DP Physics (last assessment 2024)

Question 21M.2.HL.TZ1.5b.i

Date	May 2021	Marks available	[Maximum mark: 2]	Reference code	21M.2.HL.TZ1.5b.i
Level	HL	Paper	2	Time zone	TZ1
Command term	Show that	Question number	b.i	Adapted from	N/A

b.i.

[Maximum mark: 2]

21M.2.HL.TZ1.5b.i

A device sends an impulse of electrical energy to maintain a regular heartbeat in a person. The device is powered by an alternating current (ac) supply connected to a step-up transformer that charges a capacitor of capacitance 30 μF.

The voltage across the primary coil of the transformer is 220 V. The number of turns on the secondary coil is 15 times greater than the number of turns on the primary coil.

(b.i)

Show that the maximum energy stored by the capacitor is about 160 J.

[2]

Markscheme

$V_{s} = 15 \times 220 = « 3300 V»$ ✓

$E = \frac{1}{2} C V^{2} = \frac{1}{2} \times 30 \times 10^{- 6} \times 3300^{2}$

163 «J» ✓

Allow use of 220 V as an RMS value to calculate V_s = 467 V and E = 327 J for full marks if appropriate work is provided.
Answer must be to 3 or more sf or working shown for MP2

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 21M.2.HL.TZ1.5b.i

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