Question 1: Define potential difference and explain its role in an electrical circuit.

Answer: Potential difference is the work done per unit charge to move charge between two points in a circuit. It indicates the energy transferred from the power supply to the components.

Question 2: Describe how potential difference is related to the energy transfer in a circuit.

Answer: Potential difference represents the energy transferred per unit charge as it moves through a component, affecting how much energy is supplied to each part of the circuit.

Question 1: A battery supplies 12 joules of energy to move 3 coulombs of charge. Calculate the potential difference across the battery.

Answer: PD = E / Q = 12 J / 3 C = 4 volts

Question 2: A circuit has a potential difference of 9 volts across a resistor, and 0.5 coulombs of charge pass through per second. Calculate the power dissipated in the resistor.

Answer: Power = PD × Q = 9 V × 0.5 C/s = 4.5 watts

Question 1: Identify three variables that must be controlled in an experiment measuring PD across different resistors.

Answer: Variables include the current, temperature of resistors, and the power supply voltage.

Question 2: List two safety precautions to consider when working with circuits involving potential difference measurements.

Answer: Ensure the circuit is switched off before connecting or adjusting components, and avoid touching live wires during operation.

Question 1: A circuit with a fixed current of 2 amperes shows a potential difference of 6 volts across resistor A and 12 volts across resistor B. Explain what this indicates about the resistances of the two resistors.

Answer: Since PD = IR, resistor A has a resistance of 3 ohms (6 V / 2 A), and resistor B has a resistance of 6 ohms (12 V / 2 A). Resistor B has twice the resistance of resistor A.

Question 1: Explain why a higher potential difference across a resistor results in a higher power dissipation, using appropriate equations. (6 marks)

Answer: Power dissipated in a resistor is given by P = V × I. As potential difference V increases, for a constant current I, the power P increases proportionally. Additionally, using P = V^2 / R, increasing V results in a quadratic increase in power, causing more energy to be converted into heat. Therefore, higher PD leads to higher power dissipation in the resistor.

Question 1: Explain how potential difference is utilised in electrical power transmission to reduce energy loss over long distances.

Answer: High potential difference (voltage) is used in power transmission lines to reduce current for the same power, which decreases the resistive energy loss (I^2 R). This makes long-distance transmission more efficient and reduces heat generation in the wires.