Physics · OCR H556 · Topic 6.1

Capacitors

Capacitors, energy, charging and discharging capacitors.

  • 3 sub-topics
  • Taught in Year 13
  • 49 practice questions
  • 3 animated lessons
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What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 6.1.1CapacitorsY13
  2. 6.1.2EnergyY13
  3. 6.1.3Charging and discharging capacitorsY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 49 questions in this topic, free, no account needed.

Electric and magnetic fields

Not needed for Year 12Easy

Capacitance · Year 13 · 2 marks

A 2200 μF capacitor is charged to 6 V.
  1. (a)
    Define capacitance.
    (1)
  2. (b)
    Calculate the charge stored.
    (1)

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Part (a)1 mark

  1. The charge stored per unit p.d.: C = Q/V.

Part (b)1 mark

  1. Q=0.0132 CQ = 0.0132\ \mathrm{C}

Total2

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Electric and magnetic fields

Not needed for Year 12Medium

Energy stored by a capacitor · Year 13 · 4 marks

A 4700 μF capacitor is charged to 12 V.
  1. (a)
    Calculate the energy stored.
    (2)
  2. (b)
    Explain why the energy stored is not QV.
    (1)
  3. (c)
    The p.d. is halved. State the fraction of the energy that remains.
    (1)

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Part (a)2 marks

  1. W=12CV2W = \tfrac{1}{2}CV^2
  2. W=0.338 JW = 0.338\ \mathrm{J}

Part (b)1 mark

  1. The p.d. rises from 0 to V as the capacitor charges, so the average p.d. is V/2: the area under the V–Q graph.

Part (c)1 mark

  1. One quarter (W ∝ V²).

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Electric and magnetic fields

Not needed for Year 12Hard

Exponential discharge · Year 13 · 5 marks

A 220 μF capacitor charged to 12 V discharges through a 47 kΩ resistor.
  1. (a)
    Calculate the p.d. across the capacitor after 10 s.
    (2)
  2. (b)
    Calculate the current at 10 s.
    (1)
  3. (c)
    Calculate the time for the p.d. to fall to 1.0 V.
    (2)

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Part (a)2 marks

  1. V=12e−10/10.3V = 12\mathrm{e}^{-10/10.3}
  2. V=4.56 VV = 4.56\ \mathrm{V}

Part (b)1 mark

  1. II=9.71×10−5 A{}= 9.71 \times 10^{-5}\ \mathrm{A}

Part (c)2 marks

  1. t=RCln⁡121.0t = RC\ln\frac{12}{1.0}
  2. t=25.7 st = 25.7\ \mathrm{s}

Total5

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  • B
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