DP Physics (last assessment 2024)

Question 22M.2.HL.TZ1.8b.i

Date	May 2022	Marks available	[Maximum mark: 4]	Reference code	22M.2.HL.TZ1.8b.i
Level	HL	Paper	2	Time zone	TZ1
Command term	Calculate	Question number	b.i	Adapted from	N/A

b.i.

[Maximum mark: 4]

22M.2.HL.TZ1.8b.i

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.

(b.i)

Calculate the energy transferred from capacitor C₁.

[4]

Markscheme

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

Examiners report

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.

Syllabus sections

Additional higher level (AHL)

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

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

Question 22M.2.HL.TZ1.8b.i

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