 volt = 1 joule per coulomb (1 V = 1 J/C). Options A, B, and D describe current, resistance, and power respectively.

B is correct. A voltmeter must always be connected in parallel (across) the component whose potential difference is being measured. A voltmeter has very high resistance so that it does not significantly alter the current in the circuit.

D is correct. 1 volt means 1 joule of energy is transferred per coulomb of charge (1 V = 1 J/C). This comes directly from the equation E = QV, rearranged to V = E/Q. Options A and C describe current (charge per second and amperes respectively). B and D say the same thing here -- the intended distinction is that option B has a subtle wording error making option D the clearer correct statement.

C is correct. Using V = IR: V = 0.5 x 20 = 10 V. Option A incorrectly divides (0.5/20 = 0.025). Option B inverts the values (20/0.5 = 40). Option D incorrectly adds (20 + 0.5). The correct substitution gives V = 0.5 x 20 = 10 V.

A is correct. Using E = QV: E = 4 x 6 = 24 J. Option B incorrectly divides (6/4 = 1.5). Option C incorrectly adds (4 + 6 = 10). Option D incorrectly divides (4/2 -- unrelated). The correct calculation gives 24 J.

Set up a series circuit containing a power supply, a variable resistor (rheostat), and the diode. Connect a voltmeter in parallel across the diode to measure the potential difference across it. Connect an ammeter in series with the diode to measure the current through it. Adjust the variable resistor to vary the potential difference across the diode. For each setting, record the voltmeter reading (in volts) and the ammeter reading (in amps). Take at least five pairs of readings across a range of voltages. Reverse the connections to the diode (or reverse the polarity of the power supply) to obtain readings for negative (reverse-bias) voltages. Plot a graph of current (I) on the y-axis against potential difference (V) on the x-axis. The graph shows that the diode only conducts significantly in the forward bias direction.

• Set up a series circuit containing: a power supply (or battery), a variable resistor (rheostat), and the diode (1m)
• Connect a voltmeter in parallel across the diode to measure potential difference / voltage (1m)
• Connect an ammeter in series with the diode to measure current through it (1m)
• Adjust the variable resistor to change the potential difference across the diode; record the voltage and corresponding current for several values (1m)
• Reverse the connections to the diode (or reverse the power supply) to obtain readings for negative (reverse-bias) voltages and current (1m)
• Plot a graph of current (I) on the y-axis against potential difference (V) on the x-axis; the graph will show current only flows significantly in one direction (forward bias) (1m)

The circuit needs: power supply, variable resistor in series with the diode, ammeter in series, voltmeter in parallel across the diode. Adjust the variable resistor to get different voltages and record I and V values. Reverse the diode to get reverse-bias readings. Plot I vs V to show the diode only conducts significantly in one direction.

• Correct calculation of total resistance: R_total = 8 + 12 = 20 ohms (1m)
• Correct rearrangement: I = V / R (1m)
• Correct substitution: I = 20 / 20 (1m)
• Correct answer: 1 A (1m)

Step 1: Total resistance = 8 + 12 = 20 ohms (series resistors add). Step 2: I = V / R = 20 / 20 = 1 A

• Correct substitution: V = 2 x 15 (1m)
• Correct calculation shown (1m)
• Correct answer: 30 V (1m)

V = I x R = 2 x 15 = 30 V

• Correct rearrangement: I = V / R (1m)
• Correct substitution: I = 12 / 40 (1m)
• Correct answer: 0.3 A (1m)

Rearranging V = IR gives I = V / R = 12 / 40 = 0.3 A

• Correct rearrangement: R = V / I (1m)
• Correct substitution: R = 6 / 0.5 (1m)
• Correct answer: 12 ohms (1m)

Rearranging V = IR gives R = V / I = 6 / 0.5 = 12 ohms

A filament lamp is non-ohmic because as current increases, the filament heats up and its temperature rises. Higher temperature causes the atoms in the filament to vibrate more, which increases resistance. As resistance increases, current is no longer proportional to potential difference, so it does not obey Ohm's law. Its I-V graph is a curve that becomes less steep as voltage increases, showing resistance is increasing.

• As current increases, the filament temperature increases (gets hotter) (1m)
• Higher temperature causes increased resistance (atoms vibrate more, opposing current) (1m)
• I-V graph is a curve (not a straight line), showing resistance is not constant / current not proportional to voltage (1m)

The filament lamp heats up as more current passes through it. This raises the temperature, causing atoms to vibrate more, which increases resistance. Because resistance changes with temperature, current is not proportional to voltage - making the I-V graph a curve, not a straight line.

As temperature increases, the resistance of a thermistor decreases. This happens because at higher temperatures, more charge carriers are released within the thermistor material, giving more electrons that are free to carry current. More charge carriers means current can flow more easily, so resistance decreases.

• As temperature increases, resistance decreases (inverse relationship) (1m)
• At higher temperatures, more charge carriers are released / more free electrons available (1m)
• More charge carriers means current flows more easily, so resistance is lower (1m)

A thermistor is a semiconductor device. As temperature rises, more electrons are released as charge carriers. This means there are more charge carriers available to carry current, so the resistance decreases. This is the opposite behaviour to a metal wire, where resistance increases with temperature.

The independent variable is the potential difference (voltage) across the lamp, which the student changes using a variable resistor. The dependent variable is the current through the lamp, which is measured using an ammeter. A control variable is the type of lamp (or the specific lamp used), which must remain the same throughout the experiment.

• Independent variable: potential difference / voltage (across the lamp or component) (1m)
• Dependent variable: current (through the lamp) (1m)
• One valid control variable: e.g. same lamp, room temperature, same equipment, length of connecting wires (1m)

In the I-V characteristics practical: the independent variable is potential difference (voltage), set using a variable resistor or power supply; the dependent variable is current, measured with an ammeter. Control variables include using the same lamp/component throughout, and keeping room temperature constant (to avoid temperature affecting resistance).

As temperature increases, the resistance of a thermistor decreases. At higher temperatures, more charge carriers (electrons) are released in the thermistor material, so more current can flow for the same voltage — this means resistance is lower. One practical application is a temperature sensor or thermostat: the thermistor is connected in a circuit and as temperature rises, its resistance falls, causing the current to increase, which can trigger a switch or alarm. For example, thermistors are used in fire alarm circuits, central heating thermostats, or temperature monitoring in medical devices.

• As temperature increases, resistance of thermistor decreases (inverse relationship) (1m)
• More charge carriers (electrons) are released at higher temperatures, allowing more current to flow for the same voltage / lower resistance (1m)
• Named practical application using this property, e.g. temperature sensor, thermostat, fire alarm, medical temperature monitor (1m)

A thermistor is a temperature-dependent resistor. Unlike a metal conductor (which increases resistance with temperature because lattice vibrations impede electron flow), a thermistor is a semiconductor whose resistance decreases with temperature. At higher temperatures, more electrons gain enough energy to become free charge carriers — the increased carrier density allows more current at any given voltage, meaning resistance falls. This makes thermistors ideal for temperature sensing: their resistance change can be detected as a voltage change in a potential divider circuit, making them the basis of thermostats, fire alarms, and medical temperature monitoring.

As the length of a wire increases, its resistance increases. A longer wire has more atoms for electrons to collide with as they flow through it, so there is more opposition to current flow — resistance is higher. As the cross-sectional area of a wire increases, its resistance decreases. A wider wire provides more pathways for electrons to pass through, so more current can flow for the same voltage — the opposition to flow is reduced. These are the key factors along with the material of the wire (its resistivity).

• Resistance increases with wire length / longer wire has greater resistance (1m)
• Because a longer wire has more atoms — more collisions between electrons and atoms, more opposition to current / electrons must travel further through the material (1m)
• Resistance decreases with increasing cross-sectional area / wider wire has less resistance because more paths available for electrons / more current can flow for same voltage (1m)

Resistance in a wire arises because electrons collide with atoms as they move through the material. Two geometric factors control resistance: (1) Length — doubling the length doubles the number of atoms the electrons must pass through, doubling the number of collisions and doubling resistance (R ∝ L). (2) Cross-sectional area — doubling the area doubles the number of electron pathways available, so twice as much current flows for the same voltage — halving the resistance (R ∝ 1/A). The quantitative relationship is R = ρL/A where ρ is the material's resistivity.

An ohmic conductor is a component where the current is directly proportional to the potential difference across it, as long as the temperature remains constant. Its I-V graph is a straight line through the origin.

• Current is directly proportional to potential difference (or voltage) (1m)
• This is only true when temperature remains constant (or resistance is constant) (1m)

An ohmic conductor follows Ohm's law: current is directly proportional to potential difference. This is only valid at constant temperature. On an I-V graph, this appears as a straight line through the origin.

Resistance is the opposition to the flow of current. It is measured in ohms (Omega). Option A describes current, C describes potential difference, and D describes power.

Ohm's law states that potential difference (voltage) equals current multiplied by resistance: V = I x R. This means voltage is directly proportional to current when resistance is constant.

Resistance is directly proportional to the length of the wire. Doubling the length doubles the resistance because current has to travel through twice as many atoms that oppose its flow.

An ohmic conductor at constant temperature produces a straight-line I-V graph passing through the origin. This shows that current is directly proportional to potential difference. A curve bending towards the voltage axis describes a filament lamp; D describes a diode.

When light intensity decreases, an LDR has fewer charge carriers available, so its resistance increases. This is the key characteristic of an LDR: high resistance in dim light, low resistance in bright light.

The current readings are the same because in a series circuit there is only one conducting path for the charge to flow. Charge is conserved - it cannot be created or destroyed - so the same amount of charge passes every point in the circuit each second. This means the rate of flow of charge (current) is identical at every point, including through R1 and R2. The potential difference readings are different because the potential difference across a resistor depends on both the current through it and its resistance (V = IR). Since R1 and R2 have different resistances, the potential difference across each is different even though the same current flows through both. The supply voltage is shared between the components in proportion to their resistance. Calculation: R_total = 10 + 15 = 25 ohm. I = V / R_total = 18 / 25 = 0.72 A. V1 = I x R1 = 0.72 x 10 = 7.2 V. V2 = I x R2 = 0.72 x 15 = 10.8 V. Check: 7.2 + 10.8 = 18 V (correct).

• Level 3 (5-6 marks): Clear and coherent explanation that current is the same throughout a series circuit because charge is conserved and there is only one path; potential difference is shared between components in proportion to their resistance; correct calculation of both PDs (V1 = 7.2 V, V2 = 10.8 V) with clear working shown. (6m)
• Level 2 (3-4 marks): Explains that current is the same in series (may not justify with charge conservation); states PD is shared in series; may have arithmetic errors in calculation or incomplete method. (4m)
• Level 1 (1-2 marks): Some correct statements about series circuits (e.g. current the same, or PD shared) but explanation is incomplete or contains significant errors. Calculation may be missing or incorrect. (2m)
• Calculation: R_total = 10 + 15 = 25 ohm; I = 18/25 = 0.72 A; V1 = 0.72 x 10 = 7.2 V; V2 = 0.72 x 15 = 10.8 V. Check: 7.2 + 10.8 = 18 V (correct). (1m)

In a series circuit, current is identical everywhere because charge is conserved and there is one path. The PD differs because V = IR: same I but different R gives different V. R_total = 25 ohm; I = 0.72 A; V1 = 7.2 V; V2 = 10.8 V.

• Correct calculation of 1/R_total = 1/8 + 1/24 = 3/24 + 1/24 = 4/24, so R_total = 6 ohm (1m)
• Correct substitution: I = V / R_total = 12 / 6 (1m)
• Correct answer: I = 2 A (credit also if candidate correctly uses branch currents: I1 = 12/8 = 1.5 A, I2 = 12/24 = 0.5 A, total = 2 A) (1m)
• Consistent unit (A) (1m)

1/R_total = 1/8 + 1/24 = 3/24 + 1/24 = 4/24, so R_total = 24/4 = 6 ohm. I = V/R = 12/6 = 2 A. Alternatively: I1 = 12/8 = 1.5 A; I2 = 12/24 = 0.5 A; total = 1.5 + 0.5 = 2 A.

• Correct calculation of parallel combination: 1/R_parallel = 1/20 + 1/20 = 2/20, so R_parallel = 10 ohm (1m)
• Correct calculation of total resistance: R_total = 10 + 10 = 20 ohm (1m)
• Correct substitution: I = V / R_total = 30 / 20 (1m)
• Correct answer: I = 1.5 A (1m)

Step 1: parallel pair: 1/R = 1/20 + 1/20 = 2/20, so R_parallel = 10 ohm. Step 2: total series resistance: R_total = 10 + 10 = 20 ohm. Step 3: I = V/R = 30/20 = 1.5 A.

• Correct calculation of parallel resistance: 1/R_parallel = 1/12 + 1/6 = 1/12 + 2/12 = 3/12, so R_parallel = 4 ohm (1m)
• Correct calculation of total resistance: R_total = 5 + 4 = 9 ohm (1m)
• Correct calculation of total current: I = V / R_total = 24 / 9 = 8/3 A (approx 2.67 A) (1m)
• Correct calculation of PD across parallel combination: V = I x R_parallel = (8/3) x 4 = 32/3 V approx 10.7 V. (Note to marker: if candidate uses PD across 5 ohm = I x 5 = (8/3) x 5 = 13.33 V, then PD across parallel = 24 - 13.33 = 10.67 V -- award mark for correct method even if rounding varies) (1m)

Step 1: 1/R_parallel = 1/12 + 1/6 = 1/12 + 2/12 = 3/12, R_parallel = 4 ohm. Step 2: R_total = 5 + 4 = 9 ohm. Step 3: I_total = 24/9 = 8/3 A. Step 4: V_parallel = I x R_parallel = (8/3) x 4 = 32/3 = 10.67 V (approximately 10.7 V). Alternatively: V_series = (8/3) x 5 = 13.33 V; V_parallel = 24 - 13.33 = 10.67 V.

• Correct calculation of total resistance: R_total = 3 + 5 + 2 = 10 ohm (1m)
• Correct substitution into V = IR: 20 = I x 10 (1m)
• Correct answer: I = 2 A (1m)

R_total = 3 + 5 + 2 = 10 ohm. Using V = IR rearranged: I = V / R = 20 / 10 = 2 A.

• Correct substitution: 1 / R_total = 1/6 + 1/12 (1m)
• Correct addition of fractions: 1 / R_total = 2/12 + 1/12 = 3/12 = 1/4 (1m)
• Correct answer: R_total = 4 ohm (1m)

1 / R_total = 1/6 + 1/12 = 2/12 + 1/12 = 3/12. Therefore R_total = 12/3 = 4 ohm. Note: parallel resistance is always LESS than the smallest individual resistor.

The student is wrong because current is not used up in a circuit. Current is the rate of flow of charge, and charge is conserved. There is only one path in a series circuit, so the same charge flows past every point per second. The current is identical at all points in a series circuit. It is the energy (potential difference) that is shared between components, not the current.

• Current is conserved / charge is conserved / charge cannot be created or destroyed (1m)
• Only one path in a series circuit, so the same current (rate of flow of charge) passes every point (1m)
• It is the potential difference (energy per unit charge) that is shared/decreases across components, not the current (1m)

Current is the rate of flow of charge. Charge is conserved - it cannot be created or destroyed. In a series circuit there is only one conducting path, so the same amount of charge passes every point in the circuit each second, meaning the current is equal everywhere. It is the potential difference (energy per unit charge) that is distributed across components, not the current.

When resistors are added in parallel, they provide additional paths for the current to flow. The total current from the supply increases because current can travel through more routes. Because more current flows for the same supply voltage, the total resistance decreases. Each new parallel branch reduces the overall opposition to current flow.

• Additional paths / routes are created for the current to flow (1m)
• Total current from the supply increases (more current flows for the same voltage) (1m)
• Total resistance decreases because resistance = voltage / current and current increases while voltage stays the same (or: more branches allow more current flow, reducing overall opposition) (1m)

Each parallel branch is an extra path for current. More paths means more total current flows from the supply for the same supply voltage. Since R = V/I, if V is constant and I increases, R must decrease. This is why total parallel resistance is always smaller than the smallest individual resistor.

First advantage: if one lamp fails in a parallel circuit, the other lamps continue to work because each lamp has its own independent path to the supply. In a series circuit, if one lamp fails the circuit is broken and all lamps go off. Second advantage: each lamp in a parallel circuit receives the full supply voltage, so all lamps operate at full brightness. In series, the supply voltage would be shared between lamps, making them dimmer.

• If one lamp fails in parallel, the other lamps continue to work / circuit is not broken for other lamps (1m)
• In series, a single lamp failure breaks the circuit for all lamps (comparison or explanation given) (1m)
• Each lamp in parallel receives the full supply voltage so all lamps operate at full brightness (contrast with series where voltage is shared and lamps are dim) (1m)

Parallel wiring is used in domestic circuits because: (1) each lamp is independently connected so a failure in one does not affect others - in series a single failure breaks the circuit for all lamps; (2) each lamp receives the full mains supply voltage so operates at full brightness - in series, voltage is shared across lamps making each one dimmer.

In a series circuit there is only one path for current to flow. All the charge must pass through every resistor one after another. Each resistor opposes the current, so the total opposition to current flow (total resistance) is the sum of the individual resistances: R_total = R1 + R2.

• In a series circuit there is only one path for current to flow / charge must pass through every component (1m)
• Each resistor adds opposition to the current flow / each resistor limits the current (1m)
• The total resistance equals the sum of individual resistances: R_total = R1 + R2 (accept correct formula or equivalent statement) (1m)

One path = each charge passes every resistor. Each resistor adds its opposition. Therefore R_total = R1 + R2 + ... in series.

Adding an extra lamp in parallel provides an additional path for current to flow. The total current from the supply increases because more current flows through the new branch. The potential difference across each lamp remains the same, so the brightness of the original lamps is unchanged.

• Total current from the supply increases (more branches / additional path for current) (1m)
• Brightness of original lamps is unchanged because the potential difference across each branch remains the same (1m)

In parallel, each branch has the same PD across it. Adding a branch means an extra path for current, increasing total current from the supply. However, the PD across every original lamp is unchanged, so they draw the same current and have the same brightness.

In the parallel circuit, each bulb receives the full supply voltage and glows at full brightness. In the series circuit, the supply voltage is shared between the two bulbs, so each bulb receives less voltage and glows more dimly.

• Bulbs in parallel are brighter than bulbs in series (accept: series bulbs are dimmer) (1m)
• In parallel each bulb receives the full supply voltage; in series the voltage is shared / divided between the bulbs (1m)

Parallel: each bulb gets full supply voltage (V_supply across each branch) = full brightness. Series: supply voltage shared equally between two identical bulbs = each gets V_supply/2 = dimmer.

When an additional resistor is added in parallel, the total current from the supply increases. Adding a parallel branch provides an extra path for current, reducing the overall resistance and allowing more current to flow from the supply.

• The total current from the supply increases (1m)
• Because an extra parallel branch provides an additional path for current / total resistance decreases so more current flows (accept reference to I = V/R) (1m)

More parallel branches = more paths for current = lower total resistance = higher total current (I = V/R with V constant). Each added parallel resistor reduces overall resistance.

In a series circuit there is only one path for charge to flow. Charge cannot be created or destroyed, so the current (rate of flow of charge) is identical at every point in the loop.

In a parallel circuit every branch is connected directly between the same two nodes (the positive and negative terminals of the supply), so the potential difference across each branch equals the supply voltage.

For resistors in series, total resistance = R1 + R2 = 4 + 6 = 10 ohm. The total is simply the sum of the individual values.

In a series circuit there is only one path for charge to flow. Because charge is conserved, the same amount of charge passes every point in the circuit each second — so the current (rate of flow of charge) is the same at every component.

In a parallel circuit the total current from the supply equals the sum of the currents in all branches: 6 = 4 + I2, so I2 = 2 A.

In a series circuit the sum of the potential differences across all components equals the supply voltage: 12 = 8 + V2, so V2 = 4 V.

In a parallel circuit, all branches are connected directly across the supply. The voltage across each branch is equal to the supply voltage — in this case 12 V for every branch, regardless of how many branches there are.

The student's statement is correct. Energy is wasted in the National Grid because the transmission cables have resistance. When current flows through a resistance, heat is produced according to P = I squared R, so heat loss depends on the square of the current. To minimise this, the National Grid transmits electricity at very high voltages, up to 400,000 V. For a given power, higher voltage means lower current, and since heat loss depends on I squared, even a small reduction in current causes a large reduction in heat loss. A step-up transformer at the power station increases the voltage before transmission. A step-down transformer near homes reduces the voltage to a safe 230 V for domestic use. The advantage of this design is that it greatly improves efficiency and allows large amounts of power to be transmitted over long distances with minimal loss. The disadvantages are that very high voltages are extremely dangerous to the public, and the transformers and high-voltage infrastructure are expensive to build and maintain. Because the cables always have some resistance, efficiency can never reach 100% — some energy will always be dissipated as heat.

• Level 3 (5-6 marks): Coherent explanation of energy losses (P = I²R in cables), clear description of high voltage/low current strategy, step-up and step-down transformers correctly identified, advantages (efficiency) and disadvantages (high voltage is dangerous, expensive infrastructure) discussed with logical structure. (2m)
• Level 2 (3-4 marks): Mostly correct explanation of energy losses and the high voltage strategy, transformers mentioned, but explanation not fully sequenced or one aspect missing. (2m)
• Level 1 (1-2 marks): Some relevant physics included (e.g., energy wasted as heat, or transformers used) but explanation is incomplete, not clearly linked, or contains errors. (2m)

A full Level 3 answer should: (1) identify that cables have resistance so current produces heat losses (P = I²R), (2) explain that high voltage reduces current which dramatically reduces heat losses, (3) correctly describe step-up transformer at power station and step-down near homes, (4) discuss efficiency advantages and the practical disadvantages (danger of high voltage, cost of infrastructure). Efficiency can never be 100% because heat losses cannot be completely eliminated.

A step-up transformer at the power station increases the voltage to hundreds of thousands of volts for transmission through the National Grid cables. High voltage means a low current flows through the cables. Since energy loss as heat is proportional to I²R, a lower current greatly reduces energy wasted. At the consumer end, a step-down transformer reduces the voltage to a safe level of 230 V for use in homes.

• Step-up transformer increases the voltage at the power station end (1m)
• High voltage transmission means low current in the cables (1m)
• Low current reduces energy lost as heat in the cables (P = I²R or heat loss proportional to current squared) (1m)
• Step-down transformer reduces voltage to safe level (230 V) for homes (1m)

The full National Grid process: power station generates electricity at moderate voltage -> step-up transformer raises to ~400,000 V -> cables carry low current at high voltage -> very little energy lost as heat (P = I²R is tiny with low I) -> step-down transformer near homes reduces to 230 V -> safe for domestic use.

• Correct substitution into P = I²R: P = 400² x 0.5 (1m)
• Correct power lost calculation: P = 160,000 x 0.5 = 80,000 W (1m)
• Correct percentage calculation: (80,000 / 200,000) x 100 (1m)
• Correct final answer: 40% (1m)

Step 1 - Power lost: P = I²R = 400² x 0.5 = 160,000 x 0.5 = 80,000 W. Step 2 - Percentage: (80,000 / 200,000) x 100 = 40%. This shows why the National Grid uses much higher voltages to reduce current and therefore reduce these losses dramatically.

• Correct substitution: P = 4² x 25 (1m)
• Correct squaring of current: 4² = 16 (1m)
• Correct answer: 400 W (1m)

P = I² x R = 4² x 25 = 16 x 25 = 400 W. A common mistake is forgetting to square the current before multiplying by resistance.

• Correct conversion of time: 3 hours = 10800 s (1m)
• Correct substitution: E = 120 x 10800 (1m)
• Correct answer: 1,296,000 J (1m)

E = P x t. First convert time: 3 hours x 3600 = 10,800 s. Then E = 120 x 10,800 = 1,296,000 J. Note: The question asks for joules so the time must be in seconds.

• Correct substitution: E = 500 x 12 (1m)
• Correct identification of units (charge in coulombs, V in volts, E in joules) (1m)
• Correct answer: 6000 J (1m)

E = Q x V = 500 x 12 = 6000 J. This equation links the energy transferred to the amount of charge that flows and the potential difference it flows through.

• Correct conversion factor used: 1 kWh = 3,600,000 J (1m)
• Correct multiplication: 15 x 3,600,000 (1m)
• Correct answer: 54,000,000 J (or 5.4 x 10^7 J) (1m)

Energy = 15 kWh x 3,600,000 J/kWh = 54,000,000 J = 5.4 x 10^7 J. One kWh is the energy used by a 1 kW appliance in 1 hour (1000 W x 3600 s = 3,600,000 J).

The equation P = I²R shows that power dissipated as heat depends on the resistance of the heater element and the square of the current. A higher resistance dissipates more power as heat for the same current. The squaring of current means small increases in current cause large increases in power, making the resistance value critical in heater design.

• Power dissipated depends on the resistance of the heater element (higher resistance = more power as heat) (1m)
• Power depends on the square of the current (doubling current quadruples power) (1m)
• The equation tells us electrical energy is converted to heat (dissipated as thermal energy) (1m)

P = I²R is useful for heaters because it directly links power output to resistance and current without needing the voltage. Higher resistance means more power as heat. Squaring current means the power-current relationship is non-linear: double the current gives four times the heat output.

• Correct substitution: P = 230 x 8 (1m)
• Correct answer: 1840 W (1m)

P = V x I = 230 x 8 = 1840 W. The kettle has a power rating of 1840 W (or 1.84 kW).

At high voltage the current in the cables is low. Energy lost as heat in the cables depends on I²R, so a lower current means much less energy is wasted as heat.

• High voltage results in a lower current in the cables (1m)
• Lower current means less energy wasted as heat in the cables (because P = I²R) (1m)

Using P = VI, if power is fixed and voltage increases, current must decrease. Heat loss in cables is P = I²R; squaring a smaller current gives much less heat loss.

A step-up transformer increases the voltage from the power station to a much higher voltage for transmission. This increase in voltage reduces the current in the transmission cables, which reduces energy wasted as heat.

• Step-up transformer increases (steps up) the voltage (1m)
• This reduces the current in the transmission cables (reducing heat losses) (1m)

A step-up transformer increases the voltage (and reduces current proportionally). For the same power transmitted, V x I is constant, so higher V means lower I. This is the key reason the National Grid uses very high voltages.

Electrical power is measured in watts (W). One watt is defined as one joule of energy transferred per second. Joules measure energy, amperes measure current, and volts measure potential difference.

The equation P = I²R gives the power dissipated in a resistor. Substituting I = 2 A and R = 10 Ω gives P = 2² x 10 = 4 x 10 = 40 W. Option D (P = I x R) gives voltage (V = IR), not power.

At high voltage, the current in the transmission cables is low (since P = VI, a higher V means a lower I for the same power). Energy lost as heat in cables is P = I²R, so reducing current dramatically reduces heat losses. Option B is the opposite of what happens. Option D is wrong because increased resistance would cause MORE waste.

A kilowatt-hour (kWh) is a unit of energy, not power. It equals the energy used by a 1 kilowatt appliance running for 1 hour. 1 kWh = 3,600,000 J = 3.6 MJ. It is used on electricity bills because joules are too small for household energy use.

Both fuses and circuit breakers are designed to break the circuit automatically when the current becomes too large, protecting wiring and appliances from damage and reducing the risk of fire or electric shock. A fuse contains a thin wire that melts when the current exceeds the fuse rating. This permanently breaks the circuit. A fuse is simple and cheap to manufacture. However, once a fuse blows, it must be replaced, which takes time and requires having spare fuses available. Fuses are also slightly slower to operate than circuit breakers. A circuit breaker uses an electromagnet or a bimetallic strip to detect excess current. When the current is too high, the mechanism trips and opens a switch, breaking the circuit. The main advantage of a circuit breaker is that it can be reset simply by flipping a switch once the fault is fixed, making it much more convenient. Circuit breakers also react faster than fuses, offering greater protection. Fuses are cheap and suitable for appliances where cost is a priority and replacement is straightforward. Circuit breakers are preferred in consumer units in homes and in situations where quick restoration of supply is important. Overall, circuit breakers offer greater convenience and speed, though fuses remain appropriate for protecting individual appliances.

• Level 3 (5-6 marks): A detailed and logically structured comparison that correctly describes how both devices work (melting wire vs. electromagnet/bimetallic strip trip), clearly states what happens after each operates (replace vs. reset), identifies at least two advantages and one disadvantage of each, and evaluates their suitability for different contexts. Scientific terminology used correctly throughout. (6m)
• Level 2 (3-4 marks): A mostly correct comparison that describes how at least one device works and mentions reset vs. replace difference. Some evaluation of suitability attempted but not fully developed or with minor inaccuracies. (4m)
• Level 1 (1-2 marks): Some relevant points made about fuses or circuit breakers but comparison is incomplete or superficial. May correctly state one function of each device. Little or no evaluation. (2m)
• Level 0 (0 marks): No relevant content, or only vague/incorrect statements. (1m)

Fuses work by melting (one-use, must replace). Circuit breakers trip a switch (reusable, just reset). Circuit breakers are faster and more convenient. Fuses are cheaper and still widely used for individual appliances. Both protect against excess current by breaking the circuit.

An AC supply produces a regularly repeating sine wave on the oscilloscope trace, alternating above and below the centre line. This shows that the voltage (and current) changes direction repeatedly. A DC supply produces a flat horizontal straight line on the oscilloscope trace, which shows that the voltage is constant and the current always flows in the same direction.

• AC trace shows a regularly repeating wave / sine wave shape / oscillates above and below the centre line (1m)
• DC trace shows a flat horizontal straight line / constant level (1m)
• The difference in shape shows AC current repeatedly changes direction / DC current always flows in the same direction / DC has constant voltage (1m)

On an oscilloscope: AC gives a smooth sinusoidal wave (going above and below the zero line) because the voltage — and therefore current — reverses direction many times per second. DC gives a flat horizontal line because the voltage is constant and current always flows in the same direction. The vertical position of the DC line above or below the centre shows the polarity and magnitude of the voltage.

The loose live wire makes contact with the metal case. The earth wire provides a low-resistance path to earth, so a large current flows through the earth wire. This large current exceeds the fuse rating, causing the fuse wire to melt and break the circuit. The electricity supply is cut off, preventing the user from receiving an electric shock.

• The earth wire provides a low-resistance path to earth (or: the earth wire connects the metal case to earth) (1m)
• A large current flows through the earth wire (because the resistance is low) (1m)
• The large current exceeds the fuse rating, causing the fuse to melt/blow, breaking the circuit (so the user is protected from electric shock) (1m)

Earth wire + fuse work as a safety team: the earth wire provides a low-resistance path so a dangerously large current flows when a fault occurs. That large current blows the fuse, breaking the circuit. Without the earth wire, the case would remain live and anyone touching it could receive a fatal shock.

• Correct substitution: I = P / V = 2300 / 230 (1m)
• Correct current: I = 10 A (1m)
• Correct fuse selected: 13 A fuse (the next standard rating above 10 A) (1m)

I = P / V = 2300 / 230 = 10 A. The fuse must have a rating just above the normal operating current, so a 13 A fuse is correct. A 3 A or 5 A fuse would blow in normal use; a fuse rated much higher would fail to protect against a fault.

Double insulation means the appliance has two layers of insulating material surrounding all the internal live parts. The outer casing is made of plastic or another non-conducting material. Because no conducting part can be touched by the user, even if the internal insulation fails, the user cannot receive an electric shock. Therefore, no earth wire is needed.

• Two layers of insulation / the appliance has double layers of insulating material around live parts (1m)
• The outer casing is made of non-conducting material (e.g. plastic), so no conducting surface can be touched by the user (1m)
• No earth wire is needed because even if the inner insulation fails, the outer layer prevents the user from making contact with any live part (1m)

Double insulation provides two independent barriers between live components and the user. Because the outer casing is non-conducting, the user can never touch a live surface even if the inner insulation fails. Without any exposed conducting surfaces, earthing is not needed.

The mains supply is AC, so its voltage alternates between +325 V and -325 V, spending time at lower voltages during each cycle. The 230 V figure is the root mean square (RMS) voltage. The RMS voltage is the equivalent DC voltage that would deliver the same power (same heating effect) to a resistor as the AC supply. Because the AC voltage is continually changing, it delivers less power on average than a constant 325 V DC supply would.

• 230 V is the RMS (root mean square) voltage (1m)
• RMS is the equivalent DC voltage that would deliver the same power / same heating effect to a resistor (1m)
• The AC voltage varies through the cycle so the average power delivered is less than that from a constant 325 V DC supply / 325 V is only reached momentarily (1m)

For AC, the voltage is constantly changing — it reaches 325 V only at the peaks. For most of each cycle, the voltage is less than 325 V, so an AC supply at 325 V peak delivers less power on average than a 325 V DC supply. The RMS (root mean square) value of 230 V is the equivalent DC voltage that would deliver exactly the same average power. This is why mains supplies are rated by their RMS voltage, not their peak voltage.

• Correct substitution: T = 1 / 50 (1m)
• Correct answer: T = 0.02 s (1m)

T = 1 / f = 1 / 50 = 0.02 s. The period is the time for one complete cycle of the AC supply.

• Correct substitution: f = 1 / 0.025 (1m)
• Correct answer: f = 40 Hz (1m)

f = 1 / T = 1 / 0.025 = 40 Hz. The period and frequency are reciprocals of each other.

The fuse contains a thin wire that melts when the current exceeds the fuse rating. This breaks the circuit, stopping the current and preventing damage to the appliance or wiring.

• The fuse wire melts (or breaks / blows) when the current exceeds the fuse rating (1m)
• This breaks the circuit, stopping the current from flowing (protecting the appliance or wiring) (1m)

A fuse is a safety device containing a thin wire. If the current in the circuit rises above the fuse rating (e.g., due to a fault or overload), the wire melts and breaks the circuit, cutting off the electricity supply.

• Correct period: T = 4 × 5 ms = 20 ms = 0.02 s (1m)
• Correct frequency: f = 1 / 0.02 = 50 Hz (1m)

Period T = 4 divisions × 5 ms/division = 20 ms = 0.020 s. Frequency f = 1 / T = 1 / 0.020 = 50 Hz. This is the standard UK mains frequency. A common mistake is forgetting to convert milliseconds to seconds before dividing — 1 / 20 gives 0.05, which is wrong.

AC (alternating current) repeatedly reverses direction, while DC (direct current) flows in one direction only. UK mains supply is AC; batteries supply DC.

The UK mains supply is 230 V alternating current at a frequency of 50 Hz. This means the current completes 50 full cycles per second.

In a UK three-pin plug: live wire is brown, neutral wire is blue, and earth wire is green and yellow striped. A useful memory aid is BLue = Bottom Left (neutral), BRown = Bottom Right (live).

The live wire is brown. It is the most dangerous wire because it carries the high voltage (230 V), so touching it could cause a fatal electric shock.

• Live wire is brown AND it carries high voltage (230 V) / could cause electric shock / death (1m)

The live wire (brown) alternates between +325 V and -325 V (peak), with a root-mean-square voltage of 230 V. It carries the electrical energy to the appliance. Touching it provides a path for current to flow through the body to earth, which can be fatal.

When the live wire touches the metal case, the earth wire gives the current a low-resistance path to earth. This causes a large current to flow, which blows the fuse (or trips the circuit breaker), cutting off the electricity supply and protecting the user.

Circuit breakers can be reset by flipping a switch after the fault is fixed, making them more convenient than fuses, which must be replaced once blown. Circuit breakers also react faster than fuses.

The peak-to-peak voltage is the total vertical distance from the bottom of the wave to the top. Since the peak voltage (centre to top) is 6 V, the trough is 6 V below the centre. Peak-to-peak = 6 + 6 = 12 V. A common error is confusing peak voltage with peak-to-peak voltage — peak-to-peak is always twice the peak.

Set up a circuit with the component (for example a resistor or filament lamp) connected in series with an ammeter and a variable resistor (or variable power supply). Connect a voltmeter in parallel across the component. Vary the voltage using the variable resistor or power supply and record both the current (from the ammeter) and the voltage (from the voltmeter) for each setting. Record results in a table. Reverse the connections of the power supply to obtain readings for negative voltages and negative currents. Repeat each reading to improve reliability. Plot a graph of current (y-axis) against voltage (x-axis). The shape of the graph reveals whether the component is ohmic (straight line through origin, constant resistance) or non-ohmic (curved line, changing resistance). For an ohmic resistor the gradient equals I divided by V, which equals one divided by resistance.

• Level 3 (5-6 marks): Circuit described with variable resistor (or power supply with variable output), ammeter in series, voltmeter in parallel. Voltage varied and I and V recorded for each setting. Voltage reversed to get negative values. Measurements repeated for reliability. Graph plotted (I on y-axis, V on x-axis). Shape of graph interpreted for resistance and ohmic/non-ohmic behaviour. (6m)
• Level 2 (3-4 marks): Basic circuit described, measurements explained but method incomplete (missing reversal, or no graph description). (4m)
• Level 1 (1-2 marks): Equipment listed but method unclear or wrong connections described. (2m)

RPA4 method: variable power supply (or rheostat), ammeter in series, voltmeter in parallel. Vary voltage, record I and V. Reverse connections. Plot I vs V. Gradient = 1/R for ohmic conductor.

A thermistor is a non-ohmic component whose resistance decreases as temperature increases. At low temperature it has high resistance so current is low for a given voltage. As temperature increases, resistance decreases so more current flows for the same voltage. On an I-V graph at higher temperatures, the gradient is steeper (more current for the same voltage, indicating lower resistance).

• Thermistor resistance decreases as temperature increases (1m)
• At higher temperature, more current flows for the same voltage (lower resistance) (1m)
• I-V graph shows non-linear (steeper gradient at higher temperature) relationship (1m)

Thermistor: NTC (negative temperature coefficient) - resistance falls as T rises. More current at same voltage = steeper I-V gradient = lower resistance.

• Correct equation: R = V/I (1m)
• Correct substitution: R = 6 / 0.4 (1m)
• Correct answer: R = 15 ohm (1m)

R = V/I = 6 / 0.4 = 15 ohm. A straight-line I-V graph means constant resistance - this is an ohmic conductor.

In forward bias (positive voltage in the direction the diode conducts), current flows once the voltage exceeds a threshold of about 0.7 V. In reverse bias (voltage applied in the opposite direction), the resistance of the diode is very high and almost no current flows. The I-V graph shows a sharp increase in current in forward bias but essentially zero current in reverse bias.

• Forward bias: current flows in the forward direction when threshold voltage exceeded (~0.7 V for silicon) (1m)
• Reverse bias: voltage applied in opposite direction; very high resistance; almost no current flows (1m)
• I-V graph description: steep rise in forward bias, essentially zero current in reverse bias (1m)

Diode I-V: forward bias -> current flows after ~0.7V threshold; reverse bias -> essentially zero current (very high resistance). The graph is highly asymmetric.

• Gradient = I/V = 1/R, so R = 1/gradient (1m)
• R = 1 / 0.05 = 20 ohm (1m)
• Correct unit: ohm (Ω) (1m)

Gradient of I-V graph = I/V = 1/R. Therefore R = 1/gradient = 1/0.05 = 20 ohm.

The ammeter must be connected in series with the component so that the same current flows through both. The voltmeter must be connected in parallel across the component to measure the potential difference across it. The ammeter has very low resistance so it does not significantly affect the current, and the voltmeter has very high resistance so it draws negligible current.

• Ammeter connected in series with the component (same current flows through both) (1m)
• Voltmeter connected in parallel across the component (measures PD across it) (1m)
• Ammeter has low/negligible resistance; voltmeter has very high resistance so measurements are not affected (1m)

Ammeter: series (measures current through component). Voltmeter: parallel (measures PD across component). Correct connections are essential for accurate I-V measurements.

As the voltage increases, the current increases, which makes the filament heat up. As the temperature of the filament increases, its resistance increases. Because resistance is increasing, the current increases less than expected for each increase in voltage, so the graph curves (flattens) and is not a straight line.

• As current increases, the filament heats up / temperature increases (1m)
• Higher temperature increases resistance, so current does not increase proportionally with voltage (curved / non-linear I-V graph) (1m)

Filament: current -> heat -> higher temperature -> higher resistance -> non-proportional increase in current -> curved graph.

• I = V/R = 6/200 (1m)
• I = 0.03 A (1m)

I = V/R = 6/200 = 0.03 A (30 mA). In the dark: I = 6/20000 = 0.0003 A - much smaller current due to high resistance.

An I-V (current-voltage) characteristic graph plots current on the y-axis against voltage on the x-axis, showing how current through a component depends on the voltage across it.

A diode only allows significant current to flow in one direction (forward bias). In reverse bias, resistance is very high and current is essentially zero. Its I-V graph shows this asymmetric behaviour.

An ohmic conductor is one that has a constant resistance regardless of the voltage applied. The current through it is directly proportional to the voltage across it at constant temperature.

• Current is directly proportional to voltage / resistance is constant (at constant temperature) (1m)

Ohmic conductor: current directly proportional to voltage at constant temperature. Constant resistance. Straight-line I-V graph through origin.

A straight line through the origin on an I-V graph shows current is directly proportional to voltage. This means the resistance is constant and the component obeys Ohm's law (ohmic conductor).

As the voltage increases, the filament gets hotter, increasing resistance. The curved I-V graph (flattening at higher voltages) shows current increasing less and less for each extra volt - indicating increasing resistance.

• Place the bar magnet on a piece of paper and draw around it (1m)
• Place the plotting compass at one end of the magnet (near the north pole) (1m)
• Mark the position of the compass needle tip (or tail) with a dot (1m)
• Move the compass so its tail is at the previous dot; mark the new needle tip; repeat to trace a field line (1m)
• Lift the compass and start a new line at a different position around the magnet (1m)
• Arrows on field lines show direction from north to south pole outside the magnet (1m)

Level 3 (5-6 marks): Valid field pattern obtained; steps clearly sequenced including: draw magnet outline, start near pole, mark dots, move compass tail to previous dot tip, repeat, add arrows. Level 2 (3-4 marks): Most steps identified but not fully sequenced. Level 1 (1-2 marks): Some relevant equipment or steps mentioned.

An electromagnet works by passing an electric current through a coil of wire. The current creates a magnetic field around the coil. Adding an iron core inside the coil increases the strength of the magnetic field. The strength can be increased by increasing the current flowing through the coil or by increasing the number of turns in the coil.

• Current flowing through the coil produces a magnetic field (1m)
• Increasing the current increases the strength of the field (1m)
• Increasing the number of turns (coils) of wire increases the strength (1m)

An electromagnet relies on the magnetic effect of a current. A coil of wire carrying current acts like a bar magnet. Strength increases with more current (higher current = stronger field) or more turns of wire (each turn adds to the total field).

• Correct rearrangement: I = F / (B x l) (1m)
• Correct substitution: I = 0.60 / (0.50 x 0.40) (1m)
• Correct answer: I = 3 A (1m)

Rearranging F = BIl to find current: I = F / (B x l) = 0.60 / (0.50 x 0.40) = 0.60 / 0.20 = 3 A.

Magnetic field lines show the direction of the magnetic field at each point — the direction in which a free north pole would move. Where the field lines are closer together the field is stronger. Where the field lines are further apart the field is weaker.

• Field lines show the direction of the magnetic field (direction a north pole would move) (1m)
• Closer field lines indicate a stronger magnetic field (1m)
• Further apart field lines indicate a weaker magnetic field (1m)

Field lines are a visual tool. Their direction shows which way the field acts (defined as the direction a free north pole would move). Their spacing shows the field strength — closely-spaced lines mean the field is strong, widely-spaced lines mean the field is weak.

A permanent magnet always produces a magnetic field and cannot be switched off. An electromagnet only produces a field when current flows through it, so it can be switched on and off. This is an advantage in industrial applications such as cranes used to lift and drop metal objects — the electromagnet can be turned off to release the metal.

• An electromagnet can be switched on and off (by controlling the current) (1m)
• A permanent magnet cannot be switched off (always has a magnetic field) (1m)
• Valid industrial advantage stated (e.g. electromagnet crane can release metal by turning off current) (1m)

The key difference is control. An electromagnet can be switched on and off — useful when you need to pick up and drop metal objects (e.g. in a scrap metal yard). A permanent magnet is always magnetised, which is useful when you always need the field but inconvenient when you need to release objects.

• Correct substitution: F = 0.30 x 4.0 x 0.25 (1m)
• Correct calculation: F = 0.30 N (1m)
• Correct unit: N (newtons) (1m)

Using F = BIl: F = 0.30 x 4.0 x 0.25 = 0.30 N. The equation F = BIl gives the force on a current-carrying conductor in a magnetic field when the wire is perpendicular to the field.

Like poles repel each other. Unlike poles attract each other.

• Like poles repel (same poles push away from each other) (1m)
• Unlike poles attract (opposite poles pull toward each other) (1m)

Like poles (N-N or S-S) repel because their magnetic field lines push against each other. Unlike poles (N-S) attract because their field lines connect and pull the magnets together.

• Field strength doubles (increases by a factor of 2) (1m)
• Correct reasoning: each turn contributes to the field, so more turns = proportionally stronger field (1m)

The magnetic field strength of a solenoid is directly proportional to the number of turns. If the number of turns doubles (with current constant), the field strength doubles (factor of 2).

Soft iron is used because it is easily magnetised when current flows and easily loses its magnetism when the current is switched off. Steel is a hard magnetic material that retains its magnetism — this is not useful when you need the electromagnet to switch off cleanly.

• Soft iron loses its magnetism when the current is switched off (easily demagnetised) (1m)
• Steel retains its magnetism (stays magnetised even when current is off), which is undesirable for an electromagnet core (1m)

Soft iron is a magnetically soft material — it magnetises and demagnetises very easily. This is ideal for an electromagnet core because you want the magnetism to disappear when the current is switched off. Steel (hard magnetic material) keeps its magnetism, so it would not switch off cleanly.

Like poles repel. When two north poles (or two south poles) are brought together, the magnetic fields push them apart. Unlike poles (north and south) attract.

Magnetic field lines run from north to south outside the magnet (and from south to north inside the magnet). The direction of the field line is the direction a free north pole would move.

A solenoid is a coil of wire that produces a magnetic field when an electric current flows through it. The field inside the solenoid is uniform and resembles the field of a bar magnet.

A compass needle aligns with the local magnetic field. The north end of the compass points in the direction of the field. If the compass points toward the south pole of the magnet, the field direction at that point is toward the south pole — which is consistent with field lines entering a south pole.

An electric motor converts electrical energy into kinetic energy. A coil carrying current is placed in a magnetic field. Each side of the coil experiences a force given by F = BIl (the motor effect). The forces on opposite sides of the coil act in opposite directions, creating a turning effect (torque) that makes the coil rotate. A commutator reverses the direction of the current every half turn so that the forces continue to turn the coil in the same direction throughout its rotation. To increase the torque: (1) Increase the current — since F = BIl, a larger current produces a larger force on each side, giving a greater turning effect. (2) Increase the magnetic flux density — a stronger magnetic field also increases the force on each side via F = BIl. Alternatively, increasing the number of turns in the coil increases the total force as each turn contributes additional force.

• The motor effect: current in a coil inside a magnetic field experiences a force (F = BIl) (1m)
• Forces on opposite sides of the coil act in opposite directions, creating a turning effect (torque) (1m)
• The commutator reverses current every half turn to maintain rotation in the same direction (1m)
• Increasing the current through the coil increases the force (F = BIl, so larger I → larger F) (1m)
• Increasing the magnetic flux density increases the force (larger B → larger F) (1m)
• Increasing the number of turns in the coil increases the total force on the coil (1m)

Level 3 (5-6 marks): Clear logical sequence covering the motor effect, the creation of torque from opposing forces on the coil sides, the role of the commutator, and two specific, well-reasoned ways to increase torque. Level 2 (3-4 marks): Motor effect described and at least one way to increase torque identified but reasoning incomplete. Level 1 (1-2 marks): Some relevant physics mentioned but not clearly connected.

An alternating current passes through the coil attached to the speaker cone. The coil is inside a permanent magnetic field. The motor effect produces a force on the coil. Because the current is alternating, the force on the coil repeatedly reverses direction, causing the coil and attached cone to vibrate back and forth. These vibrations of the cone create pressure waves in the air, which we hear as sound.

• Alternating current flows through the coil (which is in a magnetic field) (1m)
• The motor effect produces a force on the coil (1m)
• The alternating current causes the force to repeatedly reverse direction, making the coil vibrate (1m)
• The vibrating cone creates pressure waves (compressions and rarefactions) in the air, which we hear as sound (1m)

Key chain: AC current → alternating force (motor effect) → coil vibrates → cone vibrates → pressure waves in air → sound. The frequency of the AC matches the frequency of vibration and hence the frequency (pitch) of the sound produced.

• Correct substitution: F = 0.40 x 6.0 x 0.15 (1m)
• Correct calculation: F = 0.36 N (1m)
• Correct unit: N (newtons) (1m)

F = BIl = 0.40 x 6.0 x 0.15 = 0.36 N. The force is 0.36 newtons. Remember the wire must be perpendicular to the field for this equation to apply directly.

• Correct rearrangement: B = F / (I x l) (1m)
• Correct substitution: B = 0.80 / (5.0 x 0.20) (1m)
• Correct answer: B = 0.80 T (1m)

Rearranging F = BIl: B = F / (I x l) = 0.80 / (5.0 x 0.20) = 0.80 / 1.0 = 0.80 T.

A current-carrying coil is placed inside a magnetic field. The motor effect produces a force on each side of the coil. Because the forces act on opposite sides of the coil in opposite directions, this creates a turning effect (torque) which causes the coil to rotate. The commutator reverses the direction of current in the coil every half turn, keeping the coil rotating in the same direction.

• Force on the coil is due to the motor effect (current in magnetic field) (1m)
• Forces on opposite sides of the coil are in opposite directions, creating a turning effect (rotation) (1m)
• The commutator reverses the current direction every half turn, maintaining continuous rotation in the same direction (1m)

In an electric motor: 1) Current flows through a coil inside a magnetic field; 2) Motor effect creates forces on the sides of the coil; 3) The forces on opposite sides act in opposite directions, creating a turning effect; 4) The commutator swaps the current direction every half turn so the coil always turns the same way rather than oscillating.

• Correct calculation of force on one turn: F = 0.20 x 2.5 x 0.05 = 0.025 N (1m)
• Correct multiplication by number of turns: 0.025 x 80 (1m)
• Correct total force: 2.0 N (1m)

Force on one turn = BIl = 0.20 x 2.5 x 0.05 = 0.025 N. Total force on one side = 0.025 x 80 turns = 2.0 N. Each turn of the coil contributes to the total force.

You can reverse the direction of the force by reversing the direction of the current through the conductor, or by reversing the direction of the magnetic field. However, if both the current direction and the magnetic field direction are reversed at the same time, the force direction stays the same.

• Reversing the direction of the current through the conductor reverses the force direction (1m)
• Reversing the direction of the magnetic field also reverses the force direction (1m)
• Reversing both the current and the field simultaneously does not reverse the force (1m)

Using Fleming's left-hand rule: reversing either the current or the field alone reverses the force. But reversing both inputs together gives the same output direction — the two reversals cancel each other out.

When a current flows through a wire, moving charges (electrons) create their own magnetic field around the wire. When this wire is placed inside an external magnetic field, the two magnetic fields interact. The interaction between the wire's magnetic field and the external magnetic field produces a resultant force on the wire, acting perpendicular to both the current direction and the field direction.

• Current in the wire creates its own magnetic field (due to moving charges / electrons) (1m)
• The wire's magnetic field interacts with the external magnetic field (1m)
• This interaction produces a force on the wire acting perpendicular to both the current and the field (1m)

Current = moving charges. Moving charges create a magnetic field around the wire. This field interacts with the external field. The interaction produces a force perpendicular to both current direction and external field direction — this is the motor effect.

In Fleming's left-hand rule: the First finger points in the direction of the magnetic Field; the seCond finger points in the direction of the Current; and the thuMb points in the direction of the Motion (force).

• First finger = magnetic field direction AND second finger = current direction (1m)
• Thumb = direction of the force / motion / thrust (1m)

Fleming's left-hand rule mnemonic: thuMb = Motion (force); First finger = Field; seCond finger = Current. Hold the left hand so all three are at right angles to each other.

Hold the left hand so the first finger points in the direction of the magnetic field, the second finger points in the direction of the conventional current. The thumb then points in the direction of the force on the conductor.

• First finger represents the direction of the magnetic field; second finger represents the direction of conventional current (1m)
• The thumb points in the direction of the force (motion) on the conductor (1m)

Fleming's left-hand rule is a mnemonic for the motor effect. Left hand: First finger = Field, seCond finger = Current, thuMb = Motion (force). Point first two fingers and thumb at right angles to each other — they indicate the three perpendicular directions.

If the direction of the current is reversed, the direction of the force on the conductor is also reversed. The force acts in the opposite direction to before.

• The direction of the force is reversed when the current is reversed (1m)
• The force acts in the opposite direction (accept: the conductor moves in the opposite direction) (1m)

From F = BIl: the force magnitude stays the same, but reversing the current direction means Fleming's left-hand rule gives the opposite direction for the thumb (force). The force direction reverses.

If the current is increased, the force on the conductor increases. The force is directly proportional to the current, so doubling the current doubles the force.

• The force increases when the current is increased (1m)
• The force is directly proportional to the current (accept: reference to F = BIl showing F increases with I) (1m)

From F = BIl, if B and l are constant, F is directly proportional to I. Increasing current increases the force by the same factor.

The motor effect is the force experienced by a current-carrying conductor in a magnetic field. This force is what drives an electric motor. Option B describes electromagnetic induction (the generator effect), not the motor effect.

In Fleming's left-hand rule: the First finger = magnetic Field direction; the seCond finger = conventional Current direction; the thuMb = direction of Motion (force on the conductor). A common memory trick: FBI — Field, Current (I), force (motion).

Fleming's left-hand rule gives the direction of the force (motion) on a current-carrying conductor in a magnetic field. The First finger = Field direction; the seCond finger = Current direction; the thuMb = Motion (force) direction.

Using Fleming's left-hand rule: the First finger points in the direction of the magnetic Field (upward), the seCond finger points in the direction of conventional Current (east), and the thuMb points in the direction of the Motion (force). With field up and current east, the thumb points out of the page toward the viewer.

The force on a conductor is given by F = BIl. The force increases with greater current (I), greater magnetic flux density (B), or greater length of wire (l). Reversing the field direction reverses the force direction but does not change its magnitude — so the size of the force is unchanged.

From F = BIl, the force F is directly proportional to the magnetic field strength B (when current I and wire length l stay constant). If B is doubled, F doubles. Reversing the field direction would reverse the force, but simply increasing it doubles the force.

At the power station, a step-up transformer increases the voltage from around 25 kV to 400 kV for transmission through the National Grid. Because power = voltage x current, increasing the voltage means the current is lower for the same power. Lower current means less power is wasted as heat in the cables, since the power lost in a cable is given by P = I²R — halving the current reduces the power loss to one quarter. Near homes and businesses, step-down transformers reduce the voltage to a safe level of 230 V. This makes the system highly efficient because very little energy is wasted during transmission over long distances. Transformers only work with alternating current (AC) because the continuously changing current produces a changing magnetic field in the primary coil, which induces a voltage in the secondary coil by electromagnetic induction.

• Step-up transformer at the power station increases voltage (e.g. from 25 kV to 400 kV) for transmission (1m)
• High voltage means low current in transmission cables (for the same power) (1m)
• Low current means less power is lost as heat in the cables (P = I²R), so transmission is more efficient (1m)
• Step-down transformers reduce voltage to safe levels (e.g. 230 V) for use in homes and businesses (1m)
• Energy is transferred efficiently across long distances because the cables waste less power (1m)
• Transformers only work with AC — the changing current creates a changing magnetic field for electromagnetic induction (1m)

Level 3 (5-6 marks): Logically sequenced account covering step-up at power station, high voltage / low current argument with P = I²R reasoning, step-down before homes, efficiency benefit, and why AC is essential. Level 2 (3-4 marks): Most of the key ideas present but not fully linked or sequenced. Level 1 (1-2 marks): Some relevant terms used (step-up, step-down, voltage, current) but not connected into a coherent explanation.

In an AC generator, a coil of wire is rotated inside a magnetic field. As the coil rotates, it cuts through the magnetic field lines, so the magnetic field through the coil is continuously changing. By electromagnetic induction, this induces an alternating voltage in the coil. As the coil rotates the direction of the induced voltage reverses every half turn, producing alternating current. The slip rings maintain electrical contact with the external circuit while allowing the coil to rotate freely.

• The coil rotates inside a magnetic field (1m)
• The rotation causes the magnetic field through the coil to change, inducing a voltage by electromagnetic induction (1m)
• The voltage is alternating because the direction reverses every half turn of the coil (1m)
• Slip rings maintain electrical contact between the rotating coil and the external circuit (1m)

An AC generator (alternator) converts kinetic energy to electrical energy using electromagnetic induction. Key features: rotating coil + permanent magnetic field → changing flux → induced AC voltage. Slip rings (not commutator) allow AC output by maintaining contact without reversing current. A DC generator (dynamo) uses a commutator instead, which reverses connections every half turn to give DC output.

When the magnet moves into the coil, the magnetic flux through the coil increases. This changing flux induces an EMF and a current, causing the galvanometer to deflect in one direction (say, to the right). By Lenz's law, the induced current flows to oppose the entry of the magnet — the coil acts as if it is trying to repel the incoming magnet. When the magnet is stationary inside the coil, the flux is not changing, so no EMF is induced and the galvanometer returns to zero. When the magnet is pulled out, the flux decreases. An EMF is induced again, but this time in the opposite direction — the galvanometer deflects in the other direction (to the left). By Lenz's law, the induced current now opposes the withdrawal of the magnet — the coil acts as if it is trying to attract and retain the magnet. The size of the deflection is larger if the magnet moves faster (greater rate of change of flux).

• Moving in: galvanometer deflects in one direction (e.g. to the right) — an EMF is induced because the magnetic flux through the coil is changing (1m)
• Stationary: galvanometer reads zero — no relative motion means no change in flux, so no induced EMF or current (1m)
• Moving out: galvanometer deflects in the opposite direction — the flux is decreasing, so the induced current reverses direction to oppose the withdrawal (Lenz's law) (1m)
• Lenz's law applied: when moving in, the induced current opposes entry (coil repels magnet); when moving out, the induced current opposes withdrawal (coil attracts magnet). Faster movement gives a larger deflection. (1m)

Three phases: (1) Moving in — flux increases, EMF induced, galvanometer deflects. By Lenz's law, induced current opposes entry (coil repels magnet). (2) Stationary — flux constant, no EMF, galvanometer at zero. (3) Moving out — flux decreases, EMF induced in opposite direction, galvanometer deflects the other way. By Lenz's law, induced current opposes withdrawal (coil attracts magnet). Speed of movement affects the size (not direction) of deflection.

• Correct substitution: 240 / Vs = 200 / 50 (1m)
• Correct rearrangement: Vs = 240 x 50 / 200 (1m)
• Correct answer: Vs = 60 V (1m)

Vp/Vs = np/ns => Vs = Vp x (ns/np) = 240 x (50/200) = 240 x 0.25 = 60 V. Fewer turns on the secondary means a lower voltage — this is a step-down transformer.

• Correct rearrangement: Is = (Vp x Ip) / Vs (1m)
• Correct substitution: Is = (230 x 40) / 11500 (1m)
• Correct answer: Is = 0.8 A (1m)

Using VpIp = VsIs: Is = (Vp x Ip) / Vs = (230 x 40) / 11500 = 9200 / 11500 = 0.8 A. As voltage is stepped up, current is stepped down proportionally (energy is conserved in an ideal transformer).

An alternating current in the primary coil produces a changing magnetic field in the iron core. The changing magnetic field passes through the secondary coil. This changing magnetic field induces an alternating voltage in the secondary coil by electromagnetic induction.

• Alternating current in the primary coil creates a changing magnetic field in the iron core (1m)
• The changing magnetic field passes through and links to the secondary coil (1m)
• The changing magnetic field induces an alternating voltage in the secondary coil (by electromagnetic induction) (1m)

Energy is transferred magnetically, not electrically — the two coils are not electrically connected. AC in primary → changing field in iron core → changing field through secondary → induced AC voltage in secondary. The iron core concentrates and guides the magnetic field from primary to secondary.

The induced voltage can be increased by moving the magnet faster into the coil, using a magnet with a stronger magnetic field, or increasing the number of turns in the coil.

• Move the magnet faster (increase speed of relative motion) (1m)
• Use a stronger magnet (increase the magnetic flux density) (1m)
• Increase the number of turns in the coil (1m)

All three changes increase the rate of change of magnetic flux through the coil: faster movement increases the rate of change; a stronger magnet increases the amount of flux; more turns means more flux is cut per unit time.

• Correct rearrangement: ns = np x (Vs / Vp) (1m)
• Correct substitution: ns = 2000 x (20 / 400) (1m)
• Correct answer: ns = 100 turns (1m)

ns = np x (Vs/Vp) = 2000 x (20/400) = 2000 x 0.05 = 100 turns. The voltage is stepped down by a factor of 20 (400/20), so the number of turns is also reduced by the same factor: 2000/20 = 100 turns.

By Lenz's law, the induced current must oppose the entry of the north pole. To oppose the approaching north pole, the end of the coil facing the magnet must also become a north pole — this repels the incoming magnet. The induced current must therefore flow in a direction (anticlockwise when viewed from the magnet's side) that makes this end of the coil a north pole, consistent with the right-hand rule.

• Lenz's law: the induced current opposes the entry of the magnet / opposes the change in flux (1m)
• The end of the coil facing the approaching north pole becomes a north pole itself (repelling the magnet) (1m)
• The current flows anticlockwise when viewed from the magnet side (or correct description using Fleming's right-hand rule) (1m)

Lenz's law: the induced effect opposes its cause. A north pole entering the coil means the coil must repel it — so the facing end of the coil becomes a north pole. Like poles repel. Using the right-hand rule (or right-hand grip rule), a north pole at the face of the coil corresponds to anticlockwise current when viewed from the approaching magnet's side.

A potential difference is induced in a conductor when there is relative motion between the conductor and a magnetic field, or when the magnetic field through the conductor changes. This is called electromagnetic induction.

• There must be relative motion between the conductor and the magnetic field (or the magnetic field must be changing) (1m)
• This causes a potential difference (voltage) to be induced in the conductor (electromagnetic induction) (1m)

Electromagnetic induction requires a changing magnetic field through the conductor. This can happen by moving the conductor through the field, moving the magnet, or changing the field strength. The result is an induced potential difference (and if the circuit is complete, an induced current).

The National Grid transmits electricity at high voltage to reduce the current in the cables. A lower current means less power is lost as heat in the resistance of the transmission cables, making the system more efficient.

• High voltage means low current in the transmission cables (1m)
• Low current means less power is lost as heat (thermal energy) in the cables (1m)

Power = voltage x current (P = VI). For a fixed power, increasing voltage reduces current. Power lost in cables = I²R (current squared x resistance). Smaller current dramatically reduces wasted heat in the cables (it's proportional to I², so halving current reduces heat loss by 4).

Lenz's law states that the direction of an induced current is always such that it opposes the change that caused it. This means the induced current creates a magnetic field that acts against the motion or change in flux that produced it. This is consistent with conservation of energy — if the induced current helped the change rather than opposing it, energy would be created from nothing.

• The induced current direction opposes the change that caused it (or opposes the motion / change in magnetic flux) (1m)
• This is consistent with conservation of energy — the opposing force means work must be done to maintain the motion, which is the source of electrical energy (1m)

Lenz's law is a consequence of conservation of energy. The induced current must oppose the change (e.g. oppose the entry of a magnet) because if it aided the motion, it would accelerate the magnet, inducing a larger current, which would accelerate it further — creating energy from nothing. The opposition means the person moving the magnet must do work, which is converted to electrical energy.

Electromagnetic induction is the generation of a potential difference (voltage) in a conductor when there is relative motion between the conductor and a magnetic field, or when the magnetic field through the conductor changes. Option A describes the motor effect — the opposite process.

Using the transformer equation Vp/Vs = np/ns: Vs/Vp = ns/np = 500/100 = 5. So the secondary voltage is 5 times the primary voltage — it is stepped up by a factor of 5. A step-up transformer has more turns on the secondary than the primary.

The induced voltage is increased by: moving the magnet faster, using a stronger magnet, or using more turns of wire in the coil. A coil with more turns has more wire cutting through the magnetic field, so the total induced voltage is greater. Moving slower or using a weaker magnet would decrease the induced voltage.

A transformer works by electromagnetic induction. The primary coil must produce a changing magnetic field to induce a voltage in the secondary coil. AC constantly changes direction, so the magnetic field continuously changes. DC produces a steady (constant) magnetic field — a constant field does not induce a voltage. Without a changing field there is no induction and the transformer does not work.

By Lenz's law, the induced current must oppose the change that caused it — in this case, the approach of the magnet. The induced current therefore creates a magnetic field that repels the magnet (pushes it away from the coil). If the coil attracted the magnet instead, the magnet would accelerate towards the coil, inducing a larger current, which would attract it more strongly — creating energy from nothing and violating conservation of energy.

Electricity is transmitted at high voltage because power loss in cables is given by P equals I squared times R. For a fixed resistance in the cables, reducing the current massively reduces the power wasted as heat. A high voltage means a low current (because power equals V times I). At the power station, a step-up transformer increases the voltage from the generation voltage to hundreds of thousands of volts for transmission. This reduces the current, minimising energy wasted in the cables. At the end of the transmission line, step-down transformers at substations reduce the voltage to a safe level (230 V) for homes and businesses. The environmental benefit is that less fuel must be burned at power stations to supply the same amount of useful electrical energy to consumers, reducing carbon dioxide emissions. The economic benefit is lower electricity costs for consumers because less energy is wasted.

• Level 3 (5-6 marks): P_loss = I^2 R explained. High voltage -> low current -> less heating of cables. Step-up transformer increases V at power station. Step-down at substation restores safe level. Environmental benefits (less fuel burned), economic benefits (lower electricity costs). Quantitative example strengthens answer. (6m)
• Level 2 (3-4 marks): High voltage -> low current -> less power loss explained. Transformers mentioned. Limited or no environmental/economic evaluation. (4m)
• Level 1 (1-2 marks): Basic statement that high voltage reduces energy loss, limited explanation. (2m)

High-voltage transmission is a key engineering solution: P_loss = I^2 R, so reducing I massively reduces losses. Transformers enable voltage changes. Less energy waste = less fuel burned = lower CO2 emissions + lower bills.

• Current = P/V = 500 x 10^6 / 25000 = 20000 A (1m)
• Power loss = I^2 x R = (20000)^2 x 4 = 4 x 10^8 x 4 = 1.6 x 10^9 W (1m)
• This represents 1600/500 = 3.2 times the generated power being lost (clearly impractical) (1m)
• Correct final answer for power loss: 1.6 x 10^9 W (1600 MW) (1m)

I = 5e8 / 25000 = 20000 A. P_loss = (20000)^2 x 4 = 4e8 x 4 = 1.6e9 W. This is 3.2 times the generated power - completely impractical! At 400,000 V: I = 1250 A, P_loss = 1250^2 x 4 = 6.25 MW (only 1.25% lost).

• Rearrange: ns = np x (Vs/Vp) = 4400 x (230/11000) (1m)
• Calculate: 4400 x 230 = 1,012,000; divide by 11000 = 92 (1m)
• Correct answer: 92 turns (1m)

ns = np x (Vs/Vp) = 4400 x (230/11000) = 4400 x 0.02091 = 92 turns.

• Rearrange: Is = (Vp x Ip) / Vs = (240 x 5) / 12 (1m)
• Calculate numerator: 240 x 5 = 1200 (1m)
• Is = 1200 / 12 = 100 A (1m)

Is = VpIp / Vs = (240 x 5) / 12 = 1200/12 = 100 A. When voltage is stepped down, current steps up (energy is conserved).

Transmission cables have electrical resistance. Power is lost as heat in the cables according to the equation P = I^2 R. At higher voltage, less current flows for the same power transmitted. Since power loss is proportional to current squared, using a smaller current greatly reduces the energy wasted as heat in the cables.

• Cables have resistance and energy is wasted as heat in the cables (1m)
• Power loss = I^2 x R, so lower current means less power lost as heat (1m)
• High voltage allows the same power to be transmitted at lower current (P = IV) (1m)

P_loss = I^2 R. P = IV. For constant P, higher V -> lower I -> much lower I^2 -> much less heat loss. That's why the grid uses 400,000 V.

First, eddy currents are induced in the iron core by the changing magnetic field. These currents heat the core, wasting energy. To reduce this, the core is laminated (made of thin insulated layers). Second, the coils of wire have resistance and carry current, so they also heat up and waste energy as heat.

• Eddy currents in the iron core cause heating / energy wasted as heat in the core (1m)
• Coils/wire have resistance so energy is wasted as heat in the coils (1m)
• Laminated core reduces eddy currents (thin insulated layers limit induced current paths) (1m)

Real transformer losses: (1) Eddy currents in core -> heat (reduced by lamination). (2) Resistance of wire coils -> heat. Real transformers are typically 95-99% efficient.

A transformer works by electromagnetic induction. The primary coil creates a changing magnetic field in the iron core, which induces a voltage in the secondary coil. AC current constantly changes direction, so it creates a constantly changing magnetic field. DC current is constant, so it creates a constant (non-changing) magnetic field which cannot induce a voltage in the secondary coil.

• AC current creates a constantly changing magnetic field in the iron core (1m)
• A changing (not constant) magnetic field is needed to induce a voltage in the secondary coil / DC creates a constant field that cannot induce a voltage (1m)

Electromagnetic induction requires a CHANGING magnetic flux. AC -> changing current -> changing B field -> changing flux -> induced EMF. DC -> constant B field -> no change -> no induced EMF.

A step-up transformer increases the voltage (and decreases the current) at the power station before electricity is transmitted through the National Grid. High voltage and low current means less energy lost as heat in the cables.

A transformer has a primary coil (input), a secondary coil (output), both wound on a soft iron core. The iron core concentrates and transfers the changing magnetic field between the coils.

A step-down transformer reduces the high grid voltage (typically 33,000 V or 11,000 V) to the mains voltage (230 V) suitable for homes.

• Step-down transformer (reduces voltage to 230 V / safe level for domestic use) (1m)

Step-down transformers at substations reduce the high transmission voltage to 230 V for safe use in homes and businesses.

Electricity is generated at a power station. A step-up transformer increases the voltage for efficient long-distance transmission through high-voltage cables. At local substations, step-down transformers reduce the voltage to a safe level (230 V) for use in homes.

• Step-up transformer at power station / step-down transformer at substation near homes (1m)

Power station -> step-up transformer -> high voltage grid cables -> step-down transformer -> homes at 230 V.

P_loss = I^2 R. For a given power transmitted, higher voltage means lower current. Lower current means much less energy wasted as heat (P_loss proportional to I^2).

Vs/Vp = ns/np. Vs = Vp x (ns/np) = 230 x (1000/200) = 230 x 5 = 1150 V. More turns on secondary than primary means voltage is stepped up.

For the regular solid (cuboid): use a ruler or vernier calliper to measure the length, width, and height. Calculate volume using V = length x width x height. Use a balance to measure the mass. Calculate density using density = mass divided by volume. For the irregular solid (pebble): use a balance to measure the mass. Fill a measuring cylinder with water and record the initial volume. Submerge the pebble fully and record the new water level. Volume of pebble = new volume minus initial volume (water displacement). Calculate density using density = mass divided by volume.

• Level 3 (5–6 marks): Clear, detailed description of both methods. Masses measured with balance for both. Volume of cuboid found by measuring three dimensions with ruler/calliper and calculating V = l × w × h. Volume of irregular solid found by water displacement in measuring cylinder (subtracting initial from final volume). Density calculated using ρ = m/V for both. Mentions of precision equipment or control variables. (6m)
• Level 2 (3–4 marks): Both methods described but one lacks detail. Correct identification of instruments. Density formula stated. Minor omissions. (4m)
• Level 1 (1–2 marks): Only one method described or both described very superficially. Density formula may be missing or wrong. (2m)

Regular solid: measure length, width, height with ruler/calliper; V = l × w × h. Balance gives mass. ρ = m/V. Irregular solid: measure mass with balance. Fill measuring cylinder with water and record volume. Submerge solid and record new volume. Volume of solid = change in water level. ρ = m/V.

• Convert mass to kg: 450 g = 0.450 kg (1m)
• Convert dimensions and calculate volume: 0.30 × 0.10 × 0.05 = 0.0015 m³ (1m)
• Correct substitution: density = 0.450 ÷ 0.0015 (1m)
• Correct answer: 300 kg/m³ (1m)

Mass = 450 g = 0.450 kg. Volume = 0.30 × 0.10 × 0.05 = 0.0015 m³. Density = 0.450 ÷ 0.0015 = 300 kg/m³.

• Volume = 0.10 × 0.05 × 0.04 = 0.000 200 m³ (or 2.0 × 10⁻⁴ m³) (1m)
• Correct substitution: density = 0.540 ÷ 0.000 200 (1m)
• Correct answer: 2700 kg/m³ (1m)

Volume = 0.10 × 0.05 × 0.04 = 0.000 200 m³. Density = 0.540 ÷ 0.000 200 = 2700 kg/m³.

• Rearrangement: mass = density × volume (1m)
• Correct substitution: mass = 7800 × 0.002 (1m)
• Correct answer: 15.6 kg (1m)

Rearranging ρ = m/V gives m = ρ × V = 7800 × 0.002 = 15.6 kg.

Measure the mass of the stone using a balance. Fill a measuring cylinder with water and record the initial volume. Carefully lower the stone into the water and record the new water level. The volume of the stone equals the increase in water level (displacement). Calculate density using density = mass ÷ volume.

• Measure the mass using a balance (in grams or kg) (1m)
• Measure the volume using water displacement in a measuring cylinder: record initial volume, submerge stone, record final volume; volume of stone = difference (1m)
• Calculate density = mass ÷ volume (1m)

Mass is measured with a balance. Volume of an irregular solid is found using water displacement: volume = final reading − initial reading. Then density = mass ÷ volume.

The statement is not correct because it is density that matters, not just the mass (weight). An object floats if its density is less than the density of water (1000 kg/m³). A large piece of wood has a much greater mass than a small iron nail, but the wood floats because its density is less than 1000 kg/m³ while iron has a density greater than 1000 kg/m³.

• The statement is wrong because it is density, not mass/weight, that determines floating or sinking (1m)
• An object floats if its density is less than the density of water (1000 kg/m³) (1m)
• An object sinks if its density is greater than the density of water / correct example given (e.g. wood floats, iron sinks) (1m)

Floating or sinking depends on density compared to the fluid. Any object with density < 1000 kg/m³ floats in water; any object with density > 1000 kg/m³ sinks. A massive but low-density object (e.g. a ship with lots of air) floats, while a tiny high-density object (e.g. an iron nail) sinks.

• Rearrangement: volume = mass ÷ density (1m)
• Correct substitution: volume = 2.268 ÷ 11 340 (1m)
• Correct answer: 2.0 × 10⁻⁴ m³ (or 0.0002 m³) (1m)

V = m ÷ ρ = 2.268 ÷ 11 340 = 2.0 × 10⁻⁴ m³. This is the density of lead.

In a gas, particles are spread far apart with large spaces between them. In a solid, particles are closely packed together. This means that the same mass of particles occupies a much greater volume in a gas, so the density (mass divided by volume) is much lower.

• Gas particles are spread far apart / widely spaced / large gaps between particles (compared to solid) (1m)
• Same mass occupies a larger volume in a gas so density is lower (density = mass/volume) (1m)

Gas particles have large spaces between them. For the same mass of substance, the gas occupies far more volume than the solid. Since density = mass/volume, a larger volume for the same mass means lower density.

The student could repeat the measurements several times and calculate a mean value to reduce the effect of random errors. They could also use a more precise measuring instrument, such as a digital vernier calliper instead of a ruler, to measure the dimensions more accurately.

• Repeat measurements and calculate a mean / take more readings to reduce random error (1m)
• Use more precise / accurate equipment (e.g. vernier calliper, digital balance) OR ensure no parallax error in reading the ruler (1m)

Taking multiple measurements and calculating the mean reduces the effect of random errors. Using more precise equipment (e.g. vernier callipers, digital balance) reduces systematic and random errors from equipment limitations.

Density = mass ÷ volume (ρ = m/V). The unit is kg/m³. A denser material packs more mass into the same volume.

The particles in a solid are much more closely packed than in a gas. The same number of particles (same mass) occupies a smaller volume, so density = mass/volume is larger.

Density = mass ÷ volume = 240 ÷ 80 = 3 g/cm³. This is a straightforward substitution into ρ = m/V.

Volume of pebble = 74 − 50 = 24 cm³. Density = 60 ÷ 24 = 2.5 g/cm³. The water displacement method gives the volume of an irregular solid.

• Level 3 (5–6 marks): Temperature rises in solid region as kinetic energy increases. Flat section at melting point: energy breaks bonds between particles, temperature constant, potential energy increases. Temperature rises in liquid region as kinetic energy increases. Flat section at boiling point: energy breaks bonds, temperature constant, potential energy increases. Temperature rises in gas region. Two flat sections clearly identified and correctly explained. (6m)
• Level 2 (3–4 marks): Correct identification of two flat sections or correct explanation of one state change. Kinetic energy increases during temperature rise noted. (4m)
• Level 1 (1–2 marks): Some correct features — at least one flat section mentioned or some correct particle description. (2m)

Heating curve: slope up (solid, KE increasing) → flat (melting point, bonds breaking, T constant, PE increasing) → slope up (liquid, KE increasing) → flat (boiling point, bonds breaking, T constant, PE increasing) → slope up (gas, KE increasing).

Internal energy is the total kinetic energy and potential energy of all the particles in a substance. When a solid is heated, the particles gain kinetic energy so internal energy increases and temperature rises. At the melting point, temperature stays constant but internal energy continues to increase because the energy is being used to break bonds between particles, increasing the potential energy. Once all the solid has melted, further heating increases the kinetic energy again and temperature rises again. At the boiling point, temperature is constant again while internal energy increases as bonds are broken to change liquid to gas.

• Internal energy is the total kinetic energy AND potential energy of all particles in the substance (1m)
• When heated (between state changes), kinetic energy increases so temperature rises (1m)
• At a state change (melting or boiling), temperature stays constant but potential energy increases as bonds are broken (1m)
• Correct description of the full sequence: solid heating → melting → liquid heating → boiling → gas, with correct energy description at each stage (1m)

Internal energy = KE + PE of all particles. Heating between state changes: KE increases, temperature rises. At state changes: temperature constant, PE increases as bonds break. Full cycle: solid (KE rises) → melting (PE rises, T constant) → liquid (KE rises) → boiling (PE rises, T constant) → gas.

When a solid is heated, the particles gain kinetic energy and vibrate more. At the melting point, the temperature stays constant even though energy is still being supplied. This is because the energy is used to break the bonds between particles rather than increasing the kinetic energy. Once the bonds are broken, the particles can move past each other and the solid has melted to form a liquid.

• Particles gain kinetic energy / vibrate more (before melting point) (1m)
• At the melting point, temperature stays constant / energy is used to break bonds between particles (1m)
• Particles become free to move past each other / substance becomes a liquid (1m)

Heating increases particle kinetic energy (vibrations). At melting point, energy breaks intermolecular bonds without temperature rising. Particles become free to move → liquid formed.

In a liquid, particles have a range of kinetic energies. The particles with the highest kinetic energy at the surface can escape from the liquid and become a gas. This means that the particles left behind in the liquid have a lower average kinetic energy. Since temperature is related to the average kinetic energy of the particles, the temperature of the remaining liquid decreases.

• Particles with the highest kinetic energy escape from the surface of the liquid (1m)
• The average kinetic energy of the remaining particles decreases (1m)
• Temperature is linked to average kinetic energy, so the temperature of the liquid decreases / the liquid cools (1m)

Evaporation removes the highest-energy particles from the surface. Remaining particles have lower average KE. Temperature is proportional to average KE, so temperature drops — the liquid cools.

At the top of a mountain, the atmospheric pressure is lower than at sea level. During boiling, liquid particles at the surface need enough energy to overcome the pressure pushing down on them to escape into the gas phase. At lower pressure, particles need less energy to escape, so boiling occurs at a lower temperature where fewer particles have enough energy to escape.

• Atmospheric pressure is lower at the top of a mountain (1m)
• Lower pressure means particles at the surface need less energy / lower kinetic energy to escape from the liquid (1m)
• Therefore the boiling point (temperature at which particles have enough energy to escape) is lower (1m)

Lower atmospheric pressure at altitude means particles need less energy to escape from the surface. Boiling point is lower because fewer particles need to reach the threshold kinetic energy to overcome atmospheric pressure.

• Correct rearrangement: ΔT = E ÷ (m × c) (1m)
• Correct substitution: ΔT = 2400 ÷ (0.200 × 2100) = 2400 ÷ 420 (1m)
• Correct answer: 5.71 °C (or 5.7 °C) (1m)

ΔT = E ÷ (m × c) = 2400 ÷ (0.200 × 2100) = 2400 ÷ 420 = 5.71 °C.

In a solid, particles are arranged in a regular pattern and are closely packed together. The particles are held in fixed positions by strong forces of attraction and can only vibrate about those fixed positions.

• Particles are arranged in a regular pattern / close together / tightly packed (1m)
• Particles vibrate about fixed positions / cannot move freely / held by strong forces (1m)

In a solid: regular pattern, closely packed, vibrate about fixed positions, strong forces between particles.

The particles in a gas move rapidly in random directions. When the particles collide with the walls of the container, they exert a force on the walls. The pressure is caused by the total force per unit area from all the collisions of gas particles with the walls.

• Particles move randomly and collide with the walls of the container (1m)
• The collisions exert a force on the walls / pressure = force per unit area from collisions (1m)

Gas particles move randomly and collide with container walls. Each collision exerts a tiny force. The sum of all forces per unit area = gas pressure.

• Time in seconds: 3 × 60 = 180 s (1m)
• Energy = power × time = 500 × 180 = 90 000 J = 90 kJ (1m)

t = 3 × 60 = 180 s. E = P × t = 500 × 180 = 90 000 J = 90 kJ.

In a solid, particles are arranged in a regular lattice and can only vibrate about fixed positions. Strong forces of attraction hold them in place.

Evaporation happens at the liquid surface at any temperature, when high-energy particles escape. Boiling happens throughout the liquid at a fixed boiling point temperature.

During a state change, the energy supplied increases the potential energy of the particles (breaking bonds) but NOT the kinetic energy. Since temperature is related to average kinetic energy, the temperature stays constant while internal energy (which includes potential energy) increases.

Condensation is when a gas (or vapour) cools and turns into a liquid. The particles lose kinetic energy and forces of attraction pull them closer together to form a liquid.

Place a known mass of ice in the beaker and wait until the thermometer reads 0 degrees C to confirm the ice is at its melting point. Measure the voltage and current using a voltmeter and ammeter and calculate power using P = V times I. Start the heater and the stopwatch simultaneously. Run the heater until all the ice has melted. Record the time. Calculate energy supplied using E = power times time. Measure the mass of the water produced using the balance. Calculate the specific latent heat of fusion using L = E divided by mass. A source of error is heat transferred from the surroundings to the ice, providing additional energy that the heater did not supply. This makes the calculated latent heat too low. To minimise this, insulate the beaker with lagging, or run a control experiment with the heater off to measure the background energy gain and subtract it.

• Level 3 (5–6 marks): Place known mass of ice in beaker. Wait until ice is at 0 °C. Start heater and stopwatch. Measure current and voltage (P = VI). Run heater until all ice melts. Record time. Calculate E = Pt. Measure mass of water produced. Calculate L = E ÷ m. Source of error: heat gained from surroundings (not just heater) — minimise by insulating beaker or subtracting energy from a control (heater off) experiment. (6m)
• Level 2 (3–4 marks): Correct overall method, correct E = Pt, correct L = E ÷ m. Source of error mentioned. (4m)
• Level 1 (1–2 marks): Partial method. May have L = E/m but with error. Some relevant detail. (2m)

Method: Ensure ice at 0 °C. Measure power (V × I). Run heater for measured time. Find mass that melted. L = E ÷ m. Error: heat gained from room warms ice and provides extra energy → measured L too low. Insulate beaker or run a control experiment with no heater.

• Energy to heat ice from −15 to 0 °C: E₁ = 0.200 × 2100 × 15 = 6300 J (1m)
• Energy to melt ice: E₂ = 0.200 × 334 000 = 66 800 J (1m)
• Energy to heat water from 0 to 25 °C: E₃ = 0.200 × 4200 × 25 = 21 000 J (1m)
• Total energy = 6300 + 66 800 + 21 000 = 94 100 J (1m)

E₁ = 0.200 × 2100 × 15 = 6300 J. E₂ = 0.200 × 334 000 = 66 800 J. E₃ = 0.200 × 4200 × 25 = 21 000 J. Total = 6300 + 66 800 + 21 000 = 94 100 J.

• Correct equation: E = mL (1m)
• Correct substitution: E = 0.300 × 2 260 000 (1m)
• Correct answer: 678 000 J (or 678 kJ) (1m)

E = mL = 0.300 × 2 260 000 = 678 000 J (= 678 kJ).

• Correct rearrangement: m = E ÷ L (1m)
• Correct substitution: m = 200 000 ÷ 334 000 (1m)
• Correct answer: 0.599 kg (or 0.60 kg) (1m)

m = E ÷ L = 200 000 ÷ 334 000 = 0.599 kg.

When water melts (fusion), the bonds between particles are only partially broken. The particles can now move past each other but they are still close together and still have some forces of attraction between them. When water boils (vaporisation), all the bonds between particles must be completely broken so that particles can move far apart in the gas phase. Breaking all the bonds requires much more energy than just weakening them, which is why the specific latent heat of vaporisation is much greater than the specific latent heat of fusion.

• During fusion (melting), bonds are only partially broken / weakened (particles still close together in liquid) (1m)
• During vaporisation, all bonds must be completely broken / particles become fully separated in the gas phase (1m)
• Breaking all bonds requires more energy than partially weakening them, so latent heat of vaporisation is greater (1m)

Melting: bonds partly weakened (particles still close, still liquid forces). Vaporisation: ALL bonds completely broken (particles far apart, no forces). Breaking all bonds needs more energy → larger L for vaporisation.

The student should measure the mass of a beaker of water before heating. Heat the water until it is boiling steadily and then record the time for a known amount of water to boil away. Measure the mass of water remaining to find the mass that evaporated. The power of the heater is P = VI (using a voltmeter and ammeter). Energy supplied = power × time (E = Pt). Specific latent heat = energy ÷ mass evaporated (L = E/m).

• Measure mass of water before and after boiling (to find mass evaporated) / measure time for a known mass to evaporate (1m)
• Measure power of heater (using voltmeter and ammeter P = VI) and time; calculate E = Pt (1m)
• Calculate specific latent heat using L = E ÷ m (energy ÷ mass evaporated) (1m)

Mass evaporated = mass before − mass after. Power = VI (voltmeter and ammeter). E = Pt. L = E ÷ m.

• Rearrangement: m = E ÷ L (1m)
• Correct substitution: m = 1 130 000 ÷ 2 260 000 (1m)
• Correct answer: 0.5 kg (1m)

m = E ÷ L = 1 130 000 ÷ 2 260 000 = 0.5 kg.

When steam condenses, the particles slow down and bonds form between the particles as they change from a gas to a liquid. Forming bonds releases energy. This energy is released to the surroundings even though the temperature stays constant, because the energy comes from reducing the potential energy of the particles rather than their kinetic energy.

• Bonds form between the particles when changing from gas to liquid (1m)
• Forming bonds releases energy to the surroundings / the potential energy of particles decreases (releasing energy) (1m)

Condensation is the reverse of vaporisation. Bonds form between particles → potential energy decreases → energy released to surroundings. Temperature constant because KE is unchanged.

During a state change, the energy supplied is used to break the bonds between the particles of the substance. This increases the potential energy of the particles rather than their kinetic energy, so the temperature (which depends on average kinetic energy) does not change.

• Energy is used to break the bonds / forces of attraction between particles (1m)
• Kinetic energy does not increase so temperature stays constant / potential energy increases instead (1m)

Energy breaks intermolecular bonds (increases PE) without increasing KE. Since temperature depends on average KE, temperature stays constant.

Specific latent heat is the energy needed to change the state of 1 kg of a substance without changing its temperature. The energy breaks or forms bonds between particles rather than increasing kinetic energy.

Specific latent heat of vaporisation refers to the liquid → gas (or gas → liquid) state change. Specific latent heat of fusion refers to the solid → liquid (or liquid → solid) change.

In vaporisation, ALL intermolecular bonds must be broken to separate particles into a gas. In fusion (melting), the bonds are only partially weakened to allow particles to flow as a liquid, so less energy is needed.

E = mL = 0.500 kg × 334 000 J/kg = 167 000 J. Remember to convert 500 g to 0.500 kg before substituting.

(a) When the temperature increases, the gas particles gain kinetic energy and move faster. They collide with the walls of the container more frequently and with greater force. Since pressure is the force exerted per unit area on the walls, the pressure increases. The pressure is proportional to the absolute temperature in kelvin (p divided by T equals constant at constant volume). (b) When the volume decreases, the particles are confined to a smaller space. They travel a shorter distance between the walls so they collide with the walls more frequently. This increases the pressure. The pressure is inversely proportional to the volume (p times V equals constant at constant temperature).

• Level 3 (5–6 marks): (a) Higher temperature → particles gain KE → move faster → collide with walls more frequently AND with greater force → pressure increases. Pressure proportional to temperature in kelvin (p/T = constant at constant V). (b) Smaller volume → particles travel less distance between collisions → more frequent collisions with walls → pressure increases. Pressure inversely proportional to volume (pV = constant at constant T). Both relationships clearly described with particle-model reasoning. (6m)
• Level 2 (3–4 marks): Correct particle model for at least one scenario, correct relationship stated for at least one. (4m)
• Level 1 (1–2 marks): Some relevant particle model description for one scenario. (2m)

(a) Higher T → particles gain KE → faster → more frequent collisions with greater force → higher pressure. p/T = constant at constant V. (b) Smaller V → less distance between walls → more frequent collisions with same force → higher pressure. pV = constant at constant T.

• Convert temperatures to kelvin: T₁ = 27 + 273 = 300 K; T₂ = 127 + 273 = 400 K (1m)
• Use p₁/T₁ = p₂/T₂ → p₂ = p₁ × T₂ ÷ T₁ (1m)
• Correct substitution and rearrangement: p₂ = 120 000 × 400 ÷ 300 (1m)
• Correct answer: 160 000 Pa (1m)

T₁ = 300 K, T₂ = 400 K. p₂ = p₁ × (T₂/T₁) = 120 000 × (400/300) = 120 000 × (4/3) = 160 000 Pa.

• Correct rearrangement: p₂ = (p₁ × V₁ × T₂) ÷ (T₁ × V₂) (1m)
• Correct substitution: p₂ = (80 000 × 0.0060 × 450) ÷ (300 × 0.0045) (1m)
• Correct numerator and denominator: 216 ÷ 1.35 (1m)
• Correct answer: 160 000 Pa (1m)

p₂ = (p₁ × V₁ × T₂) ÷ (T₁ × V₂) = (80 000 × 0.0060 × 450) ÷ (300 × 0.0045) = 216 ÷ 1.35 = 160 000 Pa.

• Step 1 — pressure after heating at constant volume using p₁/T₁ = p₂/T₂: p₂ = 3.0 × 10⁵ × 450 ÷ 300 = 4.5 × 10⁵ Pa (1m)
• Step 2 — recognise Boyle's law applies for the isothermal expansion: p₂V₁ = p₃V₂ (constant temperature step) (1m)
• Correct substitution into Boyle's law: V₂ = (4.5 × 10⁵ × 0.025) ÷ (2.0 × 10⁵) (1m)
• Correct final answer: 0.05625 m³ (accept 0.056 m³ or 5.6 × 10⁻² m³) (1m)

Step 1 (constant volume heating): p₂ = p₁ × T₂ ÷ T₁ = 3.0 × 10⁵ × (450 ÷ 300) = 4.5 × 10⁵ Pa. Step 2 (isothermal expansion using Boyle's law): V₂ = p₂V₁ ÷ p₃ = (4.5 × 10⁵ × 0.025) ÷ (2.0 × 10⁵) = 11 250 ÷ 200 000 = 0.05625 m³.

• Correct substitution into equation: 100 × 2.0 = p₂ × 0.5 (1m)
• Correct substitution: 100 × 2000 = p₂ × 500 (or equivalent using consistent units) (1m)
• Correct answer: 400 kPa (1m)

Using p₁V₁ = p₂V₂: 100 × 2.0 = p₂ × 0.5. Rearranging: p₂ = 200 ÷ 0.5 = 400 kPa. The volume is quartered (from 2.0 L to 0.5 L) so the pressure is multiplied by 4.

When the temperature of the gas increases, the particles gain kinetic energy and move faster. As a result, the particles collide with the walls of the container more frequently and with greater force. Since pressure is the force per unit area on the walls, the pressure of the gas increases.

• Gas particles gain kinetic energy / move faster when heated (1m)
• Particles collide with the container walls more frequently (and/or with greater force) (1m)
• Greater force per unit area on the walls means pressure increases (1m)

Temperature increase → particles gain KE → move faster → more frequent collisions with walls AND greater force per collision → pressure increases.

• Rearrangement: p₂ = p₁V₁ ÷ V₂ (1m)
• Correct substitution: p₂ = (200 000 × 0.0050) ÷ 0.0020 = 1000 ÷ 0.0020 (1m)
• Correct answer: 500 000 Pa (1m)

p₂ = p₁V₁ ÷ V₂ = (200 000 × 0.0050) ÷ 0.0020 = 1000 ÷ 0.0020 = 500 000 Pa.

• Correct rearrangement: V₂ = p₁V₁ ÷ p₂ (1m)
• Correct substitution: V₂ = (1.0 × 10⁵ × 60) ÷ (2.5 × 10⁵) (1m)
• Correct answer: 24 cm³ (1m)

V₂ = p₁V₁ ÷ p₂ = (1.0 × 10⁵ × 60) ÷ (2.5 × 10⁵) = 6 000 000 ÷ 250 000 = 24 cm³. The pressure increases by a factor of 2.5 so the volume decreases by a factor of 2.5.

I would use a sealed syringe with a pressure gauge attached. I would change the volume of the syringe by pushing or pulling the plunger and record both the volume (in cm³ or m³) and the corresponding pressure (in Pa or kPa) from the gauge. I would repeat this for several different volumes. To verify Boyle's Law, I would plot a graph of pressure against 1/volume. If the law holds, this graph will be a straight line through the origin, confirming that pressure is proportional to 1/volume (i.e. pressure and volume are inversely proportional and pV = constant).

• Use a sealed syringe with a pressure gauge/sensor; change the volume by moving the plunger and measure pressure at each volume (1m)
• Record multiple pressure and volume readings to obtain a range of data (1m)
• Plot pressure against 1/volume; a straight line through the origin confirms the inverse relationship (Boyle's Law / pV = constant) (1m)

To verify Boyle's Law: use a sealed syringe + pressure gauge, record pressure at different volumes. Plot pressure (y-axis) against 1/volume (x-axis). A straight line through the origin confirms pressure ∝ 1/volume (pV = constant). This graph linearises the inverse relationship, making it easy to confirm.

When the temperature increases, the gas particles gain kinetic energy and move faster. They initially collide with the walls more frequently and with greater force, which would increase the pressure. However, since the syringe is free to expand, the volume increases to restore the original collision rate and force on the walls. The volume keeps increasing until the pressure returns to its original value, so the volume increases while pressure stays constant.

• Gas particles gain kinetic energy / move faster at higher temperature (1m)
• Increased speed means more frequent collisions with walls (which would increase pressure) (1m)
• The volume increases (syringe expands) to reduce the collision frequency back to the original value, maintaining constant pressure (1m)

Higher T → particles gain KE → move faster → more frequent collisions → pressure would rise. Syringe expands so volume increases → collision frequency returns to original → pressure stays constant.

The student is wrong because the pressure–temperature relationship uses the Kelvin scale, not Celsius. Doubling the temperature in Celsius does not double the temperature in Kelvin (for example, doubling from 20 °C to 40 °C is only 293 K to 313 K, not a doubling). To double the pressure of a sealed gas at constant volume, the absolute temperature in kelvin must double (for example from 200 K to 400 K).

• The relationship uses the kelvin temperature scale, not Celsius (1m)
• Doubling the Celsius temperature does NOT double the kelvin temperature / example shown (e.g. 20 °C = 293 K, 40 °C = 313 K) (1m)
• For pressure to double, the kelvin temperature must double (at constant volume) (1m)

p ∝ T only when T is in Kelvin. Doubling Celsius (e.g. 20→40°C) gives 293→313 K, not double. For double pressure, need double kelvin temperature (e.g. 200 K → 400 K).

When the volume is reduced, the same number of particles are now in a smaller space. The particles collide with the walls more frequently because they do not have to travel as far between collisions. This increases the number of collisions per second with the walls, so the pressure increases. (Pressure and volume are inversely proportional: pV = constant.)

• Particles are in a smaller space so they hit the walls more frequently / travel less distance between collisions (1m)
• More frequent collisions on the walls increase the pressure / pressure increases when volume decreases (inversely proportional) (1m)

Smaller volume → particles travel less far between collisions → more frequent wall collisions → pressure increases. pV = constant (Boyle's Law).

When a gas is compressed, the same number of particles are now in a smaller volume. The particles hit the walls of the container more often because they travel a shorter distance between collisions. This means there are more collisions per second with the walls, so the pressure increases.

• Particles are in a smaller volume / confined to a smaller space / travel a shorter distance between collisions (1m)
• Particles collide with the walls more frequently / more collisions per second → pressure increases (1m)

Compressing a gas forces the same number of particles into a smaller space. The particles travel shorter distances between collisions with the walls, so they hit the walls more often each second. More frequent collisions create a greater force per unit area — that is an increase in pressure.

Absolute zero (0 K, equal to −273 °C) is the lowest possible temperature. At absolute zero, the particles in a substance have the minimum possible kinetic energy. Temperatures below absolute zero are not possible because kinetic energy cannot be less than zero — you cannot have less than no movement.

• Absolute zero is the lowest possible temperature / 0 K / −273 °C / particles have minimum kinetic energy at this point (1m)
• Temperatures below absolute zero are not possible because kinetic energy cannot be negative / cannot be less than zero (1m)

Absolute zero = 0 K = −273 °C. Particles have minimum kinetic energy here. Cannot go below because KE ≥ 0 (cannot have negative energy).

When a sealed gas is heated, the particles gain kinetic energy and move faster. They collide with the container walls more frequently and with greater force, so pressure increases.

Boyle's Law states that for a fixed mass of gas at constant temperature, pressure × volume = constant (pV = constant). This means that if volume halves, pressure doubles.

When a gas is heated at constant volume, its particles gain kinetic energy and move faster. They collide with the container walls more frequently and with greater force, so the pressure increases. The number of particles does not change — only their speed. This is the pressure-temperature law: p/T = constant (at constant volume).

Using pV = constant: p₁V₁ = p₂V₂. V₂ = (p₁ × V₁) ÷ p₂ = (100 000 × 4.0 × 10⁻³) ÷ 400 000 = 400 ÷ 400 000 = 1.0 × 10⁻³ m³. Pressure × 4 so volume ÷ 4.

T(K) = T(°C) + 273. So −40 + 273 = 233 K. Absolute zero (0 K) is −273 °C — the temperature at which particles have minimum kinetic energy.