Full Solution of Ch 25

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NSS Physics in Life Full Solution of Textbooks(Electricity and Magnetism)

Chapter 25 Electrical Power and Domestic Electricity

Checkpoint (p.197)

1. The heating effect of current increases with the applied voltage across the wire.

Microscopically, when a voltage is applied across the ends of a metal wire, an

electric field is set up between the ends. The free electrons are accelerated by the

field and collide frequently with the metal ions in the wire. Thus, some energy is

transformed into the internal energy of the wire.

When the applied voltage increases, the electric field is stronger. The electrons

collide harder as well as more frequently with the ions. Thus, more energy is

transformed into the internal energy.

2. The heating element is usually wound into a coil to facilitate heat transfer from

the heating element to the targeted area. The coiling of the heating element

results in a long enough wire that gives a reasonable resistance. This is to protect

the source against short circuit.

3. No. The resistance of a copper wire is much smaller than that of a Nichrome wire

of the same length. If we replace the wire with a copper wire, the battery will be

short-circuited and become very hot in a short time. It may even explode.

Checkpoint (p.200)

1. (a) The equation defines the average electrical power P as the amount of

electrical energy E transformed in a time interval t. It is applicable (but not

restricted) to all kinds of electrical energy transformation.

(b) The equation defines the electrical power P supplied to or dissipated by an

electrical component. When the voltage across the component is V and the

current flowing through the component is I, the electrical power P = VI.

The equation is applicable to all kinds of electrical components.

(c) The equation defines the electrical power P supplied by a source. When a

source of e.m.f. ε delivers a current I to a component, the power supplied

P = εI. The equation is applicable to sources only.

(d) The power P dissipated by a resistor of resistance R is given by P = I2R

where I is the current flowing through the resistor. The equation is

applicable to all resistive components.

(e) The power P dissipated by a resistor of resistance R is given by

where V is the voltage across the resistor. The equation is applicable to all

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resistive components.

2. (a) Applying , the total current I passing through the circuit is

Applying , the power of the 2 Ω resistor is

.

Similarly, the power of the 4 Ω resistor is and

the power of the 6 Ω resistor is .

Thus, the total power = 2 + 4 + 6 = 12 W.

(b) Applying , we have

the power of the 2 Ω resistor ,

the power of the 4 Ω resistor , and

the power of the 6 Ω resistor .

Thus, the total power = 72 + 36 + 24 = 132 W.

Exercise (p.200)

1. A

Applying , we have

for choice A, the wire resistance ;

for choice B, the wire resistance ;

for choice C, the wire resistance ;

for choice D, since the wire has a radius of (d/2) and a length larger than 2l, the

resistance .

By , when the same voltage V is applied across the wires, the wire with

the smallest resistance has the greatest heating effect. Thus, the answer is choice

2



A.

2. B

Applying , we have

The power supplied by the cell is 1.2 mW.

3. A

Let I be the total current delivered by the battery, r be the resistance of each

resistor and PP, PQ and PR be the power dissipated in resistors P, Q and R

respectively.

Applying P = I2R, we have PP = I2r.

By symmetry, the currents through resistors Q and R are equal to .

Thus, we have and

the total power .

It is given that the total power delivered by the battery is 12 W, we have

Therefore, the powder dissipated in R is .

4. D

Let R be the total resistance of the circuit.

Under a constant supplied voltage ε, the power P consumed in the circuit is given

by . From the equation, a smaller R will result in a larger P. The smallest

total resistance is obtained when the two resistors are connected in parallel to the

battery as shown.

The smallest total resistance .

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Thus, the largest power consumed .

5. B

Since L1 and L2 are identical and are connected in parallel, they have the same

brightness. As the equivalent resistance of L1 and L2 connected in parallel is

smaller than the resistance of L3, the p.d. across L3 is larger than that across L1

and L2. Thus, L3 is the brightest.

6. A

If the filament of L2 is burnt out, the p.d. across L1 and that across L3 are the

same. Thus, they have the same brightness.

Since the equivalent resistance across the two ends of L1 increases, the p.d.

across L1 increases. As the e.m.f. of the battery is constant, L1 becomes brighter

than before while L3 becomes dimmer.

7. Let Rthicker and Pthicker be the resistance and power of the thicker wire while Rthinner

and Pthinner be the resistance and power of the thinner wire respectively.

By , .

By , .

(a) If they are connected in series to the supply, the same current I passes through the two wires. By , we have

and

Therefore, Pthicker : Pthinner = 1 : 4.

(b) If they are connected in parallel to a supply of negligible resistance, the

voltages across the two wires are the same. By , we have

and

.

Therefore, Pthicker : Pthinner = 4 : 1.

8. (a) Applying , the voltage V supplied to the heater is

(b) Applying , the operating resistance R of the heater is

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9. First, use the electronic balance to find the mass m of the water. Second, use the

thermometer to measure the initial temperature of the water. Third, connect the

immersion heater to the power supply and immerse the heater into the water.

Forth, heat the water for a period of time t with the help of the stop-watch. Fifth,

switch off the heater and use the thermometer to measure the final temperature of

the water.

Assume that there is no loss of energy to the surroundings. The energy E

transferred to the water is given by

where c is the specific heat capacity of water and ΔT is the difference of the final

and initial temperatures. Thus, the power output of the heater is given by

10. Let P1 and P2 be the power dissipated in R1 and R2 respectively.

Consider R1 and R2 connected in parallel to the source.

Since , by , we have

Consider R1 and R2 connected in series to the source. The same current passes through the resistors. By , we have

.

Thus, the power dissipated in R1 to that in R2 = 2 : 1.

11. (a) The bulb can provide a desired output power by closing different switches.

When only switch A is closed, only 1 resistor is connected to the supply to

provide a power of 50 W.

When only switch B is closed, another resistor of smaller resistance is

connected to the supply to provide a power of 100 W.

When both switches are closed, the two resistors are connected in parallel to

the supply to provide a desired power of 150 W.

(b) When switch A is closed, the resistance of the 50 W filament is

When switch B is closed, the resistance of the 100 W filament is

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(c) Yes. As the filaments are connected in parallel to the supply, the meltdown

of one filament does not affect another. The bulb can still glow.

12. (a)

Since the operating voltage of the fan is 24 V, the fan should be connected

in parallel to the 24 V d.c. power supply. The switch should be installed in

the main branch to control both the heating element and the fan.

(b) (i) Applying , the operating resistance R of the heating element is

given by

(ii) Applying P = VI, the current through the heating element is given by

.

Similarly, the current through the fan is given by

.

Thus, the total current drawn from the supply

= 6.25 + 0.8333

≈ 7.08 A

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(c) The two possible circuits are

Checkpoint (p.207)

1. (a) correct

(b) incorrect

(c) incorrect

2. No. The bulbs may have different power ratings, so they may have different

brightness.

3.

electrical

appliance

power rating

/ W

operating

voltage / V

operating

current / A

operating

resistance / Ω

electric kettle 2000 110 18.2 6.05

toaster 1100 220 5 44

ray box lamp 24 12 2 6

waffle iron 1180 220 5.37 41

torch 1.2 3 0.4 7.5

Checkpoint (p.210)

1. A

Statement (3) is incorrect because the cost also depends on the operating time of

the appliance.

2. The time taken for the fan to use up 1 kW h of electrical energy is

3. The power of the air conditioner is .

The cost of running it per month

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Exercise (p.211)

1. A

The resistance of the heater .

When the heater is connected to a 110 V main supply, the power consumed by

the heater

=

2. B

Statement (1) is correct. Applying , the operating resistance of the cooker

is

Statement (2) is incorrect. Applying P = VI, the operating current of the cooker is

Statement (3) is correct. The electrical energy consumed =

3. A

Statement (1) is correct. From the figures, the power of the hair dryer is twice as

large as that of the electric kettle. By P = VI, the current drawn by the hair dryer

is also twice as large as that drawn by the electric kettle.

Statement (2) is incorrect. The appliances will operate at a lower power output.

Statement (3) is incorrect. By , the resistance of the hair dryer is only half

of that of the electric kettle.

4. B

According to , the power output is the smallest if the equivalent resistance

of the heating elements is the largest. Thus, the heating elements should be

connected in series to the a.c. source by connecting terminals 2 and 3.

5. B

From the given energy efficiency label, 840 kW h of electrical energy is used in

1200 hours. Thus, the power P of the air conditioner is

.

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Thus, the energy used in 8 hours = Pt = 0.7 × 8 = 5.6 kW h.

6. (a) Applying P = VI, the mains voltage in Australia

(b) Applying P = I2R, the operating resistance R is

7. (a) Applying , the operating resistance R of the electric boiler is

.

(b) Applying P = VI, the current drawn is

.

(c) The energy consumed = Pt = 1000 × 5 = 5000 W h = 5 kW h

(d) The cost of running it for 5 hours = 5 × 0.90 = $4.50

8.

electrical

appliance

power / W operating time

per day / h

energy

consumed / kW h

cost /

$

lamps 40 5 6 5.4

microwave

oven

800 0.25 6 5.4

air

conditioner

1000 6 180 162

LCD TV 150 5 22.5 20.25

refrigerator 150 24 108 97.2

iron 750 0.25 5.625 5.0625

kettle 2000 0.5 30 27

Checkpoint (p.219)

1. (a)

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(b) A fault in one lamp will not affect the operation of the others.

2. (a) Kilowatt-hour meter. It measures the electrical energy supplied to a

household in kilowatt-hours.

(b) The main fuse should be connected to the live wire.

(c)

3. (a) A: neutral wire; B: earth wire; C: fuse; D: live wire

(b)

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Exercise (p.220)

1. A

2. A

3. B

4. B

5. (a)

(b) Originally, the lamps work at their normal rating of ‘220 V, 60 W’. We have

where V1 = 220 V and P1 = 60 W.

If the circuit is connected to a 110 V mains supply (i.e. V1/2), the power

consumed by each lamp is

The power consumed and hence the brightness of the lamps will decrease

by 4 times.

6. (a) P

(b) For socket Q, there is no earth wire connected.

For socket R, the live and neutral pin-holes are wrongly connected.

(c) Yes, sockets Q and R can still be used to power an appliance.

For socket Q, the live and neutral pin-holes are connected correctly. If there

is no current leakage problem, the appliance will still work properly.

For socket R, although the live and neutral pin-holes are connected wrongly,

the p.d. across the pin-holes is the same as that across a normal socket. A

normal a.c. current will flow through an appliance connected across the pin-

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holes. However, there is risk that the appliance is at high potential even

when it is switched off.

7. (a)

(b) A fault in one lamp will not affect the operation of the others.

(c) The chance of overloading the circuit is reduced as the current flows from

the consumer unit to the sockets via two paths. Thus, thinner and cheaper

wires can be used.

Checkpoint (p.228)

1. (a) Incorrect. Fuses are used to protect circuits against overloading or short

circuit.

(b) Incorrect. The rating of a fuse should be slightly higher than the operating

current of the appliance.

(c) correct

(d) correct

(e) Incorrect. Electrical appliances with double insulation do not need an earth

wire.

2. (a) The fuse and switch should be connected to the live wire instead of the

neutral wire.

(b) (i) incorrect

(ii) correct

(iii) incorrect

(iv) Incorrect. Since the neutral and earth wires are both at zero potential,

there is no current flowing between the two wires.

(v) incorrect

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3. (a) 5 A, 5 A, 10 A

For the toaster, the operating current .

Thus, a fuse of 5 A rating should be used.

For the microwave oven, the operating current

. Thus, a fuse of 5 A rating should be used.

For the electric kettle, the operating current

. Thus, a fuse of 10 A rating should be used.

(b) No. The total operating current = 4.545 + 3.636 + 8.1818 ≈ 16.4 A > 13 A.

It is unsafe to run all the appliances the same time.

Exercise (p.230)

1. B

2. B

3. C

4. C

5. (a) When the switches are open, the bulbs are not at high potentials.

(b) 0.795 A

Applying P = VI, the total current drawn from the mains is

(c) No. The fuse rating of the fuse is much higher than the total current drawn

from the mains. It is unsafe to run the lighting circuit. Even the current

drawn from the mains is much higher than the sum of the operating currents

of the bulbs, the circuit is still not broken.

6. (a) The operating current .

(b) The fuse of a rating of 5 A is the most suitable.

(c) Yes. Even there is no earth wire, an alternating current still flows to and fro

in the live and neutral wires.

(d) The appliances have double-insulation.

(e)

7. (a) Water from the wet hands may form unintended paths that short the live

parts of the appliances. Thus, we may get electric shocks if we touch

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electrical appliances when our hands are wet.

(b) The wires or conducting parts of the appliances may get loose and touch the

casing of the appliances accidentally. Electric shock will be received when a

person touch the appliances if its earth wire and insulation fails to work.

(c) The current drawn from the mains may be too high that overloading may be

resulted. This may cause a fire if the fuses or circuit breakers installed fail

to work.

Chapter Exercise

1. C

2. C

There is no difference in the heating effect whether the conductor is wound into a

coil or not. The difference is the transfer of heat from the conductor to the

targeted area.

3. C

The operating current I =

Thus, fuse of 6 A rating should be used.

4. B

The operating current I

Applying P = VI, the estimated power of the MP3 player

5. B

6. B

7. C

The circuit can be redrawn as

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When more bulbs are connected in parallel to the battery, the size of the total

current drawn from the battery increases. Thus, the potential drop across the

internal resistance of the battery increases while the p.d. across each bulb

decreases. Therefore, statement (1) is incorrect while statement (3) is correct.

According to P = VI, the total output power of the battery increases too.

Therefore, statement (2) is correct.

8. C

The lamps are connected in series to the 220 V mains.

Statement (1) is correct because each lamp shares a p.d. of

Statement (2) is incorrect as the circuit will be broken.

By , the resistance of the 60 W lamp is larger than that of the 100 W lamp.

When the lamps are connected in series to the 220 V mains, the same current

passes through them. By P = I2R, the power of the 60 W lamp is higher than that

of the 100 W lamp. Therefore, statement (3) is correct.

9. D

Let R be the resistance of the bulb and ε be the e.m.f. of the battery.

For circuit A, the total resistance of the circuit .

Thus, the total power of the circuit .

For circuit B, the total resistance of the circuit .

Thus, the total power of the circuit

.

10. (HKCEE 2000 P2 Q35)

11. (HKCEE 2003 P2 Q34)

12. (HKCEE 2006 P2 Q39)

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13. (HKALE 1999 P2 Q20)

14. (HKALE 2005 P2A Q17)

15. D

16. The operating resistance of the heater . (1M)

If the heater is connected to the mains in Hong Kong, by , the operating

current . (1M)

Therefore, the fuse will blow. (1A)

17. (a) The electrical energy consumed by the appliances on that day in kW h

(1M)

The total cost = 2.16 0.9 = $1.944 (1M + 1A)

(b) (i) A fuse protects a circuit against overloading or short circuit (1A). It

contains a thin metal strip which will melt and break the circuit if the

current through the strip is larger than the rated value (1A).

(ii) For the water heater, the operating current I is given by

Thus, a fuse of rating 13 A should be used. (1A)

For the television, the operating current I is given by

Thus, a fuse of rating 1 A should be used. (1A)

For the table lamp, the operating current I is given by

Thus, fuse of rating 1 A should be used. (1A)

18. (a) Applying , the current IABD through the 3 resistor is

. (1M)

Hence, the power dissipated in the 2 resistor in the branch ABD

(1M+1A)

Applying P = VI, the voltage VAD across A and D is

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.

Applying , the current through the branch ACD

(1M)

Hence, the power dissipated in the 2 resistor in the branch ACD

(1M+1A)

Similarly, the power dissipated in the 5 resistor

(1A)

(b) The total power output = 10 + 6.666 + 8.503 + 3.401 = 28.57 ≈ 28.6 W.

(1M)

The total energy supplied by the battery in 1 minute = 28.6 60 ≈ 1710 J.

(1A)

19. (a) live wire and neutral wire (1A+1A)

(b) earth wire (1A)

(c) An earth wire protects people who touch the metal casing of a faulty

electrical appliance against electric shock (1A). It ensures the current goes

directly into the earth by providing a low resistance path between the casing

and the earth (1A).

(d) The hair dryer has double insulation. (1A)

(e) live wire: brown; neutral wire: blue; earth wire: green and yellow (3A)

20. (a) X: lighting circuit Y: ring circuit (2A)

(b) The water heater and the air conditioner are connected in parallel. (1A)

The appliances can work under the same voltage. (1A)

The faulty appliances will not affect other appliances. (1A)

(c) A: earth wire; B: live wire; C: neutral wire (3A)

(d) The chance of overloading is reduced (1A) as the current flows from the

consumer unit to the sockets via two paths (1A). Thus, thinner and cheaper

wires can be used (1A).

(e) It ensures that the bulbs are not at high electric potential when the switch is

open. (1A)

(f) Independent fuses can be assigned to those appliances (1A) as some of the

appliances draw exceptionally large currents (1A).

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(g) circuit breakers (1A)

21. (a) Whenever the internal temperature is lower than 80°C, S2 is closed. After S1

is closed, the branch of S2 and R1 is shorted as the resistance of bulb A is

negligible (1A). Electrical energy is transformed into heat in R2 and R2 will

heat up the water and the rice (1A). When the internal temperature reaches

100°C, the water is boiled and the rice is cooked (1A). At this temperature,

S1 opens automatically to prevent the rice from overcooked (1A). Cooking is

then finished. This is the operation of the cooking mode.

(b) When the internal temperature reaches 80°C in the cooking mode, S2 is open

(1A). When the internal temperature reaches 100°C, S1 is open and the

internal temperature starts to decrease (1A). Whenever the internal

temperature is lower than 80°C, S2 is closed again and electrical energy is

transformed into heat in both R1 and R2 (1A). Since R1 and R2 are now

connected in series (1A), the total resistance of the circuit is increased. By

, the power of the cooker is lowered to a level that can keep the rice

warm. This is the operation of the warming mode (1A).

(c) In the cooking mode, the total resistance of the circuit = R2 = 50 Ω.

Applying , the power of the cooker is . (1M + 1A)

In the warming mode, the total resistance of the circuit = R1 + R2 = 550 Ω.

Applying , the power of the cooker is . (1M + 1A)

22. (a) (i) Applying , the current through bulb A is

.

Similarly, the current through bulb B is

and

the current through bulb C is

. (1M)

Since the 3 bulbs are working at their rated values, the voltage drop

across bulbs A, B and C are 2 V, 1.5 V and 1.5 V respectively. (1M)

The voltage drop VPY across P and Y = 2 − 1.5 = 0.5 V.

The voltage drop across the internal resistance of the battery

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(1M)

The voltage drop VXP across X and P = 6 − 0.5 − 2 − 1.5 = 2 V. (1M)

Thus, the maximum resistance Rmax of the rheostat

(1M+1A)

(ii) RXP = 20 Ω and the maximum resistance Rmax of the rheostat is 30 Ω.

Since the resistance of the resistance wire in the rheostat is directly

proportional to its length, the length x of the wire segment between X

and P is given by

(1M+1A)

(b) Bulb C will become dimmer (1A). When the sliding contact P is moved to Y,

the equivalent resistance across bulb A decreases while the total resistance

of the circuit increases (1A). Thus, the voltage across bulb C will decrease

(1A) and bulb C will become dimmer.

23. (HKCEE P1 Q8)

24. (HKCEE P1 Q11)

25. (HKCEE P1 Q6)

26. (OCR GCE A-level Jan 2006 Physics A 2825/03 Q7)

27. (a) (i) 0 V (1A)

(ii) no (1A)

Under normal operation, current flows to and fro between the live and

neutral wires (1A). When the potential of the live wire is normal, the

resistance of an MOV is so high that the current is not able to flow

through it (1A).

(b) (i) Both of them protect the appliances by preventing large currents from

flowing through them (1A). However, the appliances will stop working

if the fuses installed are blown while the appliances keep working if a

surge protector is used. (1A)

(ii) A fuse blows when the current passing through it is too large.

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However, it may not be sensitive enough if the duration and magnitude

of the change in current is too short and small (1A). Computers are very

sensitive to such changes in voltage and current. Thus, a fuse is

insufficient to protect a computer upon a power surge (1A).

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Full Solution of Ch 25

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