Physics questions, chapter 3

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Electricity

45 marks. Use Practice to check each answer as you go, or Exam to answer everything against the clock first.

Section A: objective questions

11 marks

Choose the best answer for each question.

Question 1

What is an electric field?

The path along which electrons move in a wire
The region around a charged particle where another charge experiences an electric force
The work done to move one coulomb of charge around a circuit
The rate of flow of electric charge

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An electric field is the region around a charged particle where any electric charge experiences an electric force (textbook p. 94). The other options describe a conductor, e.m.f. and current.

For the teacher: Remembering, low difficulty, standard 3.1.1

Textbook p. 94

Question 2

Which is the S.I. unit of electric field strength?

N C⁻¹
J C⁻¹
C s⁻¹
N m⁻¹

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E = F / q, so the unit is newton per coulomb (N C⁻¹), which equals V m⁻¹. J C⁻¹ is the volt and C s⁻¹ is the ampere.

For the teacher: Remembering, low difficulty, standard 3.1.2

Textbook p. 95

Question 3

A candle flame is placed between two charged metal plates. The flame spreads out more towards the negative plate than towards the positive plate. What is the reason?

The negative plate is hotter than the positive plate
More negative ions than positive ions are formed in the flame
Positive ions are larger and heavier than negative ions, so more of them move towards the negative plate
Negative ions are larger than positive ions and move to the negative plate

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The hot flame ionises the air. Positive ions go to the negative plate and are larger and heavier than the negative ions, so the flame spreads more in that direction.

For the teacher: Understanding, medium difficulty, standard 3.1.3

Textbook p. 97

Question 4

A potential difference of 500 V is applied across two parallel metal plates that are 2.0 cm apart. What is the electric field strength between the plates?

250 V m⁻¹
1000 V m⁻¹
10 000 V m⁻¹
25 000 V m⁻¹

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E = V / d = 500 ÷ 0.020 = 25 000 V m⁻¹. The distance must be in metres (2.0 cm = 0.020 m).

For the teacher: Applying, medium difficulty, standard 3.1.2

Textbook p. 95

Question 5

The graph of potential difference against current for a filament bulb is a curve. Why does the filament not obey Ohm's law?

The voltage across the bulb is too high
The filament is made of an insulator
The current is the same at every point in the filament
The temperature of the filament rises, so its resistance increases

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Ohm's law holds only at constant temperature. As the current increases the filament becomes hotter and its resistance increases, so the graph of V against I curves.

For the teacher: Understanding, medium difficulty, standard 3.2.1

Textbook p. 102

Question 6

A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel. This pair is then connected in series with a 4.0 Ω resistor. What is the effective resistance of the circuit?

4.5 Ω
6.0 Ω
9.0 Ω
13 Ω

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Parallel: 1/R = 1/6 + 1/3 = 1/2, so R = 2.0 Ω. In series with 4.0 Ω: 2.0 + 4.0 = 6.0 Ω.

For the teacher: Applying, medium difficulty, standard 3.2.2

Textbook p. 103

Question 7

A wire has resistance R. A second wire is made of the same material, but it is twice as long and has half the diameter. What is the resistance of the second wire?

R / 2
2R
4R
8R

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R = ρl / A. Doubling the length doubles R. Halving the diameter makes the area A one quarter, which multiplies R by 4. So R becomes 2 × 4 = 8R.

For the teacher: Applying, high difficulty, standard 3.2.4

Textbook p. 109

Question 8

Which property is needed for the material of a heating element?

High resistivity and high melting point
Low resistivity and low melting point
Low resistivity and high melting point
High resistivity and low melting point

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A heating element must produce a lot of heat (high resistivity) and must not melt when it is hot (high melting point), for example nichrome.

For the teacher: Understanding, low difficulty, standard 3.2.5

Textbook p. 111

Question 9

A voltmeter is connected across a dry cell. Which statement is correct?

The reading is the same whether the switch is open or closed
The reading with the switch open is the e.m.f. of the cell
The reading with the switch closed is the e.m.f. of the cell
The reading with the switch open is the potential difference across the bulb

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With the switch open no current flows, so there is no voltage drop across the internal resistance and the voltmeter reads the e.m.f. When the switch is closed the reading falls to the terminal potential difference.

For the teacher: Understanding, medium difficulty, standard 3.3.1

Textbook p. 116

Question 10

A cell of e.m.f. 3.0 V and internal resistance 1.2 Ω is connected to a 4.8 Ω resistor. What is the potential difference across the resistor?

0.6 V
1.8 V
2.4 V
3.0 V

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I = ε / (R + r) = 3.0 ÷ (4.8 + 1.2) = 0.50 A. V = IR = 0.50 × 4.8 = 2.4 V.

For the teacher: Applying, medium difficulty, standard 3.3.2

Textbook p. 117

Question 11

An electric heater of power 2 kW is switched on for 3 hours. The cost of electricity is 30 sen per unit. What is the cost?

RM1.80
RM0.60
RM6.00
RM18.00

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Energy = 2 kW × 3 h = 6 kWh = 6 units. Cost = 6 × RM0.30 = RM1.80.

For the teacher: Applying, low difficulty, standard 3.4.4

Textbook p. 127

Section B: structured questions

34 marks

Answer all questions. Write your answers in the spaces provided.

Question 12

A bulb labelled 12 V, 0.50 A is switched on for 10 minutes.

[1 mark] (a)Calculate the charge that flows through the bulb.

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Q = It = 0.50 × 600 = 300 C1

Tick each point you got.

[1 mark] (b)Calculate the electrical energy transferred in the bulb.

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E = VQ = 12 × 300 = 3600 J1

Tick each point you got.

[1 mark] (c)State what is meant by potential difference.

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The work done (energy transferred) in moving one coulomb of charge from one point to another1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 3.1.5

Textbook p. 99

Question 13

Diagram 1 shows three bulbs X, Y and Z connected to a 12 V supply.

12 V supply, bulb X of 3 ohm in series with bulb Y of 3 ohm and bulb Z of 6 ohm connected in parallel, with an ammeter in the circuit — Diagram 1

[2 marks] (a)Calculate the effective resistance of the circuit.

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Y and Z in parallel: 1/R = 1/3 + 1/6, R = 2 Ω1
Total resistance = 3 + 2 = 5 Ω1

Tick each point you got.

[1 mark] (b)What is the reading of the ammeter?

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I = V / R = 12 ÷ 5 = 2.4 A1

Tick each point you got.

[1 mark] (c)Calculate the potential difference across bulb X.

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V = IR = 2.4 × 3 = 7.2 V1

Tick each point you got.

[1 mark] (d)Which is brighter, bulb Y or bulb Z? Give a reason.

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Y is brighter: same potential difference (4.8 V) across both but Y has the smaller resistance, so a larger current and a larger power (P = V²/R)1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 3.2.2

Textbook p. 104

Question 14

A nichrome wire has resistivity 1.1 × 10⁻⁶ Ω m, length 2.0 m and diameter 0.40 mm.

[2 marks] (a)Calculate the cross-sectional area of the wire.

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r = 0.20 mm = 2.0 × 10⁻⁴ m1
A = πr² = π(2.0 × 10⁻⁴)² = 1.26 × 10⁻⁷ m²1

Tick each point you got.

[2 marks] (b)Calculate the resistance of the wire.

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R = ρl / A = (1.1 × 10⁻⁶ × 2.0) ÷ (1.26 × 10⁻⁷)1
= 17.5 Ω1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 3.2.6

Textbook p. 113

Question 15

Diagram 2 shows graphs of resistance R against length l for two wires P and Q. Both wires have the same diameter and the same temperature.

Graph of resistance against length: straight line P reaching 3.2 ohm at 0.8 m and straight line Q reaching 1.6 ohm at 0.8 m — Diagram 2

[1 mark] (a)Calculate the gradient of the graph for wire P.

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Gradient = 3.2 ÷ 0.8 = 4.0 Ω m⁻¹1

Tick each point you got.

[2 marks] (b)Which wire has the greater resistivity? Give a reason.

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P1
Gradient = ρ / A and A is the same, so the wire with the larger gradient has the larger resistivity1

Tick each point you got.

[1 mark] (c)State the relationship between the gradient of the graph and the resistivity of the wire.

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The gradient is directly proportional to the resistivity (for the same cross-sectional area)1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 3.2.3

Textbook p. 108

Question 16

A student uses a rheostat to obtain a graph of potential difference V against current I for a dry cell, as shown in Diagram 3.

Graph of V against I for a dry cell: points lie on a straight line from 1.5 V at 0 A down to 1.0 V at 1.0 A — Diagram 3

[2 marks] (a)State the e.m.f. of the dry cell. Explain how you obtain it from the graph.

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1.5 V1
Intercept on the V axis (when I = 0)1

Tick each point you got.

[2 marks] (b)Calculate the internal resistance of the dry cell.

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Gradient = (1.0 − 1.5) ÷ (1.0 − 0) = −0.50 V A⁻¹1
r = −gradient = 0.50 Ω1

Tick each point you got.

[1 mark] (c)Explain why V decreases when I increases.

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A larger current gives a larger voltage drop (Ir) across the internal resistance, so V = ε − Ir decreases1

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 3.3.3

Textbook p. 119

Question 17

Diagram 4 shows two dry cells connected in series and in parallel to a 2.0 Ω resistor. Each cell has e.m.f. 1.5 V and internal resistance 0.50 Ω.

Two cells in series connected to a 2 ohm resistor — Series

Two cells in parallel connected to a 2 ohm resistor — Series

[2 marks] (a)Calculate the current in the series arrangement.

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ε = 3.0 V, r = 0.5 + 0.5 = 1.0 Ω1
I = ε / (R + r) = 3.0 ÷ 3.0 = 1.0 A1

Tick each point you got.

[2 marks] (b)Calculate the current in the parallel arrangement.

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ε = 1.5 V, 1/r = 1/0.5 + 1/0.5, r = 0.25 Ω1
I = 1.5 ÷ (2.0 + 0.25) = 0.67 A1

Tick each point you got.

[1 mark] (c)Which arrangement is better for a torch bulb that needs a large current? Give a reason.

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Series, because the effective e.m.f. is higher (3.0 V), giving the larger current1

Tick each point you got.

For the teacher: Evaluating, high difficulty, standard 3.3.4

Textbook p. 121

Question 18

An electric kettle is labelled 240 V, 2000 W.

[1 mark] (a)Calculate the current when the kettle works normally.

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I = P / V = 2000 ÷ 240 = 8.3 A1

Tick each point you got.

[2 marks] (b)Calculate the resistance of the heating element.

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R = V² / P = 240² ÷ 2000 (or V / I)1
= 28.8 Ω1

Tick each point you got.

[2 marks] (c)Calculate the cost of using the kettle for 15 minutes if 1 unit costs 21.8 sen.

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Energy = 2.0 kW × 0.25 h = 0.50 kWh1
Cost = 0.50 × 21.8 = 10.9 sen1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 3.4.3

Textbook p. 126

Question 19

A company wants to design a portable electric kettle that boils water quickly and safely from a 240 V supply.

[3 marks] (a)Suggest three features of the heating element and explain each.

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Any 3 points

Made of a material with high resistivity and high melting point (e.g. nichrome), so it gets very hot without melting1
Shorter or thicker wire (smaller resistance), so a larger current flows and the power P = V² / R is larger, boiling water faster1
Wire coiled into a spiral, so a long wire fits in a small space / the heat is concentrated1
A fuse in the circuit, so the current is cut off if it is too large (safety)1

Tick each point you got.

For the teacher: Creating, high difficulty, standard 3.2.5

Textbook p. 111

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