June 2019 QP-2

Examination Overview

This examination paper is for the GCSE Combined Science: Trilogy Higher Tier Physics Paper 1H, conducted on Wednesday, 22 May 2019. It lasts for 1 hour and 15 minutes, and the maximum mark for this paper is 70. Students must adhere to the following instructions:

  • Materials required include a ruler, a scientific calculator, and the enclosed Physics Equations Sheet.
  • The exam should be completed using black ink or black ball-point pen.
  • Candidates must fill in their information at the top of the page and answer all questions in the provided spaces.
  • Rough work should be done in the book with any unwanted work crossed through before submission.
  • In all calculations, students should clearly show their working.
  • Good English and clear presentation are expected in answers.

Section 0: Current in a Resistor

Question 1

A student investigated how the current in a resistor varies with the potential difference across the resistor.

1.1

Question: Describe the correct way to connect the ammeter and voltmeter in the circuit.
Options:

  • (A) Ammeter in parallel with the resistor / Voltmeter in series with the resistor
  • (B) Ammeter in series with the resistor / Voltmeter in series with the resistor
  • (C) Ammeter in series with the resistor / Voltmeter in parallel with the cell
  • (D) Ammeter in series with the resistor / Voltmeter in parallel with the resistor
  • (E) Ammeter in parallel with the cell / Voltmeter in series with the resistor
    Correct answer correlating with circuit theory.
1.2

Question: How does increasing the resistance of the variable resistor affect the current in the circuit?
Response: With an increase in resistance, the current decreases in accordance with Ohm's law.

1.3

Question: Describe how to change the circuit to allow for negative values of current and potential difference.
This typically involves reversing the polarity of the voltage source.

1.4

Question: What is the relationship type between current and potential difference for a resistor at constant temperature?
Answer: This is a direct proportional relationship, described by Ohm's law.

1.5

Question: Write the equation linking current (I), potential difference (V), and resistance (R).
Equation: V=IimesRV = I imes R

1.6

Question: Given a current of 0.12 A and a potential difference of 3.0 V, calculate the resistance of the resistor.
Calculation:
Using Ohm's law, rearranging gives:
R = rac{V}{I} = rac{3.0 ext{ V}}{0.12 ext{ A}} = 25 ext{ Ω}

Section 1: Changes in Air Pressure and Temperature

Question 2

2.1

Question: Explain the change in pressure in the container of air when the temperature decreases from 20 °C to 0 °C while maintaining constant volume.
Explanation:
As temperature decreases, the kinetic energy of the air molecules also decreases, leading to slower motion. Consequently, molecules collide with the container walls less forcefully and less frequently, resulting in a decrease in pressure.

2.2

Question: Given that the change in internal energy of frozen water is 0.70 kJ and the specific latent heat of fusion is 330 kJ/kg, calculate the mass of ice produced.
Calculation:
Using the formula:
Q=mimesLQ = m imes L
where

  • Q is the energy change,
  • m is mass,
  • L is the specific latent heat.
    Rearranging gives:
    m = rac{Q}{L} = rac{0.70 imes 10^3 ext{ J}}{330 imes 10^3 ext{ J/kg}} = 0.00212 ext{ kg}
2.3

Question: Discuss the states of matter for oxygen, nitrogen, and carbon dioxide at -190 °C, based on their boiling/freezing points.
Oxygen is a solid, nitrogen is a solid, carbon dioxide is also a solid.

2.4

Question: Explain the changes in the arrangement and movement of argon particles as the temperature decreases from 20 °C to -190 °C.
Explanation:
Initially, aragon atoms in the gaseous state have high kinetic energy, allowing them to move freely. As temperature drops, the particles lose energy, decreasing their movement, which leads ultimately to a liquid state and then to a solid state as temperature approaches absolute zero. The arrangement changes from random, free movement to ordered, fixed positions in a solid state.

Section 2: Hybrid Car Energy Sources

Question 3

3.1

Question: Define a non-renewable energy resource.
Definition:
A non-renewable energy resource is a resource that is available in limited quantities and cannot be replenished on a human timescale once depleted, such as fossil fuels.

3.2

Question: Calculate the current used to charge a battery powered by a 230 V mains supply, if the power used is 6.9 kW.
Calculation:
Using the formula:
P=VimesIP = V imes I
Rearranging gives:
I = rac{P}{V} = rac{6.9 imes 10^3 ext{ W}}{230 ext{ V}} = 30 A

3.3

Question: Explain the difference between direct and alternating potential difference.
Explanation:
Direct current (DC) provides a constant voltage (e.g., from batteries) that flows in one direction, while alternating current (AC) alternates the direction of flow, typically used in mains electricity.

3.4

Question: Explain why a low resistance cable is preferred over a high resistance cable for connecting to mains supply.
Answer:
A low resistance cable minimizes energy loss as heat due to the resistive heating effect, improving efficiency during energy transmission.

Section 3: Circuit Components

Question 4

4.1

Question: Figure 2 shows a circuit where the lamp and resistors have a resistance of 10 Ω each. Calculate the total resistance of the circuit.
Answer Options:

  • Between 20 and 30 Ω
  • Exactly 20 Ω
  • Exactly 30 Ω
  • Less than 20 Ω
    Response: Assessment using series or parallel calculations depending on configuration.
4.2

Question: Explain your answer to the previous resistance calculation.
*Explanation needed:
This should ideally relate to series and parallel combinations based on circuit theory.

4.3

Question: Draw the circuit symbol for a thermistor.

4.4

Question: Describe how the current in a thermistor changes with temperature.
Explanation:
As temperature increases, the resistance of the thermistor decreases, resulting in an increase in current through the thermistor.

4.5

Question: Analyze how closing the switch affects the potential difference across both the resistor and the lamp.
Explanation:
The closing of the switch enables current to flow, hence the potential difference across the resistor would lower, while that across the lamp would increase, leading it to illuminate more brightly.