4.3.3

Action and use of circuit components

Cambridge IGCSE Physics (0625)  · Unit 4: Electricity and magnetism  · 6 flashcards

Action and use of circuit components is topic 4.3.3 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 6 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. Know Know that the p.d. across an electrical conductor increases as its resistance increases for a constant current
  2. Describe Describe the action of a variable potential divider Supplement
  3. Recall Recall and use the equation for two resistors used as a potential divider Supplement
Key Concept Flip

A resistor has a constant current of 0.5 A flowing through it. If the resistance of the resistor is 10 ohms, calculate the potential difference across the resistor.

Answer Flip

Formula: V = IR
Calculation: V = 0.5 A * 10 ohms = 5 V
Answer: The potential difference across the resistor is 5 V. Explanation: Ohm's Law states that voltage is directly proportional to the product of current and resistance.

Key Concept Flip

State what happens to the potential difference across a conductor if its resistance increases, assuming the current flowing through it remains constant.

Answer Flip

The potential difference across the conductor increases. Explanation: According to Ohm's Law (V=IR), if the resistance (R) increases while the current (I) remains constant, the potential difference (V) must also increase to maintain the equality.

Key Concept Flip

A potential divider circuit consists of a 12V battery connected in series with a 200Ω resistor and a variable resistor. The variable resistor is adjusted so that its resistance is 300Ω. Calculate the voltage across the 200Ω resistor.

Answer Flip

V = IR. First calculate the total resistance: R_total = 200Ω + 300Ω = 500Ω. Then calculate the current: I = V/R = 12V / 500Ω = 0.024A. Finally, calculate the voltage across the 200Ω resistor: V = IR = 0.024A * 200Ω = 4.8V. The voltage across the 200Ω resistor is 4.8V.

Key Concept Flip

Describe how a variable potential divider can be used to control the voltage supplied to a component, such as a lamp.

Answer Flip

A variable potential divider uses a variable resistor to split the voltage from a power supply. By adjusting the position of the sliding contact on the variable resistor, the proportion of the total resistance across which the lamp is connected changes. This alters the voltage across the lamp. Moving the slider towards one end of the resistor increases the voltage across that section, thus increasing the voltage supplied to the lamp. Moving the slider towards the other end decreases the voltage across the lamp.

Key Concept Flip

Two resistors, 300 Ω and 500 Ω, are connected in series to a 12V power supply. Calculate the voltage across the 300 Ω resistor.

Answer Flip

Formula: V₁ = (R₁ / (R₁ + R₂)) * V_in

Calculation: V₁ = (300 / (300 + 500)) * 12V = (300/800) * 12V = 0.375 * 12V = 4.5V

Answer: The voltage across the 300 Ω resistor is 4.5V. This is because the voltage divides proportionally to the resistance.

Key Concept Flip

Explain how a potential divider circuit allows you to obtain a voltage that is lower than the voltage of the power supply. Relate your answer to the resistance values in the circuit.

Answer Flip

A potential divider uses resistors in series to 'divide' the input voltage. The voltage across each resistor is proportional to its resistance. The higher the resistance of a resistor compared to the total resistance, the greater the proportion of the input voltage that will appear across it. Therefore, by using a resistor network, the voltage across a single resistor will always be lower than the total supply voltage unless that resistor comprises the whole resistance in the circuit.

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Test yourself

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4.3.2 Series and parallel circuits 4.4 Electrical safety

More topics in Unit 4 — Electricity and magnetism

Action and use of circuit components sits alongside these Physics decks in the same syllabus unit. Each uses the same spaced-repetition system, so progress in one informs the next.

Related Physics guides

Long-read articles that go beyond the deck — cover the whole subject's common mistakes, high-yield content and revision pacing.

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