Exam-style Questions: Capacitors

  1. The graph shows how the chanreg of a capacitor varies with the p.d. across the capacitor.

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    Use the graph to find

    a) the energy stored by the capacitor when charged to a potential difference of 8 V

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

    (2 Marks)

    b) the capacitance of the capacitor

    capacitance = ..................... F

    (2 Marks)

    (Marks available: 4)

  2. Fig. 7.1 shows a circuit diagram of a capacitor discharging through a resistor.

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    A simple mathematical model of the discharge of the capacitor is shown in Fig. 7.2. It is assumed that the current / is constant over each small time interval, Δt, The process is repeated as shown.

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    Complete the table for the discharge of the 4700μF capacitor. The small time interval used is Δt = 2.0 s.

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    (3 Marks)

    (Marks available: 3)

  3. In the circuit in Fig. 6.1, the capacitor is charged to a potential difference of 6.0 V.

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    When the switch is moved from A to B , the capacitor discharges through the resistor.

    a) Show that the initial value of the discharge current is about 1 mA.

    (2 Marks)

    b) The time constant RC of the discharge circuit is about 26 s.

    Calculate the current in the discharge circuit after the switch has been closed for a time equal to RC.

    current = .............. mA

    (2 Marks)

    (Marks available: 4)

    Answer

    Answer outline and marking scheme for question:

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

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

      b) Area under graoh (equiv to ½ QV) = ½ x 3.5 x 10-3 x 8 (1 Mark)

      = 0.014 J (1 Mark)

      (2 Marks)

      c) Grad = 3 x 10-3 / 6.8 (for example) (1 Mark) = 4.4 x 10-4 F (1 Mark)

      (2 Marks)

      (Marks available: 4)

    2. First line of table : 2.4 x 10-3, 4.8 x 10-3 (1 Mark)

      Second line of table : 2.2 x 10-3, 4.4 x 10-3 (1 Mark)

      4.7 x 10-2

      (3 Marks)

      (Marks available: 3)

    3. a) l = V/R = 6/ 5.6 x 103 (1 Mark) = 1.1 x 10-3 A = (about) 1 mA (1 Mark)

      (2 Marks)

      b) l = 1.1 x 10-3 x e-1 = 1.1 x 10-3 x 0.37 (1 Mark) = 0.4 mA (1 Mark) (accept rule of thumb third, answers using 1 mA and answers using decay equation)

      (2 Marks)

      (Marks available: 4)