4.2.4

Resistance

Cambridge IGCSE Physics (0625)  · Unit 4: Electricity and magnetism  · 10 flashcards

Resistance is topic 4.2.4 in the Cambridge IGCSE Physics (0625) syllabus , positioned in Unit 4 — Electricity and magnetism , alongside Simple phenomena of magnetism, Electric charge and Electric current.

This topic is examined in Paper 1 (multiple-choice) and Papers 3/4 (theory), plus Paper 5 or Paper 6 (practical / alternative to practical).

The deck below contains 10 flashcards — covering the precise wording mark schemes reward.

What the Cambridge 0625 syllabus says

Official 2026-2028 spec

These are the exact learning objectives Cambridge sets for this topic. Match the command word (Describe, Explain, State, etc.) in your answer to score full marks.

  1. Recall Recall and use the equation for resistance
  2. Describe Describe an experiment to determine resistance using a voltmeter and an ammeter and do the appropriate calculations
  3. State State, qualitatively, the relationship of the resistance of a metallic wire to its length and to its cross-sectional area
  4. Sketch Sketch and explain the current-voltage graphs for a resistor of constant resistance, a filament lamp and a diode Supplement
  5. Recall Recall and use the following relationship for a metallic electrical conductor: (a) resistance is directly proportional to length (b) resistance is inversely proportional to cross-sectional area Supplement
Key Concept Flip

A potential difference of 12 V is applied across a resistor. If the current flowing through the resistor is 2 A, calculate the resistance of the resistor.

Answer Flip

Resistance (R) = Voltage (V) / Current (I)
R = 12 V / 2 A
R = 6 Ω

Explanation: The formula R=V/I is used to calculate resistance from voltage and current.

Key Concept Flip

A component has a resistance of 10 Ohms. If a potential difference is applied across the component, describe the relationship between the potential difference and the current flowing through it.

Answer Flip

According to Ohm's Law (V=IR), the potential difference is directly proportional to the current, assuming the resistance remains constant. This means if you double the potential difference, the current will also double, and vice-versa.

Key Concept Flip

In an experiment to determine the resistance of a resistor, a voltmeter reads 6.0 V when connected across the resistor. The ammeter in the circuit reads 0.50 A. Calculate the resistance of the resistor.

Answer Flip

Resistance, R = Voltage, V / Current, I
R = 6.0 V / 0.50 A
R = 12 Ω

The resistance is calculated using Ohm's Law. Voltage is divided by current to find the resistance.

Key Concept Flip

Describe how you would use a voltmeter and an ammeter to determine the resistance of an unknown resistor in a circuit.

Answer Flip

1. Connect the resistor in a circuit with a power supply, an ammeter (in series), and a voltmeter (in parallel across the resistor).
2. Adjust the power supply to provide a suitable voltage and record the voltage (V) from the voltmeter and the current (I) from the ammeter.
3. Calculate the resistance (R) using Ohm's Law: R = V/I.

Key Concept Flip

State the relationship between the resistance of a metallic wire and its length.

Answer Flip

The resistance of a metallic wire is directly proportional to its length. This means that if the length of the wire increases, the resistance increases proportionally, assuming all other factors remain constant.

Key Concept Flip

Explain, qualitatively, how the cross-sectional area of a metallic wire affects its resistance.

Answer Flip

The resistance of a metallic wire is inversely proportional to its cross-sectional area. This means that as the cross-sectional area increases, the resistance decreases, and vice versa, assuming all other factors remain constant. A larger area provides more space for electrons to flow, reducing opposition.

Key Concept Flip

Sketch a current-voltage (I-V) graph for a filament lamp. Explain the shape of the graph.

Answer Flip

Sketch should show a curve, starting with a steep slope near the origin, then flattening out at higher voltages/currents.

Explanation: At higher voltages, the filament lamp gets hotter. The increased temperature causes the resistance of the filament to increase. Since resistance = voltage / current (R = V/I), for the same increase in voltage, the increase in current becomes smaller. This results in a less steep slope on the I-V graph at higher voltages.

Key Concept Flip

Describe the shape of the current-voltage (I-V) graph for a diode.

Answer Flip

The I-V graph for a diode shows very little current flowing when the voltage is negative (reverse bias). At a specific positive voltage (forward voltage), the current increases rapidly.

Key Concept Flip

A copper wire has a resistance of 2.0 Ω and a length of 5.0 m. What is the resistance of a copper wire of the same material with a length of 15.0 m and the same cross-sectional area?

Answer Flip

Resistance is directly proportional to length.

R₁/L₁ = R₂/L₂

2.0 / 5.0 = R₂ / 15.0

R₂ = (2.0 / 5.0) * 15.0

R₂ = 6.0 Ω

Answer: 6.0 Ω

Key Concept Flip

Explain why increasing the cross-sectional area of a wire decreases its resistance.

Answer Flip

Increasing the cross-sectional area provides more space for electrons to flow. With more available space, electrons experience fewer collisions with atoms in the wire. Fewer collisions mean less opposition to the current, hence a lower resistance.

(Area ∝ 1/Resistance)

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4.2.3 Electromotive force and potential difference 4.2.5 Electrical energy and electrical power

More topics in Unit 4 — Electricity and magnetism

Resistance sits alongside these Physics decks in the same syllabus unit. Each uses the same spaced-repetition system, so progress in one informs the next.

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How to study this Resistance deck

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