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Question 19M.2.HL.TZ2.4

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Date May 2019 Marks available [Maximum mark: 10] Reference code 19M.2.HL.TZ2.4
Level HL Paper 2 Time zone TZ2
Command term Calculate, Deduce, State, Suggest Question number 4 Adapted from N/A
4.
[Maximum mark: 10]
19M.2.HL.TZ2.4

Three identical light bulbs, X, Y and Z, each of resistance 4.0 Ω are connected to a cell of emf 12 V. The cell has negligible internal resistance.

(a)

The switch S is initially open. Calculate the total power dissipated in the circuit.

[2]

Markscheme

total resistance of circuit is 8.0 «Ω» ✔

P = 12 2 8.0 = 18 «W» ✔

Examiners report

Most candidates scored both marks. ECF was awarded for those who didn’t calculate the new resistance correctly. Candidates showing clearly that they were attempting to calculate the new total resistance helped examiners to award ECF marks.

(bi)

The switch is now closed. State, without calculation, why the current in the cell will increase.

[1]

Markscheme

«a resistor is now connected in parallel» reducing the total resistance

OR

current through YZ unchanged and additional current flows through X ✔

Examiners report

Most recognised that this decreased the total resistance of the circuit. Answers scoring via the second alternative were rare as the statements were often far too vague.

(bii)

The switch is now closed. Deduce the ratio  power dissipated in Y with S open power dissipated in Y with S closed .

 

[2]

Markscheme

evidence in calculation or statement that pd across Y/current in Y is the same as before ✔

so ratio is 1 ✔

Examiners report

Very few gained any credit for this at both levels. Most performed complicated calculations involving the total circuit and using 12V – they had not realised that the question refers to Y only.

(c)

The cell is used to charge a parallel-plate capacitor in a vacuum. The fully charged capacitor is then connected to an ideal voltmeter.

The capacitance of the capacitor is 6.0 μF and the reading of the voltmeter is 12 V.

Calculate the energy stored in the capacitor.

[1]

Markscheme

E = « 1 2 C V 2 = 1 2 × 6 × 10 − 6 × 12 2 = » 4.3 × 10 − 4 « J » ✔

Examiners report

Most answered this correctly.

When fully charged the space between the plates of the capacitor is filled with a dielectric with double the permittivity of a vacuum.

(di)

Calculate the change in the energy stored in the capacitor.

[3]

Markscheme

ALTERNATIVE 1

capacitance doubles and voltage halves ✔

since  E = 1 2 C V 2 energy halves   ✔

so change is «–»2.2×10–4 «J»  ✔

 

ALTERNATIVE 2

E = 1 2 C V 2  and  Q = C V  so  E = Q 2 2 C  ✔ 

capacitance doubles and charge unchanged so energy halves ✔

so change is «−»2.2 × 10−4 «J» ✔

Examiners report

By far the most common answer involved doubling the capacitance without considering the change in p.d. Almost all candidates who did this calculated a change in energy that scored 1 mark.

(dii)

Suggest, in terms of conservation of energy, the cause for the above change.

[1]

Markscheme

it is the work done when inserting the dielectric into the capacitor ✔

Examiners report

Very few scored on this question.

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