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## Notes

### Electricity current, I

• defined as the rate of flow of electric charge
• There is electric current only when there are moving electric charges
• I = Q / t
• where Q is the amount of charge flows (in C), and t is the time taken (in s)
• SI unit is ampere, A
• The ammeter, milliammeter, and microammeter are current-measuring instruments and must be connected in series in the circuit

### Potential difference V and electromotive force, emf

• The potential difference or voltage, V between 2 points is defined as the work done, W in taking 1 C of positive charge from the lower potential to the point of higher potential.
• In a circuit, it is the energy sources that supply energy, not the electric charges.
• The energy that drives the free electrons around the circuit is known as emf
• More than 1 cell can be connected in a circuit (series/parallel)

Cells in series
• the combined emf used to drive the electric charges is the sum of all the individual cell's emf
• with more cells, the circuit will have more power to drive the electric charges

Cells in parallel
• the combined emf used to drive the electric charges is the emf of one individual cell (each cell contributes an equal amount of emf)
• with more cells, the circuit will have longer time to drive the electric charges
Potential difference = work done / charge transferred

• V = E / Q
• where V is potential difference, E is energy, Q is charges flow
• SI unit is volt

Example

To transfer 2 C of charge from points X to Y in an electrical circuit 50 J of energy is needed. What is the potential difference between X and Y?

Solution

Potential difference between X and Y = 50 / 2 = 25 V

• 1 volt is defined as the potential difference between two points such that one joule of work is done in transferring 1 C of charge from one point to the other.
• The voltmeter and millivoltmeter are voltage-measuring instruments and must be connected in parallel to the component across which the potential difference is being measured.
• The emf of an electrical source like a battery is equal to the electrical energy provided by the source for every coulomb of charge which flows round the circuit.
• The emf of an electrical source can also be defined as the potential difference across the terminals of the source in an open circuit

Voltmeter connected in parallel
Ammeter connected in series

### Ohm's Law and Resistance

• Ohm's Law states that the current I, passing through a conductor is directly proportional to the potential difference, V between its ends provided that the physical conditions and temperature of the conductor remain constant.
• A resistor is a conductor with known value of resistance. It can be used to control (reduce) the size of current flowing in a circuit.
• Resistance, is therefore a measure of how difficult it is for the current to pass through the circuit.
• V / I = constant

• Conductors or resistors which obey Ohm's Law are called ohmic.
• Eg. pure metal, copper sulphate solution with copper electrodes, metal alloy
• Those which do not obey Ohm's Law are called non-ohmic.

• The resistance, R of an electrical component is defined as the ratio of the potential difference, V across the component to the current, I flowing through it.

ie. if we have a high resistance then a bigger push is needed to push the current round the circuit.

#### Experimental technique for measuring resistance

Example

The voltage across a lamp is found to be 1.4V when the current in the lamp is 0.2A. Calculate the resistance of the lamp.

Solution

Resistance of lamp, R = V / I = 1.4 / 0.2 = 7 ohm

• The resistance is also given by the gradient of the graph V vs. I.

### Rheostat

• a variable resistor used to vary the control of electric current
• A rheostat can be used to find the resistance of an unknown resistor.
• The voltmeter is connected in parallel
• Use the rheostat to adjust the size of the current to a convenient value. Hence, record the readings shown on the ammeter and voltmeter
• adjust the rheostat to take 5 sets of readings of I and V
• Calculate the resistance from the equation R = V / I

### Factors affecting resistance of a wire

1. Length
• for a wire of uniform cross-sectional area, the resistance is directly proportional to the length of the wire
• hence, the longer the wire, the higher the resistance
2. Cross-sectional area
• for a wire of fixed length, its resistance is inversely proportional to the cross-sectional area
• so, the thinner the wire, the higher the resistance
3. Material
• resistance depends on the kind of substance
• copper is a good conductor and is used for connected wires
• nichrome has more resistance and is used in the heating elements of electric heater
4. Temperature
• for metallic wires, as temperature increases, the resistance increases
• but for some materials like silicon and germanium as temperature increases, the resistance decreases

### Electric Circuits

An electric circuit is a complete or closed path through which electric charges flow from one terminal of an electrical source to the other, passing through one or more circuit components.

#### Series circuit

• It has only one path for the current to flow.
• the sum of voltages across individual components in the circuit is equal to the voltage across the terminals of the electrical source or the whole circuit.
• Application: voltage divider

#### Parallel circuit

• It has more than 1 path for the current to flow
• The sum of the currents flowing in the separate branches of a parallel circuit is equal to the current from the source.
• Application: electrical household connections

#### What actually happens in an electric circuit?

• We can measure the energy difference between the loaded lorries going into the bulb and the empty ones leaving it using a voltmeter.
• The voltmeter is connected across the bulb to measure how much energy has been transferred to the bulb by comparing the energy (Joules) carried by the lorries (Coulombs) before and after the bulb.
• Each Volt represents one Joule transferred by one Coulomb.
• The proper name of this is potential difference (because the current has more potential to do work before the bulb than after it) but is often called the voltage.

#### Energy Transfers in Series and Parallel Circuits

• A bulb converts electrical energy to thermal and light energy.
• A motor converts electrical energy to kinetic energy.
• A resistor converts electrical energy to thermal energy.
• A loudspeaker converts electrical energy to sound energy.
As energy cannot be created or destroyed all the electrical energy supplied by the cell must be converted into other forms of energy by the other components in the circuit.

• This means that in a series circuit the sum of the voltages across the components must equal the voltage across the cell.
• The current is the same through all components, the potential difference is shared between components.
• In a parallel circuit, each Coulomb of charge only passes through one component before returning to the cell.
• Therefore, it has to give all the energy it carries to that component.
• Therefore, the potential difference across each component is the same as the potential difference of the cell. Potential difference is the same across all components, current is shared between components.

### Short circuit

• A short circuit occurs when a large current flows due to the very little or negligible resistance of the circuit
• A short circuit leads to
• overheating of wires which may cause electric fires
• damage of the electrical source (eg battry) and other circuit components
• To prevent short circuits, use fuse
- fuses break the circuit if the current flowing through them exceed their respective ratings.

### Combined resistance of resistors in series or parallel

In Series: Effective resistance = R1 + R2 + R3

Total voltage = V1 + V2 + V3 + ...

• Application: voltage divider

In Parallel: Effective resistance

1/RTOTAL = 1/R1 + 1/R2 + 1/R3

Total current = I1 + I2 + I3 + ...

• Application: current divider

### Diode

- A diode allows the electric current to flow in only ONE direction
• The follow arrow on the diode symbol shows that it is forward biased - the current flows easily
• The reverse arrow shows that the diode is reverse biased - the current is nearly zero

### Rectifier

• in a direct current or d.c. circuit, the current only flows in one direction, ie from positive to negative
• in a alternating current or a.c. circuit, the power supply can be controlled in such a way that the current alternates between forward and reverse directions, ie from positive to negative for a short period, then from negative to positive for another short period
• since a diode only lets current flow in the forward direction and stops all the reverse current, an a.c. can be changed into a d.c. by using a diode
• the conversion of an a.c. into a d.c. is called rectification
• the diode used to achieve rectification is called rectifier
• in half-wave rectification, the diode conducts in the forward half cycle of the a.c. (forward biased) and cuts off the reverse half cycle of the a.c. (reverse biased)

## MCQ Questions

1. The diagram shows the magnitude and directions of the electric currents entering and leaving junction X.
What will be the magnitude and direction of the current in the wire XY?
magnitude        direction
a.        1A                    X to Y
b.        1A                    Y to X
c.         5A                    X to Y
d.         5A                    Y to X

e.         8A                    X to Y

2. The diagram shows a circuit.
What is the reading on voltmeter V2?
a. 3V
b. 6V
c. 9V
d. 15V
e. 18V

3. Which quantity can be measured in units of joule/coulomb?
a. charge
b. current
c. potential difference
d. power
e. resistance

4. A current flows in two resistors connected in series as shown. A1 and A2 are the readings on the ammeter,V1 and V2 are the readings on the voltmeters.
Which of the following correctly describes the ammeter and voltmeter readings?

a.        A1 < A2                        V1 < V2
b.        A1 < A2                        V1 > V2
c.        A1 = A2                        V1 < V2
d.        A1 = A2                        V1 = V2
e.        A1 > A2                        V1 = V2

5. The diagram shows a resistor connected to a cell of e.m.f. 2V.
How much heat energy is produced in the resistor in six seconds?
a. 0.4J
b. 2.5J
c. 4.8J
d. 10J
e. 60J

6. V represents a potential different, I a current, R a resistance, and t a time. Which of the following has units of energy?
a. IRt
b. I2R
c. V/I
d. V2/R
e. VIT

7. An electric lamp is marked '240 volts 150 watts'. It is used on a ring main socket marked '30 amps maximum'. Which fuse is best to use in series with the lamp?
a. 40 amp
b. 30 amp
c. 13 amp
d. 3 amp
e. 1/2 amp

8. A 40 W fluorescent lamp turns half the electrical energy it uses into light energy. How much light does it give out in 10 s?
a. 8J
b. 20J
c. 200J
d. 400J
e. 800J

9. The diagram shows a circuit.
What is the effective resistance of the three resistors?
a. 0.67Ω
b. 1.50Ω
c. 6.70Ω
d. 15.0Ω
e. 108Ω

10. The earth wire to an electric toaster should be connected to
a. the heating element
b. the metal case
c. the ON/OFF switch
d. the plastic legs
e. the toast

11. A battery moves a charge of 60C around a circuit at a constant rate in a time of 20 s. What is the current in the circuit?
a. 0.3A
b. 3.0A
c. 40A
d. 80A
e. 1200A

12. Which of the following changes to a wire will double its resistance?
cross-sectional area        length
a.            double                        double
b.            double                        no change
c.            no change                   halve
d.            halve                           halve
e.            halve                           no change

13. A heater which is to be used on a 250V mains circuit, has a 5A fuse in its plug. Which of the following is the most powerful heater that can be used with this fuse?
a. 50W
b. 250W
c. 1000W
d. 2000W
e. 3000W

14. What is the smallest total resistance which can be obtained using only a 6Ω resistor and a 12Ω resistor?
a. 2Ω
b. 4Ω
c. 6Ω
d. 8Ω
e. 12Ω

15. Which one of the following is a unit of potential difference?
a. Watt
b. Ohm
c. Ampere
d. Volt

16. The resistances of two wires X and Y are in the ratio 2:1, their lengths are in the ratio 1:2 and their diameters are also in the ratio 1:2. The ratio of the resistivities of X and Y is then
a. 1:2
b. 1:1
c. 2:1
d. 4:1

17. A length of resistance wire is connected to the terminals of a cell. Which of the following would decrease the current through the cell?
a. using a cell with higher output voltage
b. connecting an identical wire in parallel to the first one
c. using a thicker wire of the same material and the same length
d. using a longer wire of the same material and same thickness

18. In the circuit below, the p.d. between P and Q is 20V. The p.d. between X and Y is
a. 10V
b. 20V
c. 40V
d. 120V

19. A three-pin is connected to the lead for a 1 kW electric iron to be used on a 250V supply. Which of the following statements is not correct?
a. the fuse should be fitted in the live lead
b. the live wire is coloured brown
c. A 13A fuse is the most suitable rating to use
d. the yellow and green wire should be connected to the earth pin

20. A torch bulb takes a current of 0.4A from a 3V supply for 2 minutes. How much electrical energy is used?
a. 2.4J
b. 45J
c. 57.6J
d. 144J

21. A plug connected to a table lamp contains a 3A fuse. Why is the fuse used?
a. to reduce the voltage across the lamp
b. to protect the wiring from overheating
c. to make it easier for the current to flow
d. to reduce the current that flows through the lamp

22. Why is electricity transmitted along power lines at very high voltages?
a. to reduce resistance of the cables
b. so that transformers can be used
c. to ensure that the current is the same all the way along the power lines
d. to reduce loss of energy

23. A small heater operates at 12 V, 2A. How much energy will it use when it is run for 5 minutes?
a. 30J
b. 120J
c. 1800J
d. 7200J

24. A 1.0 Ω resistor and a 2.0 Ω resistor are connected in series across a 12 V d.c. supply. What is the current in the circuit?
a. 0.25 A
b. 4.0 A
c. 6.0 A
d. 12 A

25. In an a.c. electric circuit in a house, the switch for any device is always connected to the 'live' lead. Why is this?
a. no current ever flows in the neutral lead of the device
b. the device will be shorted if the switch is in the earth lead
c. the device can never be switched off if the switch is in the neutral lead
d. the device can only be isolated (made safe) if the switch is in the live lead

26. The p.d. between the ends of a conductor is 12 V. How much electrical energy is converted to other forms of energy in the conductor when 100C of charge flows through it?
a. 0.12 J
b. 8.3 J
c. 88 J
d. 1200 J

27. A combined bathroom unit of a heater and a lamp is controlled by one switch. The unit contains a 2kW heater and a 100 W lamp. In one week, the lamp uses 1 kWh of electrical energy. How much electrical energy is used by the heater alone?
a. 2 kWh
b. 4 kWh
c. 10 kWh
d. 20 kWh

28. An electrical kettle is plugged in and switched on. The fuse in the plug blows immediately. Which single fault could cause this?
a. the earth wire is not connected to the kettle
b. the live wire and neutral wire connections in the plug are swapped around
c. the live wire touches the metal case of the kettle
d. the wires connected to the plug are too thin

29. How much electric charge passes through a 12 V battery in one minute when the current is 0.5 A?
a. 0.5 C
b. 6.0 C
c. 30 C
d. 360 C

30. When a current of 4 A flows for 1 minute through a lamp, 480 J of energy is transformed. What is the potential difference across the lamp?
a. 2 V
b. 120 V
c. 480 V
d. 1920 V

31. A 800 W of electric toaster has been used for 12 hours in a month at a cost of \$0.20 per kWh. What is the cost of the electrical energy used in a month?
a. \$1.60
b. \$1.92
c. \$13.33
d. \$1920

32. Power losses in the grid system are reduced by
a. thin wires
b. thick wires
c. high voltages
d. direct current instead of alternating current

33. A 6V cell is connected to a 3Ω resistor. How much charge flows through the resistor in 2 minutes?
a. 4C
b. 9C
c. 240C
d. 360C

34. A battery drives 30C of charge round a circuit. The total work done is 600J. What is the electromotive force of the battery?
a. 0.05V
b. 5V
c. 20V
d. 300V

35. A piece of wire 0.4m long has a cross-section of 2mm2. Which of the following wires of the same material has half its resistance?
Length        Area/mm2
a.        0.2            1.0
b.        0.2            4.0
c.        0.8            4.0
d.        0.8            8.0

36. What is the smallest total resistance that can be obtained by using only a 3Ω resistor and a 12Ω resistor?
a. 0.07Ω
b. 2.4Ω
c. 4Ω
d. 15Ω

37. A generator produces 100kW of power at a potential difference of 10kV. The power is transmitted through cables of total resistance 5Ω. What is the power lost in the cable?
a. 50W
b. 250W
c. 500W
d. 1000W

38. The resistance of a certain circuit element is directly proportional to the current passing through it. When the current is 1.0A, the power dissipated is 6.0W. What is the power dissipated when the current is raised to 2.0A?
a. 6.0W
b. 12.0W
c. 24.0W
d. 48.0W

39. Which of the following is a correct unit for electrical energy?
a. ampere
b. coulomb
c. joule
d. volt
e. watt

40. A house-owner replaced a failed fuse for the lights of his house. When the lights were switched on the new fuse also failed. The house-owner put another fuse in with a higher rating than the previous two. Why was this not a sensible thing to do?
a. fuses only work if the rating is exactly right
b. using a fuse with too high a rating would cause electric shocks
c. higher rating fuses only work for power points
d. the fuse had already failed because the rating was too high
e. a fuse with higher rating might work but the fault would not be corrected

41. Why is a fuse used in an electrical appliance?
a. to earth the appliance
b. to protect the appliance and its cable
c. to change the efficiency of the appliance
d. to change the current rating of the appliance
e. to change the voltage of the supply to the appliance

42. When using 3-core wiring (live, neutral, earth leads), where should the fuse be fitted?
a. only in the live lead
b. only in the neutral lead
c. only in the earth lead
d. in either the live or the neutral lead
e. in any lead

43. Which of the following has volt (V) as its unit?
a. current + resistance
b. power x current
c. rate of flow of charge
d. the charge in a capacitor
e. the electromotive force of a cell

44. A 5 kW immersion heater is used to heat water for a bath. It takes 40 minutes to heat up the water. How much electrical energy has been converted into thermal energy?
a. 2.0 x 102 J
b. 1.2 x 103 J
c. 2.0 x 104 J
d. 2.0 x 105 J
e. 1.2 x 107 J

45. A resistor is used in an electronic circuit but it quickly burns out. What is the reason for this?
a. a fuse has blown in the circuit
b. the current flowing is too low
c. the resistor's power rating is too high
d. the resistor's power rating is too low
e. the voltage of the battery is too low

46. A lamp is labelled 250 V, 100 W. What is its resistance?
a. 0.400 Ω
b.50 Ω
c. 62.5 Ω
d. 625 Ω

1. b
2. a
3. c
4. c
5. c
6. e
7. d
8. c
9. b
10. b
11. b
12. e
13. c
14. b
15. d
16. b
17. d
18. a
19. c
20. d
21. b
22. d
23. d
24. b
25. d
26. d
27. d
28. c
29. c
30. a
31. b
32. c
33. c
34. c
35. d
36. b
37. c
38. d
39. c
40. e
41. b
42. a
43. e
44. e (energy = power x time)
45. d
46. d

## Structured Question Worked Solutions

1. A village is 5.00km from the nearest electricity substation. Two conductors are used to connect the village to the substation. Each metre length of each conductor has a resistance of 0.00120Ω.
a. Calculate
i. the combined resistance of the 2 conductors from the substation to the village
ii. the power loss in the conductors when the current through them is 40.0A

b. The voltage between the 2 conductors is 6000 V and the voltage to each house in the village is 240 V.
i. Name the device that is used to change the 6000 V supply to a 240 V supply
ii Explain why such a high voltage is used for transmitting the electricity

Solution

ai. Resistance of a 5.00km length of conductor
= (5.00 x 103) x 0.00120
= 6.00 Ω

The conductors must be connected in series in order that a closed circuit can be formed.
Combined resistance = 6.00 + 6.00 = 12.0 Ω

aii. Power loss in the conductors = I2R = (40)2 x 12.0 = 19200W

bi. step-down transformer

bii. Since electrical power = current x voltage, a high voltage used means that only a low current is required. For low currents the loss of electrical power as heat in the cables, being I2R is also low. Besides, cables needed to carry a low current can be relatively thin, thus reducing the cost of the conductor used.

2a. How much electric charge passes through a 12V battery in 1.0s when the current is 1.0A?
2b. How much energy is transferred by a 12V battery in 1.0s when the current is 1.0A?
2c. The figure shows a battery of e.m.f. 12V connected in series with a 0.50 Ω resistor and lamps of resistance 2.5 Ω and 2.0 ohm.

i. calculate the current in the circuit
ii. calculate the voltage across the 3.0 Ω lamp
iii. calculate the power developed in the 3.0 Ω lamp

Solutions

2a. charge = It = 1.0 x 1.0 = 1.0 C

2b. energy transferred = VIt = 12 x 1.0 x 1.0 = 12 J

2ci. total resistance = 0.5 + 3.0 + 2.5 = 6 Ω

current --> I = V/R = 12 / 6 = 2A

2cii. voltage = IR = 2 x 3.0 = 6 V

2ciii. Power = IV = 2 x 6 = 12 W

3. The figure shows a circuit containing a battery of e.m.f. 3.00V, a resistor of resistance 12.0 Ω and a switch S.
When switch S is closed, what is the
a. current through the circuit
b. charge passing through the battery in 1.00s
c. energy output in the resistor in 1.00s

Solutions

3a. current, I = V / R = 3 / 12 = 0.25 A

3b. charge =  current x time =  0.25 x 1 = 0.25C

3C. energy = VIt = 3 x 0.25 x 1 = 0.75 J

4. The diagram shows three 6-V filament lamps connected to a 12-V supply of negligible internal resistance. The resistance of each lamp is shown on the diagram. The current through the battery is 2.00A.
a. determine the current through each lamp
b. calculate the voltage across each lamp
c. lamp L is taken from its socket. State and explain what happens to the brightness of lamp M and what happens to the brightness of lamp N.
d. Lamp L is now replaced in its socket and lamp M is taken from its socket. State and explain what happens to the brightness of lamp L and what happens to the brightness of lamp N.

Solutions

4a. current through the 3 Ω lamp= 2A

current through each of the 6 Ω lamps = 1A

4b. voltage across the 3 Ω lamp = 3 x 2 = 6 V

voltage across each of the 6 Ω lamps = 6 x 1 = 6V

4c. both lamps M and N will not light up since the removal of lamp L will cause the circuit to be opened

4d. the brightness of the lamp depends on its power = I2R. Lamp L will be dimmer since the current passing through it is now 1.33 A. (current = 12/9 = 1.33A)

This decrease in current is due to the increase in the resultant resistance ( from 6 Ω to 9 Ω). However, lamp N will be brighter since the current through it is now more than 1A.

5. An isolated farmhouse has its own electrical generator which supplies an output voltage of 250V to each of the following circuits.

Circuit A: a lighting circuit containing 8 lamps each rated at 250V, 150W

Circuit B: a circuit for an electric cooker rated at 250V, 6.0kW

For each circuit,
a. determine the maximum current
b. suggest a suitable fuse rating

Solutions

Circuit A

5a. maximum current = 8(P/V) = 8(150/250) = 4.8A

Since the lamps are connected in parallel, total current = sum of individual lamps

5b. suitable fuse rating is 5 A

Circuit B

5a. maximum current = P/V = 6000/250 = 24A

5b. suitable fuse rating is 30 A

6. A battery has an e.m.f. of 4.0V and negligible resistance.
a. What does this tell you about the work done by the battery in driving 1 coulomb of charge around a closed circuit?

b. When a resistor is connected across the terminals of the battery, a current of 0.20A is passed.
i. what is the time taken for 1.0C of charge to pass a given point in the circuit?
ii. calculate the rate at which heat is produced in the resistor

Solutions

6a. work done = 6J

6bi. time, t = Q/I = 1/0.2 = 5s

6bii. rate of heat produced = energy dissipated/time = 4/5 = 0.8W

7. An electric lamp is marked "250V, 100W" and an immersion heater is marked "250V, 2kW"
a. calculate the current in each device when operating normally.

bi. explain why the filament of the lamp is made to have a larger resistance than the heating element of the immersion heater

bii. suggest a reason why the filament is made of a metal with a much higher melting point than that of the element

ci. the heat capacity of the filament of the lamp is very small. State one reason why this is an advantage

cii. explain why the wire connecting the immersion heater to the supply remains cool even when the heater has been in use for some time

Solutions

7a. current in lamp = power/V = 100/250 = 0.4A

current in heater = power/V = 2000/250 = 8A

7bi. the power of the lamp is small whereas the power of the heater is large.

7bii. so that the filament would not be easily melted at high operating temperature. the heater element will not rise above 100oC

7ci. the small heat capacity allows the filament to increase in temperature rapidly with minimal heat. In this way, the filament becomes very hot and emits light in a very short time

6cii. the connecting wires have low resistance and are relatively thick, thereby producing little heat

8. A battery is charged for 6 hours using a current of 0.50A. Calculate
a. the total charge which flows through the battery
b. the work done in passing this charge through the battery if the average voltage between the battery terminals during charging is 11.0V

Solutions

8a. charge, Q =It = 0.5 x 6 x 3600 = 10800C

8b. work done = QV = 10800 x 11 = 118800 J

9a. An electric generator is connected by cables to a small factory. Given that the output power of the generator is 40kW at 5000V and that the total resistance of the cables is 0.5 Ω, calculate

i. the current in the cables
ii. the voltage drop in the cables
iii. the power loss in the cables

What happens to this 'lost' power?

9b. if the same power had been supplied at 250V, the current through the same cables would have been 20 times greater. Calculate the power loss under these circumstances

9c. explain why power is better transmitted at a high voltage rather than a low voltage.

Solutions

9ai. current in cables, I = P/V = 40000/5000 = 8A

9aii. voltage drop in cable, V = IR = 8 x 0.5 = 4V

9aiii. power lost in cables, P = I2R = 82 x 0.5 = 32W
This 'lost' power is dissipated as heat to the surroundings

9b. power lost, P = I2R = 1602 x 0.5 = 12800W

9c. at a high voltage, a low current is required.

9ci. power loss in transmission cables is small

9cii. only thin cables are needed, thus it is more economical

10. A number of 8 Ω resistors are available. In the spaces below, draw diagrams to show how you could connect a suitable number of these resistors to give an effective resistance of
a. 24 Ω
b. 4 Ω
c. 18 Ω

Solutions

10a.

10b.
10c.

11. The element of an electrical heater has a power rating of 1150W when used on a 230V supply. Calculate the cost of operating the heater for 3.0 hours if the cost of 1kWh of energy is 6.0p.

Measurements indicated that 92000J of energy were given out by the heater element in a particular period of time. What quantity of charge passed through the element during that time?

Solutions

electrical energy dissipated in 3 hours = Pt = 1.15 x 3 = 3.45 kWh

cost at 6p per kWh = 3.45 x 6 = 20.7p

Energy = QV
92000 = Q x 230
Q = 92000/230 = 400C

12. A torch uses 3 cells, each of e.m.f. 1.5V and negligible internal resistance, to light a lamp rated 4.5V, 0.5A. In the space below draw a circuit diagram of the cells and lamp when the torch is switched on. Calculate
a. the resistance of the filament of the lamp when lit
b. the charge flowing through the filament of the lamp per minute

Solutions

12a. resistance = V/I = 4.5/0.5 = 9 Ω

12b. Quantity of charge = It = 0.5 x 60 = 30C

13. Electrical power may be transmitted through a system using high alternating voltages. State the advantages gained by using
a. high voltages
b. alternating voltages

Solutions

13a. the loss of energy as heat in the cable is small. the use of thin cables is more economical

13b. the voltage can be stepped up at the power station and stepped down at the consumer end, using transformers

14. An electric heater is connected, through a correctly wired 3-pin plug, to a mains supply socket. Explain briefly
a. the function of the earth wire
b. why the fuse is connected to the live wire rather than the neutral wire

Solutions

14a. the earth wire is connected to the metal casing. Should an electrical fault develop and the live wire is now connected to the metal casing, a high current now flows to earth. This will cause the fuse to blow and a person would not receive any electric shock from touching the casing

14b. the live lead is at a high alternating voltage whereas the neutral wire is at 0V. If the fuse is connected to the live lead and blows, the circuit will be disconnected from the high voltage. If the fuse is connected to the neutral wire and blows, the circuit is still "live"

15. The battery in the circuit below has an e.m.f. of 16V and negligible internal resistance

Calculate
a. the combined resistance of the two resistors connected in parallel
b. the current flowing through the 8 ohm resistors

Solutions

15a. combined resistance = 1/(1/36 + 1/18) = 12 Ω

15b. current, I = V/R = 16/(8+12) = 0.8A

16. A battery of e.m.f. 9.0V and internal resistance 1.5 Ω is connected in series with a resistor and a current 0.5A passes through the resistor. Calculate

a. the resistance of the resistor in the circuit
b. the total rate at which chemical energy is transformed by the battery

Solutions

16a. p.d. across the internal resistance = 0.5 x 1.5 = 0.75 V
p.d. across the resistor = 9 - 0.75 = 8.25 V
Resistance = V/I = 8.25/0.5 = 16.5 Ω

16b. total rate of heat transformation = IV = 9 x 0.5 = 4.5W

17. the figure shows a battery of e.m.f. 6.0V connected to a switch S and to two resistors in parallel, each of resistance 3.0 Ω.
The switch S is closed for 5.0 minutes. Calculate
a. the current through each resistor
b. the current through the battery
c. the total charge which passes through the battery
d. the energy supplied by the battery

Solutions

17a. current, I = V/R = 6.0/3.0 = 2.0A

17b. current through battery = total current through resistors = 2 x 2.0A = 4.0A

17c. total charge which passes through the battery = total current x time = 4.0 x (5.0 x 60) = 1200C

17d. energy supplied by battery = VIt = 4.0 x 6.0 x (5.0 x 60) = 7200 J

18. The figure shows the three conductors of a 240V a.c. supply cable, a fuse, a switch and a lamp

T
The cable is rated at 240V, 5A continuous working

The lamp is rated at 240V, 500W.

a. complete the figure to show how the fuse, switch and lamp should be connected to the supply

b. what fuse rating should be used?

Solutions

18a.

18b. fuse rating = 2.5A

(current drawn by lamp = P/V = 500/240 = 2.08A)

19. The figure shows a circuit consisting of a battery of e.m.f. 6.0V and two pairs of 3.0 Ω resistors in series, these pairs of resistors being connected in parallel.

ai. what is the total resistance of the path KLM
aii. what is the total resistance of the path KNM
aiii. what is the resistance of the circuit between K and M?

b. Calculate
i. the current through the battery
ii. the power developed in the battery

Solutions

19ai. total resistance KLM = 3.0 + 3.0 = 6.0 Ω

19aii. total resistance KNM = 3.0 + 3.0 = 6.0 Ω

19aiii. resistance between K and M = (6.0 x 6.0)/(6.0 + 6.0) = 3.0 Ω

19bi. current through battery = V/R = 6.0/3.0 = 2.0A

19bii. Power = IV = 3.0 x 6.0 = 18W

20. An electric iron reaches its steady working temperature 300s after being switched on. The average current flowing through the heating element during this time is 1.3A.

Calculate the energy drawn from the 240V mains supply whilst the iron is heating up

Explain why this quantity of energy is greater than the heat retained by the iron

Solutions

Energy = Power x time = VIt = 240 x 1.3 x 300 = 93600J

This quantity is greater than the heat retained by the iron because heat is also lost to the surroundings

21. The diagram shows XY, part of a circuit into which is connected an ammeter of resistance 5.0 ohm. A current flows through the ammeter. A resistor of resistance 0.010 Ω is now connected across the ammeter terminals. Calculate the combined resistance of the ammeter and the resistor.

What is the effect of connecting the resistor across the meter on
i. the current through the ammeter
ii. the total current in the circuit?

State a practical advantage of using an ammeter and a resistor connected in this way.

Define the coulomb.

The current indicated by the ammeter was 4.2 A and it flowed for 20s. Calculate the total charge passing through the ammeter.

Solutions

21. Combined resistance = 1/(1/5 + 1/0.01) = 0.010 Ω

i. the current through the meter decreases because some of the previous current is now diverted through the resistor.

ii. the total current in the circuit will be larger because the effective resistance is lower

The ammeter can be used to measure a larger current in the circuit

A coulomb is the charge which flows in 1 second past any point in a circuit in which there is a steady current of 1 ampere.

Total charge = It = 4.2 x 20 = 84C

22a. The voltage across a 3Ω resistance wire is 6V. How large is the current?

22b. What is the resistance of a filament bulb when a voltage of 3V across it causes a current of 0.5A?

22c. Find the voltage across a manganin wire of resistance 6Ω carrying a current of 2A.

Solutions

22a. 2A

22b. 6Ω

22c. 12V

23. Two resistance wires P and Q of the same material and length but of different thickness are connected in parallel to a battery. The cross-sectional area of P is twice that of Q. What is the ratio of:
a. the resistance of P to the resistance of Q
b. the current in P to the current of Q

Solutions

23a. 1:2
23b. 2:1

24. Two torch bulbs, both marked '0.2A, 3.0V' are connected (a) in series, (b) in parallel, across a 3.0V battery. Assume that the resistance of the filament in the bulbs does not change. In each case of (a) and (b)
i. describe the brightness of the bulbs.
ii. calculate the currents through each bulb
iii. calculate the current supplied by the battery

Solutions

24aii. 0.1A
24aiii. 0.1A

24bii. 0.2A
24biii. 0.4A

25. A radio takes 0.1A of current from a 6V battery.
a. what is the overall resistance of the radio?
b. what is the power of the radio?
c. how much energy would be used if the radio is switched on for 30 minutes?

Solutions

25a. 60Ω

25b. 0.6W

25c. 1080J

26. if you watched a 120W television for 2 hours and used a 20W table lamp for 4 hours every day for 30 days, how much would you have to pay at the end of 30 days, assuming that electrical energy costs 15 cents per kWh?

Solutions

\$1.44

27. An immersion heater has a rating of 3.0kW. What would it cost to use it for 5 hours at the rate of 15 cents per kWh?

Solutions

\$2.25

28a. How much electric charge passes through a 12 V battery in two minutes when the current is 0.5 A?

b. How much energy is transferred by a 12 V battery in two minutes when the current is 0.5A?

c. When a 1.5V, 6 W lamp is connected to a 1.5 V battery, calculate the amount of charge passing through the bulb in 10 minutes.

di. A consumer buys a 250 V 100 W reading lamp. If the lamp is connected to a 250 V mains supply, what is the current passing through the lamp?

dii. If fuses of rating 2 A, 5 A and 13 A are available, which fuse should be used for the lamp?

Solution

28a. charge =  current x time = 0.5 x (2 x 60) = 60 C

28b. energy = charge x voltage = 60 x 12 = 720 J

28c. I = P/V = 6 / 1.5 = 4 A
charge = current x time = 4 x (10 x 60) = 2400 C

28di. I = P / V = 100 / 250 = 4 A

28dii. 2 A fuse should be used