Practical Electricity

Practical Electricity (O Level)

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Question 1
What is a possible material for the heating element in an electric iron?
A
Copper
B
Steel
C
Nichrome
D
Stainless steel
Question 1 Explanation: 
Nichrome wire is used as the heating element because its resistance is high. It does not oxidise easily and has high melting point.
Question 2
What is the current flowing through an appliance marked “240 V, 40 W”?
A
0.17 A
B
3.00 A
C
6.67 A
D
6.00 A
Question 2 Explanation: 
When an appliance is marked “240 V, 40 W”, it means that a 240 V across the appliance will produce a power of 40 W. P = VI => 40 = 240I => I = (40/240) A = 0.17 A
Question 3
What is the power produced by an appliance marked “240 V, 2 A”?
A
8.3 mW
B
60 W
C
120 W
D
480 W
Question 3 Explanation: 
When an appliance is marked “240 V, 2 A”, it means that a 240 V across the appliance will produce a current of 2 A. P = VI = 240 x 2 = 480 W
Question 4
What is the effective resistance of an appliance marked “240 V, 2 A”?
A
0.017 Ω
B
60 Ω
C
120 Ω
D
960 Ω
Question 4 Explanation: 
When an appliance is marked “240 V, 4 A”, it means that a 240 V across the appliance will produce a current of 4 A. V = IR => R = V ÷ I = 240 ÷ 4 = 60 Ω
Question 5
What is the effective resistance of an appliance marked “240 V, 60 W”?
A
0.25 Ω
B
4 Ω
C
960 Ω
D
1440 Ω
Question 5 Explanation: 
When an appliance is marked “240 V, 60 W”, it means that a 240 V across the appliance will produce a power of 60 W. P = VI => 60 = 240I => I = (60/240) A = 0.25 A V = IR => R = V ÷ I = 240 ÷ 0.25 = 960 Ω
Question 6
What is the power dissipated by the lamp in the diagram below? 6
A
2 W
B
3 W
C
6 W
D
18 W
Question 6 Explanation: 
Current flowing through the lamp = 3 A Potential difference across the lamp = 6 V Power dissipated by the lamp = VI = 6 x 3 = 18 W
Question 7
What is the power dissipated by the lamp of resistance 6 Ω in the diagram below? 7
A
2 W
B
12 W
C
24 W
D
72 W
Question 7 Explanation: 
Current flowing through the lamp = V ÷ R = 12 V ÷ 6 Ω = 2 A Power dissipated by the lamp = VI = 12 x 2 = 24 W
Question 8
What is the power dissipated by each lamp of resistance 6 Ω in the diagram below? 8
A
1 W
B
2 W
C
6 W
D
12 W
Question 8 Explanation: 
Effective resistance of the circuit = 6 Ω + 6 Ω = 12 Ω Current flowing through each lamp = V ÷ R = 12 V ÷ 12 Ω = 1 A Power dissipated by the lamp = I2R = (1)2 x 6 = 6 W
Question 9
Which of the following circuit dissipates the greatest power? 9
A
A
B
B
C
C
D
D
Question 9 Explanation: 
The circuit with the least effective resistance will dissipate the greatest power as the e.m.f. of the circuits are the same.(P = V2 ÷ R) Let effective resistance of choice A = R Effective resistance of choice B = 2R Effective resistance of choice C = (½)R Effective resistance of choice D = (2/3)R
Question 10
The lamp is switched on for 5 mins. How much energy has been dissipated from the lamp? 10
A
120 J
B
480 J
C
600 J
D
36000 J
Question 10 Explanation: 
Power dissipated by the lamp = VI = 240 V x 0.5 A = 120 W Energy dissipated in 5 minutes = Pt = 120 W x 5 x 60 s = 36000 J
Question 11
In the circuit shown below, how much energy is used up if both lamps are operated normally for 1 h? 11
A
72 J
B
4320 J
C
57600 J
D
259200 J
Question 11 Explanation: 
Effective resistance of circuit = 2 Ω Current flowing out of the cell = V ÷ R = 12 V ÷ 2 Ω = 6 A Power dissipated from the cell = VI = 12 V x 7 A = 72 W Energy dissipated in 1 hour = Pt = 72 W x 3600 s = 259200 J
Question 12
How much energy does a spot light rated at 2 kW consume in 6 h?
A
3 kWh
B
4 kWh
C
8 kWh
D
12 kWh
Question 12 Explanation: 
Energy dissipated in 6 hours = Pt = 2 kW x 6 h = 12 kWh
Question 13
How much energy does a lamp rated at 40 W consume in 30 min?
A
0.02 kWh
B
1.2 kWh
C
20 kWh
D
1200 kWh
Question 13 Explanation: 
Power of lamp = 40 W = 0.04 kW Duration of operation = 30 min = 0.5 h Energy used = Pt = 0.04 kW x 0.5 h = 0.02 kWh
Question 14
If 1 kWh of electricity cost $0.20, how much does it cost to switch on a spot light rated at 2 kW for 6 h?
A
$ 0.60
B
$ 2.40
C
$ 15.00
D
$ 60.00
Question 14 Explanation: 
Energy used = Pt = 2 kW x 6 h = 12 kWh Cost of 1 kWh = $ 0.20 Cost of 12 kWh = 12 x $ 0.20 = $ 2.40
Question 15
If 1 unit of electricity cost $0.20, how much does it cost to switch on a heater marked “120 V, 3 A” for 90 min?
A
$ 0.11
B
$ 2.70
C
$ 64.80
D
$ 108.00
Question 15 Explanation: 
Power of heater = VI = 120 V x 3 A = 360 W = 0.36 kW Duration of operation = 90 min = 1.5 h Energy used = Pt = 0.36 kW x 1.5 h = 0.54 kWh Cost of 1 kWh = $ 0.20 Cost of 12 kWh = 0.54 x $ 0.20 = $ 0.11
Question 16
Three lamps P, Q and R are connected in parallel to a 240 V source as shown. How much does 1 unit of electricity cost if the operating cost of lamp R for 100 min is $0.10? 16
A
$ 0.20
B
$ 0.25
C
$ 0.30
D
$ 0.35
Question 16 Explanation: 
Current flowing through lamp P = V ÷ R = 240 V ÷ 120 Ω = 2 A Current flowing through lamp R = 6 A – 2 A – 3A = 1 A Power of lamp R = VI = 240 V x 1 A = 240 W = 0.24 kW Duration of operation = 100 min = (5/3) h Energy used = Pt = 0.24 kW x (5/3) h = 0.4 kWh Cost for 0.4 kWh = $0.10 Cost for 1 kWh = $0.10 ÷ 0.4 = $ 0.25
Question 17
What is the most suitable fuse rating for fuse P in the circuit below? 17
A
2 A
B
5 A
C
10 A
D
13 A
Question 17 Explanation: 
Current flowing through lamp = V ÷ R = 6 V ÷ 3 Ω = 2 A The function of a fuse is to safeguard the wiring and the appliances in the circuit. It melts and breaks the circuit to cut off the current flow once the current flowing through the circuit is too high. Choice A is not a suitable fuse rating because the operating current is calculated to be 2 A. 2 A fuse will blow when a current of 2 A or higher flows through it. A suitable fuse for a circuit is a fuse rated at a current slightly higher than the operating current.
Question 18
What is the most suitable fuse rating for fuse R in the circuit below? 18
A
4 A
B
5 A
C
8 A
D
10 A
Question 18 Explanation: 
Current flowing through each lamp = V ÷ R = 12 V ÷ 3 Ω = 4 A Current flowing through the cell = 4 A + 4 A = 8 A Since a suitable fuse for a circuit is a fuse rated at a current slightly higher than the operating current, 10 A is the only choice available.
Question 19
What is the colour of the wire indicated by P, Q and R? 19    

 

P

Q

R

A

Yellow Green Red

B

Blue Green /Yellow Brown

C

Red Green Black

D

Black Blue Brown
A
A
B
B
C
C
D
D
Question 19 Explanation: 
Earth wire is green and yellow, Live wire is brown and Neutral wire is blue.
Question 20
Which of the wire shown below is the earth wire? 20
A
A
B
B
C
C
D
D
Question 20 Explanation: 
Earth wire connects the conducting casing of the appliance to the ground. It is used to conduct any leaked charges from the conducting casing of the appliance to the ground. This prevents people from getting an electric shock should there be any leakage of charges.
Question 21
Which of the wires shown below are the Live wire, Neutral wire and the Earth wire? 21    

 

P

Q

R

A

Live Earth Neutral

B

Live Neutral Earth

C

Neutral Earth Live

D

Neutral Live Earth
A
A
B
B
C
C
D
D
Question 21 Explanation: 
Earth wire connects the conducting casing of the appliance to the ground. The wire with a fuse must be the Live wire and therefore the remaining wire must be the Earth wire which is always attached to the metal casing of the appliance.
Question 22
What is the reason for the fuse and switch to be located at the live wire?
A
To cut off the current from the appliance.
B
To cut off the high voltage from the appliance.
C
To break the circuit when there is a sudden current surge.
D
To indicate to the user that the wire is Live wire when fuse and switch is seen on it.
Question 22 Explanation: 
Basically the fuse and the switch are placed along the Live wire to disconnect the high voltage supply from the appliance. If the fuse or switch is placed on the Neutral wire, when it is opened, it may have cut the current supply but the appliance is still connected to the high voltage supply. Someone who accidently touched the “Live” appliance may get an electric shock even though there is an open circuit.
Question 23
Yi Lin connects the switch along the neutral wire of a fan. Which of the following statements is/are correct?
  1. The fan would still operate when the switch is opened.
  2. The fan would not operate when the switch is closed.
  3. The fan would still operate when the switch is opened.
A
1 only
B
3 only
C
1 and 2 only
D
1 and 3 only
Question 23 Explanation: 
When the switch is connected along the neutral wire, it will be able to act as a circuit breaker (i.e. when the switch is closed, current can flow through the circuit and when the switch is opened, current cannot flow through). The only difference is that it disconnects the appliance from the neutral (0 V) instead of the live (approx. 240 V) wire. The appliance may cause electric shock to the user.
Question 24
In which of the following situations will a fuse possibly melt?
  1. The earth wire is broken.
  2. There is a short circuit in the electrical circuit.
  3. The fuse is fixed along the neutral wire instead of the live wire.
A
1 only
B
2 only
C
1 and 2 only
D
1 and 3 only
Question 24 Explanation: 
1. The earth wire is broken. (Incorrect) Current will flow as long as the neutral wire and the live wire are not broken. Earth wire is used only when there is a leakage of charge. 2. There is a short circuit in the electrical circuit. (Correct) A short circuit will cause the current to bypass some resistance in the circuit. This will increase the current flow because the effective resistance in the circuit is lowered. This increase in current may melt the fuse. 3. The fuse is fixed along the neutral wire instead of the live wire. (Incorrect) As long as the current in the circuit does not exceed the fuse rating, the fuse will not melt. Placing the fuse along the neutral wire will not affect the current flow.
Question 25
What is the problem when a multi-plugs adaptor is used to connect many appliances to the same socket?
A
The current drawn from the mains gets higher and overheating may occur.
B
The voltage across the live and neutral wires increases and overheating may occur.
C
The flow of the current will be slowed down and the power to each appliance will be reduced.
D
The appliances will be damaged due to the higher current that flows through each appliance.
Question 25 Explanation: 
A. The current drawn through from the mains gets higher and overheating may occur. (Correct) Every additional plug increases the current drawn from the socket. The current will generate heat and eventually cause overheating. B. The voltage across the live and neutral wires increases and overheating may occur. (Incorrect) The voltage does not increase with additional plug on the multiplug adaptor. C. The flow of the current will be slowed down and the power to each appliance will be reduced. (Incorrect) The flow of current into each appliance will not be affected by the use of multiplug adaptor as the appliances are connected in parallel. D. The appliances will be damaged due to the higher current that flows through each appliance. (Incorrect) The flow of current into each appliance will not be affected by the use of multiplug adaptor as the appliances are connected in parallel.
Question 26
What are the possible risks when an electrical appliance is used in wet places?
  1. The fuse blows
  2. The user gets an electric shock.
  3. Fire occurs.
A
1 and 2 only
B
2 and 3 only
C
1 and 3 only
D
1, 2 and 3
Question 26 Explanation: 
1. The fuse blows (Correct) Water is a good conductor of electricity. When water gets into the appliance, it may cause a short circuit and a high current is drawn into the appliance causing the fuse to blow. 2. The user gets an electric shock. (Correct) Water is a good conductor of electricity. When water gets into the appliance, it may get in contact with the live wire and conduct the charge out of the wire to the user who may touch the water and get a shock. 3. Fire occurs. (Correct) Water is a good conductor of electricity. When water gets into the appliance, it may cause a short circuit and a high current is drawn into the appliance. The high current will cause intense heating in the wires and the appliance will melt and catch fire.
Question 27
If a heater draws 5 A current from a 240 V supply, what is the current in the live wire?
A
0 A
B
5 A
C
10 A
D
240 A
Question 27 Explanation: 
The electrical circuit of a heater drawing 5 A current from a 240 V supply is shown below. The current in the live wire and the neutral wire is the same, as live and neutral wires are connected in series.
Question 28
If a heater draws 5 A current from a 240 V supply, what is the current in the neutral wire?
A
0 A
B
5 A
C
10 A
D
240 A
Question 28 Explanation: 
The electrical circuit of a heater drawing 5 A current from a 240 V supply is shown below. The current in the live wire and the neutral wire is the same, as live and neutral wires are connected in series.
Question 29
If a heater draws 5 A current from a 240 V supply, what is the potential on the neutral wire?
A
0 V
B
5 V
C
120 V
D
240 V
Question 29 Explanation: 
The electrical circuit of a heater drawing 5 A current from a 240 V supply is shown below. The live wire has an alternating voltage of 240 V and the neutral is always at 0 V.
Question 30
X lamps are connected in parallel in the circuit shown below. Given that the fuse rating is 10 A and the resistance of each lamp is 480 Ω, what is the maximum value of x? 30
A
2
B
10
C
19
D
20
Question 30 Explanation: 
The current flowing into each parallel branch is V ÷ R = 240 V ÷ 480 Ω = 0.5 A. Number of lamps connect in parallel to cause 10 A fuse to blow = 10 A ÷ 0.5 A = 20 lamps. Maximum number of lamps to connect in parallel before fuse blow = 19
Question 31
Two resistors of 10 Ω and 20 Ω are connected in parallel to a battery. The ratio of the power dissipated by the 10 Ω resistor to the power dissipated by the 20 Ω resistor is
A
1 : 2
B
1 : 4
C
2 : 1
D
4 : 1
Question 31 Explanation: 
Power = V2 ÷ R Power given out by 10 Ω resistor : Power given out by 10 Ω resistor = (V2 ÷ 10) : (V2 ÷ 20) = (1 / 10) : (1 / 20) = 2 : 1
Question 32
Many domestic hair-dryers have no earth wire. Why?
A
The fan prevents the heating coil from becoming too hot.
B
The current is small.
C
The casing of the hair-dryer is made of plastic material.
D
The plug is installed with a fuse.
Question 32 Explanation: 
The plastic casing is an insulator. The earth wire is redundant in such appliance.
Question 33
Which is the most appropriate fuse rating for a 240 V, 1.2 kW heater?
A
3 A
B
5 A
C
10 A
D
13 A
Question 33 Explanation: 
I = V ÷ R = 1200 ÷ 240 = 5 A The operating current of the heater is 5 A. The fuse should be slightly higher than the operating current. The function of the fuse is to allow operating current to flow through but melts when the current becomes too high so as to protect the equipment from being damaged by the undesirably high current. The next higher rate fuse is the 10 A fuse.
Question 34
Why is the filament in a light bulb made from tungsten?
A
Tungsten has a melting point higher than 3000 deg C.
B
Tungsten has high resistance to allow high current to pass through.
C
Tungsten has high mass so that more heat can be stored in the filament.
D
Tungsten has high specific heat capacity so that small amount of heat can increase the temperature to a large extent.
Question 34 Explanation: 
Choice A: Tungsten has a melting point higher than 3000 deg C. (Correct) When the filament is so hot that it produces white light, its temperature is must be greater than 2500 deg C. The material of the filament must have a melting point higher than 2500 deg C. Choice B: Tungsten has high resistance to allow high current to pass through. (Incorrect) When a material has high resistance, the current flowing through it is low (with other factors held constant) Choice C: Tungsten has high mass so that more heat can be stored in the filament. (Incorrect) When a material has high mass, more thermal energy is required to increase its temperature. The filament would need more energy to produce the same brightness. Choice D: Tungsten has high specific heat capacity so that small amount of heat can increase the temperature to a large extent. (Incorrect) When the material has high specific heat capacity, it requires more energy to increase its temperature.
Question 35
What is 1 kWh?
A
1000 J of energy
B
1000 W of power
C
$1 worth of energy
D
36,00,000 J of energy.
Question 35 Explanation: 
Energy = Power x time 1 kWh is an amount of energy with the same effect as 1 kW x 1 h. It is a measurement of the product of power and time. 1 kW x 1 h = 1000 W x 3600 s = 36,00,000 J
Question 36
36   The circuit below shows three identical resistors R1, R2, R3. What is the ratio of the current I1 : I2 : I3 flowing in the above resistors?
A
1 : 1 : 1
B
1 : 2 : 3
C
2 : 1 : 1
D
1 : 2 : 2
Question 36 Explanation: 
The current flowing through R1 is the main current from the source. This current splits equally and flow through R2 and R3 because R2 and R3 have the same resistance. The ratio of current I1 : I2 : I3 is 2 : 1 : 1
Question 37
The circuit below shows three identical resistors. What is the ratio of the power dissipated by R1, R2, R3? 37
A
1 : 1 : 1
B
1 : 2 : 2
C
2 : 1 : 1
D
4 : 1 : 1
Question 37 Explanation: 
The current flowing through R1 is the main current from the source. This current splits equally and flow through R2 and R3 because R2 and R3 have the same resistance. The ratio of current in R1, R2 and R3 is 2 : 1 : 1. (2 parts flow in R1 and 1 part each in R2 and R3) Power dissipated by resistor = (I x I)R Ratio of power dissipated in R1, R2 and R3 is (2 x 2) R : (1 x 1) R : (1 x 1) R = 4 : 1 : 1
Question 38
The circuit below shows three identical resistors R1, R2 and R3 of resistance 1 Ω, 2 Ω and 3 Ω respectively. What is the ratio of the power dissipated by R1, R2, R3? 38
A
1 : 2 : 3
B
5 : 2 : 3
C
5 : 3 : 2
D
25 : 18 : 12
Question 38 Explanation: 
The current, I, flowing through R1 is the main current from the source. This current splits through R2 and R3 in the ratio of [3 / (2+3)] I : [2 / (2+3)] I = (3/5) I : (2/5) I Ratio of current in R1, R2 and R3 is I : (3/5) I : (2/5) I = 5 : 3 : 2 Power dissipated by resistor = (I x I)R Ratio of power dissipated in R1, R2 and R3 is (5 x 5)(1) : (3 x 3)(2) : (2 x 2)(3) = 25 : 18 : 12
Question 39
An appliance marked “240 V, 2 A” is being connected to a 120 V supply. What is the output power of the appliance?
A
60 W
B
120 W
C
240 W
D
480 W
Question 39 Explanation: 
Resistance of the appliance = V ÷ I = 240 V ÷ 2 A = 120 Ω Current flowing through the appliance at 120 V supply = V ÷ R = 120 V ÷ 120 Ω = 1 A Power of appliance = (I x I)R = (1)x(120) = 120 W
Question 40
Two appliances marked “240 V, 2 A” and “120 V, 240 W” are being connected in parallel to a 240 V supply. Assuming that the appliances did not blow, what is the total power consumption of the appliances? 40
A
720 W
B
960 W
C
1220 W
D
1440 W
Question 40 Explanation: 
Resistance of the appliance marked “120 V, 240 W” = (V x V) ÷ P = (120 x 120) ÷ 240 = 60 Ω Current flowing through the appliance at 240 V supply = V ÷ R = 240 V ÷ 60 Ω = 4 A Power of appliance marked “120 V, 240 W” = (I x I)R = (4 x 4) x (60) = 960 W Power of appliance marked “240 V, 2 A” = VI = 240 x 2 = 480 W Total power = 960 W + 480W = 1440 W
Question 41
N identical resistors are connected in series followed by connected in parallel. The ratio of the effective resistance of parallel connection is _____________.
A
N : 1
B
1 : (N x N)
C
(N x N) : 1
D
1 : N
Question 41 Explanation: 
Effective resistance of series connection: R + R + R + … = NR Effective resistance of parallel connection: {1 / [(1/R) + (1/R) + (1/R) + …]} = 1 / (N/R) = R/N NR : (R/N) (N x N) : 1
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