OCR (A) specifications: 5.2.1f,g,n,o,p; 5.2.2d
Chapter 11
Voltage, energy and power
Worksheet
Worked examples
Practical: The energy dissipated by an electric heater
End-of-chapter test
Marking scheme: Worksheet
Marking scheme: End-of-chapter test
Worksheet
Intermediate level
1
A cell has an electromotive force (e.m.f.) of 1.5 V. Calculate the chemical energy
transferred when the following charges flow through the cell:
a
1 C;
[2]
b
600 C.
[1]
2
The potential difference across a filament lamp is 6.0 V. Explain what this means in
terms of energy transfer and charge.
[1]
3
Calculate the potential difference across a component that transfers 15 J
of energy when a charge of 4.2 C flows through it.
4
5
6
[2]
A 12 V, 36 W lamp is operated for 1 hour (3600 s). Calculate:
a
the energy dissipated by the lamp;
[2]
b
the current in the lamp.
[2]
Show that 1 kW h is equal to 3.6 MJ.
[2]
An electric heater of rating 900 W is operated for a total time of 2.0 hours.
a
How much energy is transferred in joules and in kilowatt-hours?
[3]
b
What is the cost of operating the heater if the cost per kilowatt-hour is 7.5p?
[2]
Higher level
7
8
9
A 100 Ω resistor can safely dissipate 0.25 W. Calculate the maximum current in
the resistor.
[3]
A filament lamp in a small torch is labelled as ‘1.5 V, 400 mA’. The filament lamp
transfers 5.0% of the electrical energy into light and the remainder is dissipated
as heat. Calculate:
a
the power rating of the lamp;
[2]
b
the power radiated as light;
[2]
c
the resistance of the filament lamp.
[2]
A 60 W table lamp is operated for a total time of 6.0 hours.
a
How much energy is transferred in kW h?
[2]
b
For how long can a dishwasher of rating 800 W be operated for the same
cost as operating the 60 W lamp for 6.0 hours?
[2]
11 Voltage, energy and power
© Cambridge University Press 2005
103
10 The diagram shows an electrical circuit.
6.0 V
X
12 W
Y
36 W
a
Calculate the current in lamp X.
b
Calculate the ratio:
[2]
resistance of lamp X
resistance of lamp Y
[3]
Extension
11 The coiled filaments in a mains lamp and a car headlamp are made of the same
material and have the same length. Use the information below to calculate the ratio:
cross-sectional area of mains lamp filament
cross-sectional area of headlamp filament
Mains lamp: 230 V, 100 W
[4]
Car headlamp: 12 V, 36 W
12 The diagram shows two resistance wires connected in series to a power supply.
+
iron
–
nickel
The resistance wires have the same length and diameter. The resistivity of nickel is
six times that of iron.
a
Which of these two wires will be hotter? Explain your reasoning.
[3]
b
The two wires are now connected in parallel to the same power supply.
Explain which of these two wires will be hotter.
[3]
Total: ––– Score:
45
104
© Cambridge University Press 2005
%
11 Voltage, energy and power
Worked examples
Example 1
A solar cell delivers a constant current of 30 mA for a period of 2.0 minutes. During this
interval, the e.m.f. of the cell is 0.90 V. Calculate the total energy transformed by the
solar cell.
The energy W transferred is given by:
W = VQ
In this case, we use the value of the e.m.f. for V. The flow of charge Q is:
Q = It = 0.030 × 120 = 3.6 C
In this calculation, you must
convert the time into seconds
and the current into amperes.
Hence the energy transformed by the solar cell is
W = 0.90 × 3.6 = 3.2 J
Tip
An alternative route to the answer would be to use the equation:
W = IVt
Example 2
The diagram shows an electrical circuit.
Calculate the power dissipated by the 10 Ω resistor.
Use this answer to deduce the power dissipated by
the 40 Ω resistor.
The power dissipated P is given by:
0.20 A
10 Ω
40 Ω
P = I2R
Therefore:
The current in a series circuit is the
same at all points in the circuit.
P = 0.202 × 10 = 0.40 W
The current in the 40 Ω resistor is the same. The power P dissipated is
directly proportional to the resistance R. Hence, the power dissipated by
the 40 Ω resistor will be a factor of 4 times more than the 10 Ω resistor.
P = 0.40 × 4 = 1.60 W
Tip
You can also calculate the power dissipated by first calculating the p.d. across the
resistor using:
V = IR
and then using the equation:
P = VI
11 Voltage, energy and power
© Cambridge University Press 2005
105
Practical
The energy dissipated by an electric heater
Safety
Always take sensible precautions when using mains-operated supplies. Teachers and
technicians should follow their school and departmental safety policies and should
ensure that the employer’s risk assessment has been carried out before undertaking any
practical work.
Apparatus
•
•
•
•
•
•
•
•
•
immersion heater
100 ml well-lagged calorimeter
water
digital ammeter
digital voltmeter
variable d.c. supply
switch
stopwatch
thermometer
S
I
V
A
water
Introduction
lagging
The rate at which heat is dissipated by a heater
(assuming its resistance remains constant) is
directly proportional to the square of both
voltage and current. The heater should
therefore take four times as long to supply a
fixed amount of energy when either the current or the voltage is halved. Is this true? In
this experiment, you will investigate the heating effect of a heater as the voltage (and
hence the current) is varied.
calorimeter
A small amount of water in a well-lagged calorimeter is heated by a small immersion
heater. The amount of energy W supplied to the water (assuming no losses to the
environment, calorimeter, etc.) is given by:
W = VIt
or
W = I 2Rt
or
W=
V2
t
R
where I is the current in the heater, V is the p.d. across the heater and t is the period of
heating. For a given change in the temperature of the water:
I2 Rt = constant
or
V2
t = constant
R
Assuming that the resistance of the heater does not change significantly due to its
temperature, we have:
I2t = constant
or
V 2t = constant
If I (or V) is halved, then t would need to be four times as great to deliver the same
amount of energy.
Procedure
1
2
3
106
Set up the apparatus as shown in the diagram.
Set the p.d. across the heater to 12 V.
Close switch S. It takes some time for the heater itself to ‘warm up’. To get consistent
results, start the stopwatch when the temperature of the water has increased by
about 2 °C. (Remember to stir the water before measuring the temperature.)
© Cambridge University Press 2005
11 Voltage, energy and power
4
Measure the current and the time taken to change the temperature of the water
by 4.0 °C.
5
6
7
Record your results in a table.
Repeat this procedure for different values of voltage.
Do your results confirm that I 2t = constant and V 2t = constant?
Guidance for teachers
For 120 ml of water in a well-lagged copper calorimeter, the following results were
obtained for a temperature change of 4.0 °C:
V = 6.0 V
I = 8.2 A
t = 48 s
V = 3.0 V
I = 4.1 A
t = 185 s
11 Voltage, energy and power
© Cambridge University Press 2005
107
End-of-chapter test
Answer all questions.
1
2
3
State one difference and one similarity between potential difference (p.d.) and
electromotive force (e.m.f.).
[2]
The p.d. across a filament lamp is 230 V. A charge of 31 C flows through the
lamp in a time interval of 60 s. Calculate:
a
the energy transferred by the lamp;
[2]
b
the power dissipated by the lamp.
[2]
The diagram below shows an electrical circuit.
V
0.25 A
100 Ω
2.0 W
4
5
a
Calculate the power dissipated by the resistor.
[3]
b
Calculate the p.d. across the filament lamp.
[2]
a
Define the kilowatt-hour (kW h).
[1]
b
The power rating of a computer is 85 W. It is operated for a period of 5.0 hours
per week. For a period of one week, calculate:
a
the energy transferred by the computer in kW h;
[2]
ii
the cost of operating the computer if the cost per kW h is 7.1p.
[2]
A resistor of resistance R has a potential difference V across its ends. Write
an equation for the power P dissipated by the resistor.
b
6
i
The p.d. across a resistance wire is 12 V. It dissipates energy at a rate of 6.0 W.
i
Calculate the resistance of the metal wire.
[2]
ii
State and explain what will happen to the power dissipated by the
resistance wire when the p.d. across it is halved.
[2]
A metal wire of cross-sectional area 7.0 × 10–8 m2 and of length 5.0 m is coiled.
The wire dissipates energy at a rate of 40 W when the current in it is 2.5 A.
Calculate the resistivity of the metal.
Total: ––– Score:
25
108
[1]
© Cambridge University Press 2005
[4]
%
11 Voltage, energy and power
Marking scheme
Worksheet
1 a W = VQ [1]; W = 1.5 × 1 = 1.5 J [1]
b
W = 1.5 × 600 = 900 J [1]
2 A p.d. of 6.0 V across the lamp means that 6.0 J of electrical energy is transferred
into heat and light per coulomb of charge flowing through the lamp. [1]
3 V=
W
[1];
Q
4 a P=
b
V=
W
[1];
t
P = VI [1];
15
= 3.57 V ≈ 3.6 V [1]
4.2
W = Pt = 36 × 3600 = 1.3 × 105 J [1]
P 36
=
= 3.0 A [1]
V
12
I=
5 1 kW h is the energy transferred by a 1 kW device operated for 1 hour.
P = 1 kW = 1000 W, t = 1 hour = 3600 s [1]
1 kW h = Pt = 1000 × 3600 = 3.6 × 106 J (3.6 MJ) [1]
6 a Energy = 0.9 kW × 2.0 h [1]; energy = 1.8 kW h [1];
energy = 1.8 × 3.6 × 106 ≈ 6.5 MJ [1]
b
Cost = 1.8 × 7.5p [1];
cost = 13.5p [1]
7 P = I 2R [1]
I=
P
0.25
=
[1]
R
100
I = 5.0 × 10–2 A (50 mA) [1]
8 a P = VI [1]; P = 1.5 × 0.40 = 0.60 W [1]
b
Plight = 0.05 × 0.60 [1]; Plight = 3.0 × 10–2 W [1]
c
R=
V
[1];
I
R=
1.5
= 3.75 Ω ≈ 3.8 Ω [1]
0.40
9 a Energy = 0.06 kW × 6.0 h [1]; energy = 0.36 kW h [1]
b
0.36 = 0.8 × t (t = time in hours) [1]
t=
10 a I =
b
P=
0.36
= 0.45 h (t = 27 minutes) [1]
0.8
P
[1];
V
I=
V2
R
so
12
= 2.0 A [1]
6.0
R=
V2
[1]
P
V2
PX PY 36
RX
[1];
= 2 = =
Ratio
RY V
PX 12
PY
11 R = ρl
A
so
A=
ρl
V2
[1]; P =
R
R
ratio = 3.0 [1]
so
R=
V2
[1]
P
ρl
ρ lP P
∝ 2 [1]
2 =
V
V2
V
P
100
Amains
2302 100 × 122
Ratio =
=
=
= 7.6 × 10–3 [1]
Acar
36 2
2302 × 36
12
A=
11 Voltage, energy and power
© Cambridge University Press 2005
109
12 a In a series circuit the current through the wires is the same. [1]
P = I 2R = I 2 (
ρl
) ∝ ρ [1]
A
The power dissipated is directly proportional to the resistivity ρ of the material.
Therefore, the nickel wire gets hotter. [1]
b
The p.d. across the wires is the same when they are connected in parallel. [1]
P=
V2
V2
V 2A 1
=
=
∝ [1]
R
ρl
ρl
ρ
A
The power dissipated is inversely proportional to the resistivity of the material.
Therefore, the iron wire will be hotter. [1]
110
© Cambridge University Press 2005
11 Voltage, energy and power
Marking scheme
End-of-chapter test
1
Difference: Charges are losing energy for p.d. and gaining energy for e.m.f. (for p.d.,
energy is transformed from electrical to heat, whereas for e.m.f., energy is
transformed into electrical from other forms like chemical, etc.). [1]
Similarity: Both are measured in volts (both are defined as: ‘energy transferred per
unit charge’). [1]
2
3
4
5
6
W = 230 × 31 = 7.13 × 103 J ≈ 7.1 × 103 J [1]
a
W = VQ [1];
b
P=
a
P = I 2R [1];
b
P = VI [1];
a
1kWh is the energy transferred by a 1kW device operating for a period of
1hour. [1]
b
i
Energy = 0.085 × 5.0 [1]; energy = 0.425 kW h ≈ 0.43 kW h [1]
ii
Cost = 0.425 × 7.1 [1]; cost = 3.0p [1]
a
P=
V2
[1]
R
b
i
R=
ii
For a given resistance, the power is directly proportional to the square of
V2
the p.d. ( P =
∝ V 2 ). [1] Hence the power dissipated by the resistor will
R
decrease by a factor of 4 if the p.d. is halved. [1]
P = I 2R
ρ=
W
[1];
t
P=
7.13 × 103
≈ 120 W [1]
60
P = 0.252 × 100 [1];
V=
P 2.0
=
= 8.0 V [1]
I 0.25
V 2 122
=
[1];
P 6.0
so
R=
R = 24 Ω [1]
P
40
=
[1];
I 2 2.52
RA 6.4 × 7.0 × 10–8
=
[1];
l
5.0
11 Voltage, energy and power
P = 6.25 W ≈ 6.3 W [1]
R = 6.4 Ω [1]
ρ = 8.96 × 10–8 Ω m ≈ 9.0 × 10–8 Ω m [1]
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111