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    PracticeAQA GCSEAQA GCSE Physics Foundation Tier Paper 1Question 07.8
    Medium3 marksStructured
    ElectricityFoundationElectricityOhms LawResistance

    AQA GCSE · Question 07.8 · Electricity

    3.6 V A

    There is a potential difference of 3.6 V across the lamp in Figure 11.
    The current through the lamp is 0.80 A.
    Calculate the resistance of the lamp.

    How to approach this question

    1. Start with the equation V = I × R. 2. Rearrange the equation to make resistance (R) the subject. 3. Identify the values given: Potential difference (V) = 3.6 V and Current (I) = 0.80 A. 4. Substitute these values into your rearranged equation. 5. Calculate the resistance. The unit is Ohms (Ω).

    Full Answer

    Equation: resistance = potential difference / current Substitution: R = 3.6 V / 0.80 A Answer: R = 4.5 Ω
    We use Ohm's Law, which links potential difference (V), current (I), and resistance (R): V = I × R To find the resistance, we need to rearrange the equation by dividing both sides by current (I): R = V / I Now, substitute the given values: - V = 3.6 V - I = 0.80 A R = 3.6 / 0.80 R = 4.5 Ω

    Common mistakes

    ✗ Incorrectly rearranging the equation (e.g., multiplying V and I). ✗ Dividing current by voltage instead of voltage by current. ✗ Calculation errors.
    Question 07.7All questionsQuestion 08.1

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