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Date	May 2022	Marks available	2	Reference code	22M.2.SL.TZ1.4
Level	Standard level	Paper	Paper 2	Time zone	1
Command term	Calculate	Question number	4	Adapted from	N/A

Question

A power supply is connected to three resistors P, Q and R of fixed value and to an ideal voltmeter. A variable resistor S, formed from a solid cylinder of conducting putty, is also connected in the circuit. Conducting putty is a material that can be moulded so that the resistance of S can be changed by re-shaping it.

The resistance values of P, Q and R are 40 Ω, 16 Ω and 60 Ω respectively. The emf of the power supply is 6.0 V and its internal resistance is negligible.

All the putty is reshaped into a solid cylinder that is four times longer than the original length.

Calculate the potential difference across P.

[2]

The voltmeter reads zero. Determine the resistance of S.

[3]

Deduce the resistance of this new cylinder when it has been reshaped.

[3]

c.i.

Outline, without calculation, the change in the total power dissipated in Q and the new cylinder after it has been reshaped.

[2]

c.ii.

Markscheme

ALTERNATIVE 1

attempt to use potential divider equation or similar method ✓
« $6.0 \times \frac{40}{(40 + 60)}$ »= 2.4 «V» ✓

ALTERNATIVE 2

«current = $\frac{6}{60 + 40}$ » = 0.06 «A» ✓

40 x 0.06 = 2.4 «V» ✓

ALTERNATIVE 1

Pd across Q = 2.4 V so I = 0.15 « A » ✓

and pd across S is 6.0 – 2.4 = 3.6 « V » ✓

$R = \frac{V}{I} = \frac{3.6}{0.15} = 24$ «Ω» ✓

ALTERNATIVE 2

pd at PR junction = pd at QS junction ✓

so $\frac{S}{Q} = \frac{R}{P}$ OR $S = \frac{16 \times 60}{40}$ ✓

24 «Ω» ✓

ALTERNATIVE 3

$I = \frac{40 \times 0.06}{16} = 0.15$ «A» ✓

$6.0 = (16 + S) (0.15)$ OR $2.4 = (6) (\frac{16}{S + 16})$ ✓

R = 24 «Ω» ✓

Allow ECF for MP3 from incorrect MP1 or MP2.

recognition that $4 L$ leads to $4 R$ / « $\frac{ρ 4 L}{A}$ » = $4 R$ ✓

«because the volume of S is constant new area is» $\frac{A}{4}$ ✓

16 x 24 = 384 «Ω» ✓

c.i.

«total» power has decreased ✓

Because current in the branch has decreased «and P=I²R »

Because resistance has increased in branch «and $P = \frac{V^{2}}{R}$ » ✓

Allow opposite argument as ECF from (c)(i) (if candidate deduces a lower resistance).

Allow “power doesn’t change” if candidate has no change of resistance from (b) to (c)(i).

c.ii.

Examiners report

This was generally well done at the higher level. Some SL candidates struggled to calculate the correct current but earned the second marking point through ECF.

Many candidates struggled with this question. A very common mistake was to assume the current was the same in each branch, leading to a resistance of 84 Ohms. The placement of the voltmeter may have caused some confusion for candidates, and they may not have understood what the zero reading was indicating. It is important that candidates understand what voltmeters are actually reading and are familiar with different placements in circuits.

Most candidates recognized that increasing the length of the conductive putty would increase the resistance by a factor of four, but very few considered that if the volume of the putty remained constant that the cross-sectional surface area would decrease as well.

c.i.

This was an item that caused a bit of confusion for candidates. The prompt asks for a comparison of the power in the branch before and after changing the length of the conductive putty. Many candidates correctly identified that the power in Q would decrease, but either did not discuss the power in the whole branch or were not clear that the power in the putty would decrease as well.

c.ii.

Syllabus sections

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

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