DP Physics (last assessment 2024)

Question 22N.2.SL.TZ0.5b.ii

Date	November 2022	Marks available	[Maximum mark: 2]	Reference code	22N.2.SL.TZ0.5b.ii
Level	SL	Paper	2	Time zone	TZ0
Command term	Determine	Question number	b.ii	Adapted from	N/A

b.ii.

[Maximum mark: 2]

22N.2.SL.TZ0.5b.ii

Resistor R is connected in a circuit with a cell that has internal resistance.

The ammeter and the voltmeter are ideal.

The cell has an emf of 1.49 V. The resistance of R is 50.0 Ω. The voltmeter reads 1.47 V.

(b.ii)

Determine the total power dissipated in the circuit.

[2]

Markscheme

$1.49 \times 2.94 \times 10^{- 2}$

$\frac{1 . 49^{2}}{50.0 + 0.68}$

${(2.94 \times 10^{- 2})}^{2} (50.0 + 0.68)$ ✓

$0.0438$ «W» ✓

Accept use of 0.7 Ω in MP1.

Examiners report

The calculation of total power dissipated was not usually well done, as it often failed to include the internal resistance therefore calculating the power dissipated only in the resistor.

Syllabus sections

Core

Core » Topic 5: Electricity and magnetism » 5.2 – Heating effect of electric currents

Core » Topic 5: Electricity and magnetism

Question 22N.2.SL.TZ0.5b.ii

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