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Date	May 2022	Marks available	1	Reference code	22M.1.AHL.TZ2.13
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 2
Command term	Write down and Hence	Question number	13	Adapted from	N/A

Question

An electric circuit has two power sources. The voltage, $V_{1}$ $V_{1}$ , provided by the first power source, at time $t$ $t$ , is modelled by

$V_{1} = Re (2 e^{3 t i})$ $V_{1} = Re (2 e^{3 t i})$ .

The voltage, $V_{2}$ $V_{2}$ , provided by the second power source is modelled by

$V_{2} = Re (5 e^{(3 t + 4) i})$ $V_{2} = Re (5 e^{(3 t + 4) i})$ .

The total voltage in the circuit, $V_{T}$ $V_{T}$ , is given by

$V_{T} = V_{1} + V_{2}$ $V_{T} = V_{1} + V_{2}$ .

Find an expression for $V_{T}$ $V_{T}$ in the form $A cos (B t + C)$ $A \cos (B t + C)$ , where $A, B$ $A, B$ and $C$ $C$ are real constants.

[4]

Hence write down the maximum voltage in the circuit.

[1]

Markscheme

METHOD 1

recognizing that the real part is distributive (M1)

$V_{T} = Re (2 e^{3 t i} + 5 e^{3 t i + 4 i})$ $V_{T} = Re (2 e^{3 t i} + 5 e^{3 t i + 4 i})$

$= Re (e^{3 t i} (2 + 5 e^{4 i}))$ $= Re (e^{3 t i} (2 + 5 e^{4 i}))$ (A1)

(from the GDC) $2 + 5 e^{4 i} = 3.99088 \dots e^{- 1.89418 \dots i}$ $2 + 5 e^{4 i} = 3.99088 \dots e^{- 1.89418 \dots i}$ (A1)

Note: Accept arguments differing by $2 π$ $2 π$ e.g. $4.38900 \dots)$ $4.38900 \dots)$ .

therefore $V_{T} = 3.99 cos (3 t - 1.89) (3.99088 \dots cos (3 t - 1.89418 \dots))$ $V_{T} = 3.99 \cos (3 t - 1.89) (3.99088 \dots \cos (3 t - 1.89418 \dots))$ A1

Note: Award the last A1 for the correct values of $A, B$ $A, B$ and $C$ $C$ seen either in the required form or not. If method used is unclear and answer is partially incorrect, assume Method 2 and award appropriate marks eg. (M1)A1A0A0 if only $A$ $A$ value is correct.

METHOD 2

converting given expressions to cos form (M1)

$V_{T} = 2 cos 3 t + 5 cos (3 t + 4)$ $V_{T} = 2 \cos 3 t + 5 \cos (3 t + 4)$

(from graph) $A = 3.99 (3.99088 \dots)$ $A = 3.99 (3.99088 \dots)$ A1

$V_{T} = 3.99 cos (B t + C)$ $V_{T} = 3.99 \cos (B t + C)$

either by considering transformations or inserting points

$B = 3$ $B = 3$ A1

$C = - 1.89 (- 1.89418 \dots)$ $C = - 1.89 (- 1.89418 \dots)$ A1

Note: Accept arguments differing by $2 π$ $2 π$ e.g. $4.38900 \dots$ $4.38900 \dots$ .

(so, $V_{T} = 3.99 cos (3 t - 1.89) (3.99088 \dots cos (3 t - 1.89418 \dots))$ $V_{T} = 3.99 \cos (3 t - 1.89) (3.99088 \dots \cos (3 t - 1.89418 \dots))$ )

Note: It is possible to have $A = 3.99$ $A = 3.99$ , $B = - 3$ $B = - 3$ with $C = 1.89$ $C = 1.89$ OR $A = - 3.99$ $A = - 3.99$ , $B = 3$ $B = 3$ with $C = 1.25$ $C = 1.25$ OR $A = - 3.99$ $A = - 3.99$ , $B = - 3$ $B = - 3$ with $C = - 1.25$ $C = - 1.25$ due to properties of the cosine curve.

[4 marks]

maximum voltage is $3.99 (3.99088 \dots)$ $3.99 (3.99088 \dots)$ (units) A1

[1 mark]

Examiners report

The crucial step in this question was to realize that $Re (2 e^{3 t i})+Re (5 e^{(3 t + 4) i})=Re (2 e^{3 t i} + 5 e^{(3 t + 4) i})$ $Re (2 e^{3 t i})+Re (5 e^{(3 t + 4) i})=Re (2 e^{3 t i} + 5 e^{(3 t + 4) i})$ . Candidates who failed to do this step were usually unable to obtain the required result.

[N/A]

Syllabus sections

Topic 1—Number and algebra » AHL 1.12—Complex numbers introduction

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