# Chapter 12 DC circuits - plannerLIVE!

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7 Οκτ 2013 (πριν από 4 χρόνια και 7 μήνες)

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Worksheet
Worked examples
Practical: Resistors in series and parallel
End-of-chapter test
Marking scheme: Worksheet
Marking scheme: End-of-chapter test
Chapter 12
DC circuits
OCR (A) specifications: 5.2.2a,b,h,i,j
12 DC circuits
113
Worksheet
Intermediate level
1
Two resistors are connected in series to a d.c. supply. The current drawn from the
supply is 2.0A. What is the current in each resistor?[1]
2
Two identical resistors are connected in parallel. Each resistor has a resistance R.
Determine the total resistance of the combination in terms of R.[2]
3
Calculate the total resistance of each circuit below.
a b c
[1]
[2] [2]
4
In the electrical circuit shown here, the
battery has e.m.f. 6.0 V and may be assumed to
have negligible internal resistance.
Calculate:
a the total resistance of the circuit;[1]
b the current in each resistor;[2]
c the potential difference across the
220Ω resistor.[2]
Higher level
5
In the parallel circuit shown here, the cell has e.m.f. 1.5V and
may be assumed to have negligible internal resistance.
Calculate:
a the total resistance of the circuit;[2]
b the current shown by the ammeter.[2]
6
The diagram shows a number of identical resistors, each of resistance R, connected
between points A and B.
Determine the total resistance between A and B in terms of R.[3]
10 Ω 30 Ω
30 Ω
10 Ω
10 Ω
10 Ω10 Ω
A
220 Ω
100 Ω
6.0 V
A
120 Ω
1.5 V
68 Ω
R R
R
R
A B
114
12 DC circuits
7
Semiconductor diodes are easily
damaged if the current in them is too
large. To protect a diode from accidental
damage, it must have a ‘safety’ resistor
connected in series. The diagram shows
a circuit in which a diode is protected
by a resistor of resistance 100Ω.
The battery has negligible internal
resistance. Calculate:
a the potential difference across the resistor;[2]
b the current in the diode;[2]
c the rate of energy supplied by the battery.[2]
8
Six identical lamps are connected in parallel. The power dissipated by each lamp
is 60W and the total current drawn from the supply is 1.57A. Calculate:
a the potential difference across each lamp;[3]
b the resistance of each lamp.[2]
Extension
9
The resistance value of a cheap fixed resistor is often known to an accuracy or
tolerance of ±10%. Two resistors of resistances 22Ωand 10Ω, each having a
tolerance of ±10%, are connected in series to a 12V d.c. supply of negligible
internal resistance.
What are the possible maximum and minimum values of the current drawn
from the supply?[4]
10
In circuit calculations, we often assume that a voltmeter has an infinite
resistance. In practice, however, this is not the case. Voltmeters have a finite
but high value of resistance.
A student connects up the circuit shown below.
Calculate the reading expected by the student, who presumes that the voltmeter
has an infinite resistance. What is the actual reading on the voltmeter, given
that it has a resistance of 220kΩ? What effect does the voltmeter have on
the circuit?[5]
100 Ω
diode
0.65 V
9.0 V
V
100 kΩ
6.0 V
220 kΩ500 kΩ
Total: –––
40
Score: %
12 DC circuits
115
Worked examples
Example 1
In the circuit shown, calculate the total resistance between A and B.
The resistors of resistances 40Ω, 30Ω and 20Ω are connected in series. Their total
resistance is:
R = R
1
+ R
2
+ R
3
R = 40 + 30 + 20 = 90Ω
This total series resistance of 90Ω is in a parallel combination with the 10Ω resistance.
Hence:
R
1
R
2
R =
R
1
+ R
2
where now R
1
= 90Ω and R
2
= 10Ω
R =
90 × 10
= 9.0Ω
90 + 10
The total resistance between A and B is 9.0Ω.
10 Ω
30 Ω
40 Ω 20 Ω
A B
Tip
Remember that the equation
R
1
R
2
R =
R
1
+ R
2
can only be used for two resistors in parallel. You can also use the ‘reciprocal formula’:
1
=
1
+
1
+ ...
R R
1
R
2
In our case, we have:
1
=
1
+
1
=
100
so R =
900
= 9.0Ω
R 90 10 900 100
The most common error in examinations is failing to take the reciprocal in the last
stage of the calculation. As another alternative, you can always use the ‘x
–1
’ button
R = (90
–1
+ 10
–1
)
–1
= 9.0Ω
For a parallel combination,
the total resistance is always
going to be less than the
smallest resistance value.
116
12 DC circuits
Example 2
The points A and B of the circuit in example 1 are connected to a supply of e.m.f. 5.0V.
The supply has negligible internal resistance. Calculate the currents in the 10Ω and
30Ω resistors.
The p.d. across the 10Ω is 5.0V. Hence:
I =
V
=
5.0
= 0.50A
R 10
The current in the 40Ω, 30Ω and 20Ω resistors is the same because they are connected
in series. The total p.d. across them is 5.0V. Hence, the current in the 30Ω resistor is:
I =
V
=
5.0
≈ 5.6 × 10
–2
A
R 90
Tip
You can always quickly redraw the circuit so that you can visualise the series and
parallel sections of the circuit.
Remember to use the total resistance
of 90Ωhere and not 30Ω.
30 Ω
10 Ω
40 Ω 20 Ω
+ –
5.0 V
series
12 DC circuits
117
Practical
Resistors in series and parallel
Safety
There are not likely to be any major hazards in carrying out this experiment. However,
teachers and technicians should always refer to the departmental risk assessment before
carrying out any practical work.
Apparatus
• digital multimeter (with resistance facility)
• various resistors or resistance substitution box
Introduction
This experiment is based on the information given on pages 102 and 103 of Physics 1. You
will be connecting up series and parallel circuits, predicting the total resistance of the
combination and confirming experimentally whether or not your predictions were correct.
Procedure
Set up the experiments as shown below. You do not need to use an external supply. A
digital multimeter, set on the resistance setting, will instantly ‘measure’ the total
1
Measure the resistance of each resistor accurately using the multimeter.
2
Set up a circuit with your choice of resistors.
3
Apply the series rule (R = R
1
+ R
2
+ ...) or the parallel rule (R =
R
1
R
2
) to predict the
R
1
+ R
2
total resistance.
4
Measure the total resistance by using the multimeter. How good is your prediction?
5
Repeat this process for a range of resistors. Record your results as shown below.
6
Now connect the resistors in more complex combinations. Predict the resistance
before you measure it.
Guidance for teachers
Measuring the individual resistance values with the multimeter improves the accuracy
of the results. Agreement between theory and experiment can be well within ±1%. The
experiments also work better with resistances in the range 100Ω to 100kΩ.
series
R
2

R
1
parallel

R
1
R
2
multimeter as an ohmmeter
Series
R
1
(Ω) R
2
(Ω) R = R
1
+ R
2
(Ω) R (Ω)
(theoretical) (measured)
Parallel
R
1
(Ω) R
2
(Ω)
R =
R
1
R
2
(Ω)
R (Ω)
R
1
+ R
2
(measured)
(theoretical)
118
12 DC circuits
End-of-chapter test
1
Two resistors are soldered together in parallel and then connected to a battery of
e.m.f. 3.0 V. The battery has negligible internal resistance. The resistance of each
resistor is 68Ω.
a Draw a diagram for this circuit.[2]
b Calculate:
i the total resistance of the circuit;[2]
ii the current drawn from the battery.[2]
2
Calculate the total resistance of the following circuit.[3]
3
The diagram shows an electrical circuit.
a What is the total resistance between X and
Y when both switches are open? Explain
why this is so.[1]
b Determine the total resistance between X
and Y when:
i switch S
2
is open and switch S
1
is closed;[1]
ii both switches S
1
and S
2
are closed.[3]
4
The diagram below shows a length of wire of unknown resistance R connected in
a circuit.
The battery of e.m.f. 6.0V has negligible internal resistance. The ammeter shows a
current of 1.2A. Calculate the resistance R of the wire.[3]
10 Ω
20 Ω
60 Ω
100 Ω
120 Ω
100 Ω
220 Ω
S
2
S
1
X
Y
6.0 V
A
20 Ω
R
Total: –––
17
Score: %
12 DC circuits
119
Marking scheme
Worksheet
1
The same and equal to 2.0A. [1]
R
1
R
2
2
R =
R
1
+ R
2
and R
1
= R
2
= R [1]
R =
R
2
=
R
The total resistance is half that of one of the resistors. [1]
2R 2
3
a R = R
1
+ R
2
= 10 + 30 = 40Ω [1]
R
1
R
2
10 × 30
b R =
R
1
+ R
2
[1];R =
10 + 30
= 7.5Ω [1]
c Resistance of the parallel section: R =
10 × 10
= 5.0Ω [1]
10 + 10
Combining with the 10Ω resistance in series: R = R
1
+ R
2
= 10 + 5.0 = 15Ω [1]
4
a R = R
1
+ R
2
= 100 + 220 = 320Ω [1]
V 6.0
b I =
R
=
320
[1];
I = 1.88 × 10
–2
A ≈ 1.9 × 10
–2
A [1]
c V = IR = 1.875 × 10
–2
× 220 [1];V = 4.13V ≈ 4.1V [1] ( I = same)
R
1
R
2
68 × 120
5
a R =
R
1
+ R
2
=
68 + 120
Ω [1];R = 43.4Ω ≈ 43Ω [1]
V 1.5
b I =
R
=
43.4
[1];I = 3.46 × 10
–2
A ≈ 3.5 × 10
–2
A (≈ 35mA) [1]
1 1 1 1
6
R
total
=
2R
+
R
+
R
[1]
1
=
1 + 2 + 2
=
5
[1]
R
total
2R 2R
R
total
=
2R
[1]
5
7
a 9.0 = 0.65 + V [1];V = 8.35V [1]
b Current in diode = current in resistor [1]
I =
V
=
8.35
= 8.35 × 10
–2
A [1]
R 100
c P = VI = 9.0 × 8.35 × 10
–2
[1];P = 0.75W [1]
8
a Current in each lamp =
1.57
= 0.262A [1]
6
P 60
P = VI so V =
I
=
0.262
[1];V = 230V [1]
V 230
b R =
I
=
0.262
[1];R ≈ 880Ω [1]
120
12 DC circuits
9
Maximum resistance, R
max
= 1.1 × (22 +10) = 35.2Ω [1]
Minimum resistance, R
min
= 0.9 × (22 +10) = 28.8Ω [1]
V 12
I
min
=
R
max
=
35.2
= 0.341A [1]
V 12
I
max
=
R
min
=
28.8
= 0.417A [1]
The current can be any value between 0.34A and 0.42A.
10
Voltmeter has infinite resistance:
6.0
I =
(100 + 500) × 10
3
= 1.0 × 10
–5
A [1]
V = IR = 1.0 × 10
–5
× 500 × 10
3
= 5.0V [1]
The student would expect a reading of 5.0V.
Voltmeter has resistance of 220kΩ:
R =
220 × 500
= 153kΩ [1]
220 + 500
The resistance of the parallel combination is 153kΩ.
6.0
I =
(100 + 153) × 10
3
= 2.37 × 10
–5
A [1]
V = IR = 2.37 × 10
–5
× 153 × 10
3
= 3.63V ≈ 3.6V [1]
The ‘actual’ reading on the voltmeter is 3.6V. In practice, voltmeters ‘load’ circuits
and give a reading less than the ‘expected’ value.
12 DC circuits
121
Marking scheme
End-of-chapter test
1
a Correct circuit [1];correct symbols [1]
R
1
R
2
68 × 68
b i R =
R
1
+ R
2
[1];R =
68 + 68
= 34Ω [1]
ii I =
V
=
3.0
[1];I = 8.8 × 10
–2
A [1]
R 34
2
R
series
= 10 + 20 = 30Ω [1]
R
1
R
2
30 × 60
R
parallel
=
R
1
+ R
2
=
30 + 60
= 20Ω [1]
R
total
= 100 + 20 = 120Ω [1]
3
a The circuit is ‘open’, hence the total resistance is infinite. [1]
b i R = R
1
+ R
2
= 100 + 220 = 320Ω [1]
ii R
parallel
=
220 × 120
[1];R
parallel
= 77.65Ω ≈ 78Ω [1]
220 + 120
R = R
1
+ R
2
= 100 + 78 = 178Ω [1]
4
R
total
=
V
=
6.0
= 5.0Ω [1]
I 1.2
R
1
R
2
20R
R
total
=
R
1
+ R
2
therefore 5.0 =
20 + R
[1]
5(20 + R) = 20R so 100 = 15R
R = 6.7Ω [1]
3.0 V
68 Ω
68 Ω