Exam-style Questions: Power and Energy

  1. The heating element of an electric kettle operates at 230 V and has a power rating of 2.52 kW.

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    a) calculate the working resistance R of the heating element.

    working resistance = ................. Ω

    (2 Marks)

    b) A digital multimeter is used to measure the resistance of the heating element when it is cold. The value of the resistance measured by this method is considerably lower than the working resistance.

    Suggest a reason for this.

    (1 Mark)

    (Marks available: 3)

  2. Fig. 3.2 shows a negative temperature coefficient (NTC) thermistor connected to 24 V power supply of negligible internal resistance. The ammeter had negligible resistance.

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    a) When the switch S is closed, the ammeter reading is 28 mA. Calculate the power dissipated by the thermistor.

    power = ...................... W

    (3 Marks)

    b) A few minutes after closing the switch, the current has increased to a constant value of 40 mA. Explain why the current increases.

    (2 Marks)

    (Marks available: 5)

  3. A simple cell may be constructed by inserting into a fresh lemon two electrodes made from different metals. The juice of the lemon acts as an electrolyte (conducting liquid). Positive and negative ions within the lemon move towards the metal electrodes. Fig. 2.1 shows such a lemon-cell. It has an e.m.f. of 1.32 V and can provide enough electrical energy to activate a digital lock for many days.

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    a) On Fig. 2.1, indicate with an arrow the direction in which negative charge moves within the lemon.

    (1 Mark)

    b) The lemon-cell is capable of providing a steady current of 1.2 mA for eight days (6.9 x 105 x). Calculate

    i) the charge passing through the clock during eight days

    charge = ........... C

    (3 Marks)

    ii) the power delivered by the lemon-cell

    power = ..................... unit ............

    (3 Marks)

    (Total =7)

Answer

Answer outline and marking scheme for question:

  1. a) Sum of the currents = zero (at junction)OR

    sum of the currents in = sumof currents out (at junction)

    b) R = V2 / P / I = 10.96 A (1 Mark) m ; = 21.0 Ω (1 Mark) e

    (2 Marks)

    c) reference to element being colder and so lower resistance AW ora (NOT it is colder)

    (1 Mark)

    (Marks available: 3)

  2. a) P = VI / I2R /Copyright S-cool

    P = 24 x 0.028

    power = 0.672 ≈ 0.67 (W) (-1 for 10n error therefore 670 (W) scores 2/3)

    (3 Marks)

    b) (Thermistors) temperature increases (due to electrical heating) [AW] Resistance of thermisistor decreases.

    (2 Marks)

    (Marks available: 5)

  3. a) Arrow (within the lemon) and towards the negative terminal

    (1 Mark)

    b) i) Δ Q = lΔ t (Allow other subject. Δ is not necessary)

    charge = 1.2 x 10-3 x 6.9 x 105

    charge = 828 ≈ 830 (C) (-1 for 10n error tand -1 for t = 8 days)

    (3 Marks)

    ii) P = Vl

    P = 1.32 x 1.2 x 10-3 (ECF for current from b(i))

    P = 1.58 x 10-3 ≈1.6 x 10-3

    unit: W / Js-1 / VA

    (3 Marks)

    (Marks available: 7)