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Advancing Physics at Taunton’s
Questions 2.1a: Revision questions (Questions 20S)
Name:
T.G.:
Date due in:
These questions are on electric current, potential difference and power and should help
you revise some of the GCSE physics you will need for this chapter
Estimating your electricity bill
Sandra wants to budget for paying the electricity bill on her small new flat. For cooking she
mainly uses a microwave oven rated at 1 kW to reheat chilled meals, 'bake' potatoes etc.
On average she will use this cooker for 10 minutes a day. She also assumes she will use a
kettle (rated at 3.3 kW) and a toaster (rated at 500 W) daily for the about the same period
of time.
1. Calculate how many kilowatt-hours Sandra uses daily for preparing meals and snacks.
2. She remembers to include the cost of lighting: she has 100 W bulbs throughout the flat
and expects to have one of the lights on in the evening for 3 hours.
Work out the number of kilowatt-hours Sandra uses daily on lighting.
3. Estimate Sandra's quarterly bill assuming one quarter = 90 days and the cost of
electricity is 8 p / kilowatt-hour.
4. If you wish, add on your own estimates for weekly ironing and vacuuming
Cost of an electrically heated shower
5. In your student 'digs', you have to put a 10 p coin in a slot if you want to have a five
minute shower. You note that 'Power Rating = 9 kW' is marked on the shower fitting.
The last electricity bill to your home stated that one unit of electricity cost 8 p. How
much does the shower actually cost and is your 10p good value?
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Torch Bulb
3 V; 0.5 A is written on the packet of torch bulbs.
6. Use your ideas about electrons to describe the mechanism of the energy transfer when
the torch is 'on'.
7. Calculate the power conversion for the bulb in normal use.
8. The life of the bulb is approximately 10 hours.
How much energy will it have dissipated in its lifetime?
People use electric light bulbs for many purposes, from a torch used to light up a path
home, to aircraft searchlights. These lamps differ tremendously in the power they use.
9. All bulbs are stamped with two different values, for instance 36 W, 12 V. What do these
numbers tell you?
10. You can also use these values to calculate the current and the resistance of the bulb
filament. The table below shows these values for five different bulbs. Use a suitable
formula to calculate the missing values.
Bulb
p.d. / V
Headlamp
Power /
W
36
12
Current / Resistance /
A

4
Torch bulb
0.09
3
0.03
Filament bulb
100
230
Flashlight bulb
4.5
9
Energy Saving
bulb
24
230
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Fuse protection
11. Explain why appliances are protected by a fuse and explain how the fuse provides this
protection.
12. The table shows the power rating and voltage as marked on a number of appliances.
Calculate the operating current of each appliance. Suggest a suitable fuse value for
each appliance choosing from the fuse values given.
Appliances
Power p.d. / V
rating
Iron
1200 W 230
Vacuum
cleaner
Headlamp
900 W 230
Jug kettle
2.4 kW 230
Radio
100 W 230
Travel kettle
340 W 120
Microwave
cooker
1.4 kW 230
48 W
Operating Suggested fuse values
current / choosing from 3 A; 13 A
A
12
Measuring potential difference
A pupil wants to measure the potential difference across a battery connected to a circuit:
A
B
C
10 V
4 k
2 k
F
E
D
13. What instrument should he/she use?
14. The pupil notices that when the meter is put across the terminals AF, BE, CD in turn,
the reading is always the same. Why is that so?
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15. State and account for the voltmeter readings when placed across FE or AC.
A portable radio
You buy a new portable radio. It is powered by eight cells and there is a diagram printed
on the battery chamber to show you how to fit the cells:
Battery supply
1.5 V  8 R14
16. What is the total potential difference of this arrangement of cells?
17. This radio can also be connected to the 240 V a.c. mains supply which is far too large
for this radio to be used directly. What component must be included inside the radio to
change the incoming supply to 12 V?
18. Battery and mains supplies vary in potential difference. State one other significant
difference.
Hints
1. One unit of electricity (or kilowatt-hour) is the amount of electricity energy used by a 1 kW appliance
running continuously for 1 hour.
2. Cost of electricity = number of units (or kilowatt-hours)  price per unit.
3. Remember to convert the time in minutes into time in hours.
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Advancing Physics at Taunton’s
Questions 2.1b: Calculating current and power in an ion beam
Reading 30S: Text to Study
Name:
T.G.:
Date due in:
Calculating the current
Rail trucks carry coal, and the coal carries energy. Ions carry electric charge and their charges
carry energy. So you can think about a beam of moving charged particles as being like a train of
coal trucks all moving together.
Question: A coal truck carries 2 tonnes of coal. A train of 100 coal trucks takes a total time of 5
minutes to pass you. How much coal passes you each minute?
Question: An ion carries 1.6  10–19 coulombs of electric charge. 1021 ions pass you in 100
seconds. How much electric charge passes you per second? What is the electric current in
amperes?
Calculating the power
Coal is delivered to power stations because it carries energy (if there is oxygen to burn it with). The
power provided to the power station is the amount of energy provided per second.
Question: A conveyer belt carries powdered coal to the furnace of a power station. The belt
delivers 30 kg of coal every minute. One kilogram of coal provides 30 MJ of energy. What is the
power delivered to the furnace?
Question: A beam of ions has been accelerated by a potential difference of 1000 volts. The beam
current is 10 milliamperes. What power does the beam deliver? These notes aim to help you to see
how to calculate the rate of flow of charge and the rate at which energy is delivered, by a beam of
moving charged particles.
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Advancing Physics at Taunton’s
Help with calculating current and power in an ion beam
Reading 30S: Text to Study
These notes aim to help you to see how to calculate the rate of flow of charge and the rate at
which energy is delivered, by a beam of moving charged particles.
Calculating the current
Rail trucks carry coal, and the coal carries energy. Ions carry electric charge and their charges
carry energy. So you can think about a beam of moving charged particles as being like a train of
coal trucks all moving together.
Question: A coal truck carries 2 tonnes of coal. A train of 100 coal trucks takes a total time of 5
minutes to pass you. How much coal passes you each minute?
Answer: 20 coal trucks pass you in one minute (100 in 5 minutes). Each carries 2 tonnes of coal.
So 40 tonnes of coal pass you per minute. If the train arrives at a power station, 40 tonnes of coal
are delivered per minute.
The calculation can be written as an equation:
Rate of flow of coal ('coal current') = coal carriers per second  coal in each carrier
Question: An ion carries 1.6  10–19 coulombs of electric charge. 1021 ions pass you in 100
seconds. How much electric charge passes you per second? What is the electric current in
amperes?
Answer: If 1021 ions pass in 100 seconds then 1019 ions pass in 1 second (1021 / 102 = 1021– 2
= 1019 ). The electric charge passing per second is the number of ions passing per second
multiplied by the electric charge on each ion. Thus the electric current (charge per second) is 1.6 
10–19  1019 = 1.6 coulombs per second, or 1.6 amperes.
The calculation can be written as an equation:
Rate of flow of charge (electric current) = charge carriers per second  charge on each carrier
Calculating the power
Coal is delivered to power stations because it carries energy (if there is oxygen to burn it with). The
power provided to the power station is the amount of energy provided per second.
Question: A conveyer belt carries powdered coal to the furnace of a power station. The belt
delivers 30 kg of coal every minute. One kilogram of coal provides 30 MJ of energy. What is the
power delivered to the furnace?
Answer: The rate of flow of coal is 30 kg every 60 seconds, which is 0.5 kg per second. 1 kg of
coal carries 30 MJ so 0.5 kg carries 15 MJ of energy. Thus the rate of delivery of energy, or power,
is 15 MJ per second, or 15 MW, or 15  106 W.
The calculation can be written as an equation:
Rate of flow of energy (power) = rate of flow of coal in kg s–1  energy carried by coal in J kg–1
Question: A beam of ions has been accelerated by a potential difference of 1000 volts. The beam
current is 10 milliamperes. What power does the beam deliver?
Answer: The potential difference is the energy given to each coulomb of charge. From a potential
difference of 1000 volts, one coulomb of charge would have an energy of 1000 joules.
A current of 10 milliamperes, or 10–2 A, delivers 10–2 coulombs per second. If each coulomb
carries 1000 joules, the power delivered is 1000  10–2 joules per second, that is 10 watts.
The calculation can be written as an equation:
Rate of flow of energy (power) = rate of flow of charge in C s–1  energy carried by charge in J C-1
= electric current in A  potential difference in V.
10 watts would barely light up your television set, spread over the screen. Clearly the electron
beam in a television set must be accelerated by more than 1000 V, or carry a current of more than
10 mA, or both.
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Advancing Physics at Taunton’s
Questions 2.2a: Some circuit problems Question 100S: Short Answer
Name:
T.G.:
Date due in:
1. Combine the expressions P = I V and V = I R to find the useful formula for calculating power in
terms of current and resistance, and use it to fill in the table.
Resistor value / 
7.5
47
3.3 k
27 k
680 k
Power rating / W
Working current
0.42 A
0.5
2
9 mA
0.25
1
30 mA
In an electrical supplies catalogue you will find that resistors are specified according to the
maximum power they can dissipate. You need to use a 5.6 k resistor for a project where the
operating current is 15 mA. It is available with either 1 W or 2 W power rating.
2. Calculate the maximum desirable current for each resistor and decide which one to use.
3. Suggest a reason why the 2 W resistor is physically larger than the 1 W resistor.
Now think about three resistors in parallel.
2
5
10
4. What is the conductance of each resistor?
5. Which resistor will carry the largest current? (give your reasons but do not use a calculation at
this stage)
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6. What is the combined conductance of this arrangement of resistors?
7. The battery is made up of two 1.5 V cells of negligible resistance. Calculate the current through
the cell.
8. A potentiometer with a length 80 mm is connected to a 12 V supply.
+ 12V
12
rheostat
80 mm
sliding contact
V
– 0V
80
position of sliding contact / mm
Sketch a graph to show how the output voltage varies as the slider is moved along the
potentiometer. Label the diagram to make the position of the slider clear.
9. A variable resistor and a fixed resistor of 100  are in series across a 12 V supply.
Sketch the circuit.
Calculate the power that is dissipated by the 100  resistor when the variable resistor is set in
turn at 100 , 20 , 50 




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Advancing Physics at Taunton’s
Questions 2.2b:Combining resistors
Name:
Question 140S: Short Answer
T.G.:
Date due in:
Resistor series
Resistors are manufactured in limited values. You will find that the numbers range from 1.0
to 10 in 24 steps each differing from the next by about 10%. i.e.:
1.0
1.5
2.2
3.3
4.7
6.8
1.1
1.6
2.4
3.6
5.1
7.5
1.2
1.8
2.7
3.9
5.6
8.2
1.3
2.0
3.0
4.3
6.2
9.1
Two or more of these resistors can be combined to give other values of resistance.
Finding useful combinations
In the laboratory there are resistors with the values 1 k, 2.2 k, 3.3 k, 4.7 k, 5.6 k
and 6.8 k
How can you combine two or more of these resistors when you need a resistance of:
1. 3 k




2. 9 k




3. 500 




4. 5 k




5. 4 k





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Advancing Physics at Taunton’s
Circuit resistance Question 130S: Short Answer
Simplifying circuits
These are questions about replacing many resistors with one resistor which draws the
same current from the cell. Study the circuit diagrams and try to simplify sections of the
circuit by putting in an equivalent value resistor. Redraw the diagram for each step until
you are reduced to one equivalent resistor before calculating the current. Many of these
problems are easier if you think about conductance rather than resistance.
For each circuit find the current drawn from the power source.
1.
3.
I
5
I
6V
4
5
I
12 V
10
12 V
6
12
2.
.
8
5
4
12 
4
In this circuit calculate:
4.
6
12
4V
5.
The potential difference across the 12 
resistor.
3
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The current through the 6  resistor.
3
Advancing Physics at Taunton’s
Questions 2.3: Tapping off a potential difference (170S)
Name:
T.G.:
Date due in:
6V
50
100
A
B
A series circuit is connected as shown in the diagram.
1. What is the potential difference between A and B?
2. An additional resistor of 100  is connected between the 50 resistor and the cells.
What is the potential difference between A and B now?
3. The additional 100  resistor is now connected in parallel with the first 100  resistor.
What is the potential difference between A and B now?
4. A potential divider is made from a 4 k and a 6 k resistor connected in series with a
20 V supply. Draw a diagram of the arrangement. What four values of potential
difference can be tapped off?
5. A student puts a 12  variable resistor in series with a 6 V battery, expecting to get a
variable potential difference.
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12
6V
V
The voltmeter is a high resistance digital multimeter. Explain why the circuit won't work.
Draw a circuit which would work.
6. B is the wiper of a 100  rotary potentiometer.
300 
12 V
A
100 
B
What is the full range of the potential difference that can be tapped off between A and
B?
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Questions 2.4: Internal resistance of power supplies (Questions 220S)
Name:
T.G.:
Date due in:
Answer the following questions for practice in making calculations about internal
resistance and emf.
1. A cell has an e.m.f. of 1.5V and an internal resistance of 0.5Ω. This cell is connected in
series with an external resistance of 2.5Ω and an ammeter of negligible resistance. A
voltmeter is connected across the terminals of the cell.
(a) Draw a circuit diagram of the above circuit and label each part clearly.
(b) Show how to calculate the reading of the ammeter
(c) Explain why the reading of the voltmeter is less than 1.5V
(d) Calculate the actual reading of the voltmeter
2. In a similar circuit, using a cell of unknown e.m.f. and internal resistance, various
external resistances were used in turn to produce different currents, I, and
corresponding values of the voltage readings, V, shown in the table.
I / amps
1.0
1.5
2.0
2.5
3.0
V / volts
1.8
1.7
1.6
1.5
1.4
(a) Plot a graph of V against L
(b) Use your graph to determine the e.m.f. of the cell.
(c) Calculate the value of the internal resistance of the cell.
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Answer the following questions for practice in making calculations about the internal
resistance of real power supplies.
Torch batteries, car batteries, EHT supplies and solar cells
3. A typical hand-held torch runs off two 1.5 V cells, yet has a lamp rated at 2.5 V, 0.5 A.
Explain how the potential difference across the lamp can actually be 2.5 V as rated.
What is the internal resistance of each cell, supposing them to be identical?
4. A typical car battery has an emf of 12 V, and must provide a current of 80 A to the
starter motor. Why must the car battery have a very low internal resistance? If the
internal resistance is 0.05 , find the potential difference across this internal resistance
when the starter motor is running. Why is starting the car with the headlights on likely to
affect their brightness?
5. Some school laboratories have EHT (Extra High Tension) power packs giving up to
3000 V. For safety, they are provided with a 50 M resistor in series with the supply.
What is the maximum current able to be drawn from the supply? Approximately what
potential difference would there be across a torch bulb connected across such a
supply?
6. A student experimenting with a solar cell connects a 1000  voltmeter across it and
observes a potential difference of 1.0 V. Using a different, extremely high resistance
digital voltmeter, the reading is larger, 1.2 V. Why the difference? What is the internal
resistance of the solar cell?
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