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Date	May 2022	Marks available	1	Reference code	22M.2.HL.TZ1.8
Level	Higher level	Paper	Paper 2	Time zone	1
Command term	Explain	Question number	8	Adapted from	N/A

Question

Two capacitors C₁ and C₂ of capacitance 28 µF and 22 µF respectively are connected in a circuit with a two-way switch and a cell of emf 1.5 V with a negligible internal resistance. The capacitors are initially uncharged. The switch is then connected to position A.

The switch is moved to position B.

A cell is now connected by a switch to a coil X. A second coil Y of cross-sectional area 6.4 cm² with 5 turns is looped around coil X and connected to an ideal voltmeter.

The graph shows the variation with t of the magnetic flux density B in coil Y.

Show that the charge stored on C₁ is about 0.04 mC.

[1]

Calculate the energy transferred from capacitor C₁.

[4]

b.i.

Explain why the energy gained by capacitor C₂ differs from your answer in (b)(i).

[1]

b.ii.

The switch is closed at time t = 0. Explain how the voltmeter reading varies after the switch is closed.

[2]

c.i.

Determine the average emf induced across coil Y in the first 3.0 ms.

[3]

c.ii.

Markscheme

$Q =$ « $C V = 1.5 \times 28 \times 10^{- 6}$ » = 0.042 «mC» ✓

Award MP for full replacement or correct answer to at least 2 significant figures.

$E_{i} = \frac{1}{2} \times (28 \times 10^{- 6}) {(1.5)}^{2} = 3.15 \times 10^{- 5}$ «J» ✓

total capacitance = 50 «μF»

pd = « $\frac{42 \times 10^{- 6}}{50 \times 10^{- 6}} =$ » 0.84 «V»

charge on C₁ after switch moved to B = 0.0235 «mC» ✓

$E_{f} = \frac{1}{2} \times (28 \times 10^{- 6}) {(0.84)}^{2} = 9.9 \times 10^{- 6}$ «J» ✓

energy lost $= 3.2 \times 10^{- 5} - 9.9 \times 10^{- 6} = 22$ «μJ» ✓

b.i.

energy transferred to electromagnetic radiation «to environment»

energy is transferred as thermal energy / heat «to circuit components» ✓

b.ii.

initial deflection by voltmeter falling to zero reading ✓

emf is induced «only» while the field / flux is changing ✓

c.i.

attempted use of $ε = \frac{∆ Φ}{∆ t}$ OR $N A \frac{∆ B}{∆ t}$ ✓

$ε = \frac{5 \times 7.5 \times 10^{- 3} \times 6.4 \times 10^{- 4}}{3 \times 10^{- 3}}$ ✓

8.0 «mV» ✓

c.ii.

Examiners report

This was a "show that" question, and it was very well done by most candidates.

This question was challenging for many candidates. A large number successfully calculated the initial energy of C1, but then seemed confused about the next steps. Few candidates successfully calculated the energy in C1 after the switch was closed. There was an ECF opportunity for candidates who recognized that the final answer was the difference between these two values.

b.i.

This question used an "explain" command term, so examiners were looking for more than a generic "energy was lost". Candidates needed to specify a form of energy that was lost (such as thermal energy) for the mark. A very common incorrect response was simply stating that the difference was due to the capacitors having a different capacitance.

b.ii.

This question was well answered by some candidates who recognized that it was an electromagnetic induction question and understood that eventually the current in coil X would hit a steady state and the voltmeter reading would return to zero. Common issues were candidates thinking that the potential would fluctuate in a manner similar to an alternating current, candidates discussing this more as a transformer, and candidates who missed that there were two separate coils and wrote responses suggesting that a simple circuit had been formed and the voltmeter would read the potential of the cell.

c.i.

This question was well approached with most candidates recognizing that this was a Faraday's law question. Many made an attempt to use the correct equation, but common errors were choosing incorrect values from the graph and incorrectly converting the given area. Examiners were generous with ECF for candidates who clearly showed work leading to an incorrect result.

c.ii.

Syllabus sections

Additional higher level (AHL) » Topic 11: Electromagnetic induction » 11.3 – Capacitance

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