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Question 18N.2.HL.TZ0.7

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Date November 2018 Marks available [Maximum mark: 6] Reference code 18N.2.HL.TZ0.7
Level HL Paper 2 Time zone TZ0
Command term Comment, Determine Question number 7 Adapted from N/A
7.
[Maximum mark: 6]
18N.2.HL.TZ0.7

A small electric motor is used with a 12 mF capacitor and a battery in a school experiment.

When the switch is connected to X, the capacitor is charged using the battery. When the switch is connected to Y, the capacitor fully discharges through the electric motor that raises a small mass.

(a)

The battery has an emf of 7.5 V. Determine the charge that flows through the motor when the mass is raised.

[1]

Markscheme

charge stored on capacitor = 12 × 10−3 × 7.5 = 0.09 «C» ✔

(b)

The motor can transfer one-third of the electrical energy stored in the capacitor into gravitational potential energy of the mass. Determine the maximum height through which a mass of 45 g can be raised.

[2]

Markscheme

energy stored in capacitor « 1 2 CV2 or 1 2 QV =»  1 2  × 12 × 10−3 × 7.52 «= 0.338 J» ✔

height = « 1 3 × 0.338 9.81 × 4.5 × 10 − 2 = » 0.25/0.26 «m»

 

Allow use of g = 10 m s−2 which gives 0.25 «m»

(c)

An additional identical capacitor is connected in series with the first capacitor and the charging and discharging processes are repeated. Comment on the effect this change has on the height and time taken to raise the 45 g mass.

[3]

Markscheme

C halved ✔

so energy stored is halved/reduced so rises «less than» half height ✔

discharge time/raise time less as RC halved/reduced ✔

 

Allow 6 mF

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