Chapter 18 Practical Electricity

Chapter 18 Practical Electricity

Learning Outcomes

After completing this chapter, students should be able to:

1. describe the use of the heating effect of electricity in appliances such as electric kettles, ovens and heaters

2. recall and apply the relationships P = VI and E = VIt to new situations or to solve related problems 3. calculate the cost of using electrical appliances where the energy unit is the kWh 4. state the hazards of using electricity in the following situations

(i) damaged insulation (ii) overheating of cables (iii) damp conditions

5. explain the use of fuses and circuit breakers in electrical circuits and of fuse ratings 6. explain the need for earthing metal cases and for double insulation 7. state the meaning of the terms live, neutral and earth 8. describe the wiring in a mains plug 9. explain why switches, fuses, and circuit breakers are wired into the live conductor

18.1 Electric Power and Energy 1. Electricity is useful because it changes easily into other form of energy.

Page 287

2. Emphasise that electrical energy can be obtained from the definition of potential difference. Energy transferred .

Potential difference = Charge ' so Energy = Charge x Potential dIfference or E = Q x V

3. An appliance marked '240 V, 2 kW' means the operational voltage of the appliance is 240 V. When 240 V supply is connected to the appliance, it will consumed 2 kJ of energy in 1 s.

Answers to Think Time question page 290

V2 We can useE = VIt or E = PRtor E = - t

R

Answers to Think Time question page 291

The kilowatt-hour is a convenient unit because the power of many electrical appliances is in the range of kilowatt and they are being used for hours. If joule is used, the value will be a very big number.

Answers to Section Review questions page 292

1. E = VIt = 2.5 V x 0.3 A x (2 x 60) s = 90 J

2. Appliance Power P = VI Voltage V = P -:- I Current I = P -:- V

Lamp 60W 240 V 0.25 A Television 120W 240 V 0.5 A Hair dryer 500W 250 V 2A Air-conditioner 2000W 250 V 8A

3. Energy = 0.75 kW x 4 h = 3.0 kWh Cost of electricity = 3.0 kWh x $0.21 = $0.63

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Chapter 18 Practical Electricity

Physics and the Environment: Producing Electrical Energy page 293 Answers to Q

1. Pollution of the environment through the discharge of radioactive materials, hot polluted water or chemicals and gaseous discharge. Large number of people may be relocated a result of the construction of darns.

2. It is most unlikely as the basic building blocks of the universe consist of positive and negative charges. As they interact with each other, electrical energy is stored up or dissipated in destructive or constructive manner. The easy availability and the pervasiveness of electrical energy in modem consumer society make it difficult to replace electrical energy with other sources of energy.

3. By protecting the environment and ensuring that it is not polluted when we generate electrical energy.

18.2 Dangers of Electricity Page 294 1. What causes electric shock, current or voltage? [Answer: Electric shock occurs when current flows

through the body, which will happen when there is a potential difference across the body. So the initial cause of electric shock is the voltage, but the current does the damage.]

2. Emphasise the possible hazards of using electricity in the following situations • damaged insulation • overheating of cables • damp conditions

Answers to Think Time question page 294

When the wires are coiled up, the heat produced cannot dissipate so easily. This may cause overheating of the wires. The will damage the wire insulations and can cause a fIre.

Answers to Section Review questions page 296

1. Damaged insulation, overheating of cables and damp conditions

2. The insulation is damaged, resulting in a short circuit; the wire used is too thin; the power outlet is overloaded.

3. A wet body will give a lower electrical resistance.

18.3 Safe Use of Electricity in the House Page 296 1. Ask students the features incorporated in the household electrical wiring for safe use of electricity in

the house. The features include: fuse, ELCB, MCB, switch, earth wire and double insulation.

2. Emphasise to students the importance of having earth wire for appliance with metal case. Describe the wiring of the mains plug.

Answers to Think Time question page 298

(a) If any of the bulbs is faulty, the other bulbs can still light up. (b) They are inserted along the live wires of the circuits so that when the switch is off or the fuse has

'blown', the live wire is disconnected from the high potential of 240 V. (c) Total power = (60 + 15 + 100) W = 175 W

l=!... = 175W = 0.73A V 240V

(d) The 5 A fuse is suitable.

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Chapter 18 Practical Electricity

Answers to Section Review questions page 302

1. The most suitable fuse is the 3 A fuse .

2. They are inserted along the live wires of the circuits so that when the switch is off or the fuse has 'blown', the live wire is disconnected from the high potential of 240 V.

3. The three wires in a power circuit are live, neutral and earth wire. The heating element should be connected to the live and neutral wire. The metal case of the kettle should be connected to the earth wlIe.

4. live wire: brown neutral wire: blue earth wire: yellow and green stripes

IT Link page 302

2. Bedroom Living Room Kitchen

Appliance PowerfW Appliance PowerfW Appliance PowerfW Incandescent light 60 Paddle Fan 48 Fluorescent Lamp 30 Printer 300 Large Lamp 150 Microwave oven 1200 Computer 108 Stereo System 150 Dishwasher 1290 Small Fan 24 Television 300 Electric Stove 4950 Electric Blanket 132 Reading light 60 Refrigerator 330 Halogen lamp 300 Air conditioner 900 Kettle 1500

3. Electric stove uses the highest power and small fan uses the least power.

Answers to Misconception Analysis page 303

1. 2. 3. 4. 5. 6. 7. 8. 9.

10.

True False. False. True True True False. True False.

False.

Electric fans and radios use the magnetic effect of current. Kilowatt-hour is the unit for energy.

The fuse should take a current slightly larger than the maximum current.

The rating of the fuse should be slightly larger than the maximum current allowed in the appliances . Fuses should be inserted along the live wires of the circuits.

Answers to Multiple Choice Questions page 304

1. D E = VIt = 3 V x 0.4 A x (2 x 60) s = 144 J

2. D

3. C 4. B 5. D 6. C 7. C

8. B 9. C

10. A

R = ~ (The voltage for fan should be 240 V, typo error in the text printed as 'air'.) p

E = 1.5 kW x 8 h = 0.5 kW x t; t = 24 h Cost = 1.5 kW x 8 h x $0.21 = $2.52

The maximum current in the heater is less than lOA. So highest power rating < 250 V x 10 A

The fuse used should be slightly above the maximum current allowed. In this case maximum P 1000W

current I = V = 250 V = 4 A, so the fuse should be 5 A.

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Chapter 18 Practical Electricity

Answers to Structured Questions page 304, 305

1. (a) Energy = [(0.12 kW x 2 h) + (0.02 kW x 4 h)] x 30 = 9.6 kWh (b) Cost of electricity = $0.20 x 9.6 = $1.92

2. (a) P = VI = 6 V x 0.1 A = 0.6 W (b) E = Pt = 0.6 W x (30 x 60) s = 1080 J

3. (a) 1= P = 120 W = 0.5A V 240 V

V2 (b) P= -

R

120 = 2402

R R =480n

(c) The filament is made of tungsten because of its high melting point and high resistivity (d) Cost of electricity = $0.18 x (0.12 kW x 5 h) = $0.l1 (e) If the lamp is connected to a 110 V supply, the current flowing through the lamp will be lower

and the lamp will not light up or light up with lower brightness.

4. (a) and (c)

} fuse L

'---t[3---t--o 240 V mains

N

(b) Total power = 3 x 60 W = 180 W P 180W

I=V =240V=0.75A

1 A fuse should be used because the rating of the fuse should be slightly higher than the maximum current flowing in the circuit.

5. (a) (i) The fuses protects the wiring and the appliances. It also protects us. (ii) They are inserted along the live wires of the circuits so that when the switch is off or the

fuse has 'blown', the live wire is disconnected from the high potential of 240 V.

(b) The earth wire serves as a safety device. It is joined to the metal case of an appliance. If the live wire of the appliance touches the case, a large current will flow to the Earth and blows the fuse in the live wire. In this way, the case is prevented from being 'live' and users are prevented from getting an electric shock.

P 1000 W (c) Current through the kettle, I = V = 240 V = 4.17 A (5 A fuse)

P 120W Current through the two lamps, I = V = 240 V = 0.5 A (3 A fuse)

Total current = (4.17 + 0.5) A = 4.67 A (15 A fuse)

(d) Fuse X will blow because current is greater than 5 A. Fuses Y and Z will not be affected because current through the lamps remains to be 0.5 A and the current through the 15 A fuse is 10.5 A

(e) Each lamp takes a current of 0.25 A. 3A

For 3 A, number of lamps = 0.25 = 12

Since 3A is slightly higher than the maximum current, therefore number of lamps allowed is 11.

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Chapter 18 Practical Electricity

Answers to Critical Thinking Questions page 305

1. For 240 V supply Advantage Disadvantage

• Lower current in the wire and hence less • Higher voltage which is more dangerous to power loss consumers (electric shock)

• Thinner wires may be used for power transmission than 120 V supply which is more economical

For 120 V supply Advantage Disadvantage

• Lower voltage which is safer for consumers • Higher current in the wire and hence more (electric shock) power loss

• Need to use thicker wire than 240 V supply

2. When the wires are in series, current through the wires are the same

EI [ 2RI

E2 = [2R2

R = _ 1= R2

= ~ = Al

d 2 = _ 2_

d 2 I

I p-~

I p -~

1d{ r Jr(i J

For wire with twice the diameter d 2 = 2d 1

So ..s. = 4 E2

The heat produced by the thinner wire is 4 times the heat produced by the thicker wire.

3. When the wires are in parallel, potential difference across the wires are the same.

V 2 I p -

..s.=!!L = ~ = ~= ~ E2 V 2 R I A

I p_ 2

R2 Al

For wire with twice the diameter d 2 = 2d 1

So ..s. = ~ E2 4

The heat produced by the thinner wire is ~ times the heai produced by the thicker wire. 4

4. For an electrical device, its power and working voltage is fixed. So it is either rated with 'voltage, power ' or 'voltage, current' . For a battery, the current supplied depends on the load. So it is only rated with voltage since current can be any value.

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Chapter 18 Practical Electricity

Extension page 305 Assuming the 1.5 V AAA cell with rating 1200mAh cost $1.30 per piece. Total energy = QV

= 1200 rnA x 1 h x 1.5 V = 1800 mW xl h = 1.8 kWh

So 1.8 kWh cost $1.30. Therefore 1.0 kWh cost $0.72 which is much higher than the costs from the mains which is about $0.20 per kWh.

Size Cost per battery Rating Q / mAh Voltage/V Total energy / kWh Cost per kWh AA $1.30 2700 (alkaline) 1.5

AAA $1.30 1200 (alkaline) 1.5

D $2.50 19500 (alkaline) 1.5

Note: 1. Check price Of battery at the local supermarket or grocery store 2. Rating = typical capacity of the battery

Refer website: data.energizer.com http://en.wikipedia.orglwikilLiscoLbattery_sizes

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4.05 $0.32 1.8 $0.72

29.25 $0.09

Science in Focus: Physics '0 ' Level - -