· H2 Physics — Ten-Year Series, worked ← All 7 questions

2022 A-Level · Paper 2 · Question 7

An electric car, from the cost per kilometre to the cells in its battery

Data-based question · electric vehicles 21 marks D.C. circuits Rectification Electromagnetism

The twenty-one-mark data question: a passage on electric vehicles in Singapore and a table of data for a typical one. Running cost against petrol, the fewest cells the battery can have, what 141 Wh kg⁻¹ means and what it says about where the battery goes, rectifying a.c., the distance gained from regenerative braking, why wireless charging needs a.c., and the field the motor needs. Below is the full solution, and beside it the battery, the rectifier and the motor.

The model, live

Cells / output / torqueInputOver the limitWithin the limit

Part (a)(i)

1 mark

Electricity costs S$0.23 per kWh. Find the ratio of the cost per km of the typical EV to the cost per km of an ICE car.

Approach

Find each cost per kilometre from the passage and the table. A full charge is 72 kWh72\,\text{kWh} for 400 km400\,\text{km}; a full tank is S$80 for 800 km800\,\text{km}.

EV: 72×0.23400=0.0414 dollars per kmICE: 80800=0.100 dollars per km\text{EV: } \frac{72 \times 0.23}{400} = 0.0414\text{ dollars per km} \qquad \text{ICE: } \frac{80}{800} = 0.100\text{ dollars per km}
ratio=0.04140.100=0.41\text{ratio} = \frac{0.0414}{0.100} = 0.41

ratio =0.41= 0.41

Part (a)(ii)

1 mark

State why EVs in Singapore typically need charging less than once a week.

A car in Singapore typically travels 290 km290\,\text{km} a week, which is less than the EV's range of 400 km400\,\text{km} on one charge.

Part (b)(i)

3 marks

Each cell can take at most 2.0 A2.0\,\text{A} and 3.0 V3.0\,\text{V} while charging. Using the passage, find the minimum number of cells in the battery.

Approach

The home charger delivers 7.2 kW7.2\,\text{kW} at 32 A32\,\text{A}, which fixes the charging p.d. Cells in series share the p.d.; parallel strings share the current. Find the fewest of each.

V=PI=720032=225 VV = \frac{P}{I} = \frac{7200}{32} = 225\,\text{V}
cells in series=2253.0=75strings in parallel=322.0=16\text{cells in series} = \frac{225}{3.0} = 75 \qquad \text{strings in parallel} = \frac{32}{2.0} = 16
minimum number=75×16=1200\text{minimum number} = 75 \times 16 = 1200

The charging p.d. is 7200/32=225 V7200/32 = 225\,\text{V}, so each string needs at least 225/3.0=75225/3.0 = 75 cells in series; the current of 32 A32\,\text{A} needs at least 32/2.0=1632/2.0 = 16 strings in parallel. Minimum number =75×16=1200= 75 \times 16 = 1200 cells.

Part (b)(ii)

1 mark

Suggest what the manufacturer means by a "charging speed" of 40 km h−140\,\text{km h}^{-1}.

Each hour of charging adds enough energy for the EV to travel 40 km40\,\text{km} (a full charge of 400 km400\,\text{km} in 10 hours).

Part (b)(iii)

1 mark

Suggest what is meant by a specific energy of 141 Wh kg−1141\,\text{Wh kg}^{-1}.

Each kilogram of the fully charged battery stores 141 Wh141\,\text{Wh} of energy.

Part (b)(iv)

1 mark

Find the mass of the battery.

m=72 000 Wh141 Wh kg−1=511 kgm = \frac{72\,000\,\text{Wh}}{141\,\text{Wh kg}^{-1}} = 511\,\text{kg}

mass =510 kg= 510\,\text{kg}

Part (b)(v)

1 mark

Use (b)(iv) to suggest a physics-based reason for putting the battery under the floor.

The battery is about a third of the EV's 1685 kg1685\,\text{kg} mass, so placing it under the floor keeps the centre of gravity low, making the EV more stable and less likely to topple.

Part (c)(i)

1 mark

Complete the circuit with a single component that converts the a.c. input into a d.c. output across the load.

A diode connected in series between one a.c. input terminal and the load (the other input terminal joined directly to the other end of the load).

Part (c)(ii)

2 marks

Sketch the d.c. output voltage for two cycles of the a.c. input of period TT.

Half-wave rectification: positive half-sine humps from 00 to T/2T/2 and from TT to 3T/23T/2, with the output zero from T/2T/2 to TT and from 3T/23T/2 to 2T2T.

Note

If you draw the humps in the other half-cycles instead, that is also correct: it just means the diode is the other way round. What matters is one polarity only, and zero for alternate half-cycles.

Part (d)

3 marks

The EV brakes from 25 m s−125\,\text{m s}^{-1} to rest, and all its kinetic energy goes into the battery. Find the distance it can travel on this energy.

Approach

Find the kinetic energy, convert it to kWh, then use the EV's distance per kWh: 400 km400\,\text{km} from 72 kWh72\,\text{kWh}.

Ek=12(1685)(25)2=5.27×105 J=5.27×1053.6×106=0.146 kWhE_k = \tfrac12(1685)(25)^2 = 5.27\times10^{5}\,\text{J} = \frac{5.27\times10^{5}}{3.6\times10^{6}} = 0.146\,\text{kWh}
distance=0.146×40072=0.81 km\text{distance} = 0.146 \times \frac{400}{72} = 0.81\,\text{km}

distance =0.81 km= 0.81\,\text{km}

Watch

1 kWh=1000×3600=3.6×106 J1\,\text{kWh} = 1000 \times 3600 = 3.6\times10^{6}\,\text{J}.

Part (e)(i)

1 mark

Explain why a wireless charger must use an a.c. voltage rather than a d.c. voltage.

An e.m.f. is induced in the EV's coil only when the magnetic flux linking it changes. An alternating current in the ground coil gives a continually changing flux; a steady direct current gives a constant flux, and no e.m.f. would be induced.

Part (e)(ii)

1 mark

Explain why the coil in the ground pad and the coil in the EV should be as close together as possible.

The magnetic field of the ground coil spreads out and weakens with distance. The closer the coils, the more of its flux links the EV's coil, so the induced e.m.f. is larger and less energy is wasted.

Part (f)(i)

3 marks

The coil has 1200 turns, area 6.1×10−3 m26.1\times10^{-3}\,\text{m}^2 in the field, and carries 96 A96\,\text{A}. Find the flux density BB needed for the maximum output torque of 395 N m395\,\text{N m}.

Approach

Each long side of length ll feels F=NBIlF = NBIl, and the two forces form a couple of arm dd when the plane of the coil is parallel to the field. The torque is Fd=NBI(ld)=NBIAFd = NBI(ld) = NBIA, its maximum.

τ=NBIA⟹B=τNIA=395(1200)(96)(6.1×10−3)=0.56 T\tau = NBIA \quad\Longrightarrow\quad B = \frac{\tau}{NIA} = \frac{395}{(1200)(96)(6.1\times10^{-3})} = 0.56\,\text{T}

B=0.56 TB = 0.56\,\text{T}

Part (f)(ii)

1 mark

A practical motor has four pairs of poles around the coil instead of one. Suggest an advantage.

The torque is more nearly constant as the coil rotates. With one pair of poles the torque falls to zero twice each revolution; with four pairs there is always a pair close to giving maximum torque, so the motor turns more smoothly and has a larger average torque.

What this question was really testing

  1. Pulling data from a passage. Every number you need is in the text or the table; the skill is choosing the right ones.
  2. Series and parallel. Series shares the p.d., parallel shares the current.
  3. Energy units. kWh, Wh kg⁻¹ and joules, converted with care.
  4. Induction needs change. No changing flux, no induced e.m.f.
  5. Torque on a coil, NBIANBIA. And why several pole pairs smooth it out.

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