Exam-style Questions: Alternating Currents

  1. The diagram shows the field shape around two wires carrying current into or out of the plane of the page. Which currents in P and Q could produce the effect shown?


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    1. P:Out of the screen Q: Out of the screen
    2. P: Into the screen Q: Into the screen
    3. P: Into the screen Q: Out of the screen
    4. P: Out of the screen Q: Into the screen

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    You need to recognise this field shape. It is created when two wires carry current in the same direction, i.e.into or out of the page. To work out which of theses options is correct, use the corkscrew rule (see the quick learn notes on magnetism if you've not heard of it) on P and see which current direction through P gives the correct field shape to the left hand side of P (rather than in between the wires where the shape is more confusing).

  2. Look at this circuit.
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    Which of the graphs describes the flux in the coil when the switch in the circuit is closed and then, a few seconds later, opened?

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    1. A
    2. B
    3. C
    4. D

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    Flux is another way of saying magnetic field.

    When a current flows through a wire, a magnetic field is set up around the wire and in the coil. In this case, when the switch is closed a field is set up. (So the graph must go from zero to a value.) When the switch is opened at the end, the field disappears. (So the graph must go from a value to zero.) In between, as it is a d.c. circuit, there is a steady current flowing so the field has a steady value.

  3. a) Name an instrument used to measure

    i) electric current,

    ii) potential difference

    b) The electric charge delta Q passing through a point in a circuit is given by the equation

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    State what is represented by the symbols I and Δt.

    c) A 1.2kW water heater is switched on for 1500s. During this time a charge of 7.5 x 103C passes. Calculate

    i) the electric current,

    Current =..........................................A

    ii) the p.d. across the heater,

    p.d. = .........................................V

    iii) the electrical energy transformed by the heater,

    energy = ..........................................J

    iv) the cost of using the heater given that the cost of 1kWh is 6.4p.

    Cost =..........................................p

    (Marks available: 13)



    Answer outline and marking scheme for question:

    3. a) i) Ammeter

    ii) Voltmeter

    (2 marks)

    b) I = current

    Δt = time (interval/duration)

    (2 marks)

    c) i) I = 7.5 x 103 / 1500

    I = 5.0 (A)

    (2 marks)

    ii) V = P / I (allow other variants

    V = 1.2 x 103 / 5.0 (possible e.c.f)

    V = 240 V (-1 for missing the kW or 103 factor)

    (3 marks)

    ii) E = 1.2 x 103 x 1500

    E = 1.8 x 106J (-1 for missing the kW or the 103 factor, penalise once only in (ii) & (iii))

    (2 marks)

    iii) units = 1500 / 3600 x 1.2 = 0.5 units = 1.8 x 106 / 3.6 x 106 = 0.5

    cost = 0.5 x 6.4 = 3.2 p (possible (e.c.f if (iii) used)

    (2 marks)

    (Marks available: 13)