CHAPTER 10
Direct current circuits
LEARNING OBJECTIVES
The learning objectives for this chapter are listed below. By ticking each objective
when you have met it, you can track your progress over the course.
Identify concepts from the list below (‘Achievement’ criterion one):
parallel circuits with resistive components in series with the source
circuit diagrams
voltage
current
resistance
energy and power
voltage–current characteristics for diodes.
Describe situations involving concepts from the list below (‘Achievement’/
‘Merit’ criterion one):
parallel circuits with resistive components in series with the source
circuit diagrams
voltage
current
resistance
energy and power
voltage–current characteristics for diodes.
Explain situations involving concepts from the list below (‘Merit’/‘Excellence’
criterion one):
parallel circuits with resistive components in series with the source
circuit diagrams
voltage
current
resistance
energy and power
voltage–current characteristics for diodes.
236
DIRECT CURRENT CIRCUITS
parallel circuits with resistive components in series with the source
circuit diagrams
voltage
current
resistance
energy and power.
INTRODUCTION
The components used in NCEA Level 2 Physics are shown in the following table, along with
their circuit symbols.
Circuit symbol
Components
Circuit symbol
Components
switch
single cell
CHAPTER 10
Solve problems involving concepts from the list below (‘Achievement’/‘Merit’/
‘Excellence’ criterion two):
lamp or bulb
power pack or battery
resistor
fuse
diode
variable resistor
LED
A
ammeter
LDR
V
voltmeter
thermistor
ELECTRIC CURRENT
An electric current flows when charges move. The charges
may be electrons (as in a wire) or ions (as in an electrolyte).
When one coulomb of charge passes a point in one second
then the current at that point is one ampere. When a charge
of q coulombs passes a point in t seconds then the current I at
that point is given by:
Chapter 10
Direct current circuits
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BEGINNING PHYSICS WORKBOOK
I=
q
t
Current is the rate of flow of charge.
Current has the symbol, I, and the unit A, ampere.
Example
A heater draws a current of 5 A.
a How many coulombs of charge flow
through the heater in 2 hours?
2 hours = 2 × 60 × 60 = 7200 seconds.
q = I × t = 5 × 7200 = 36 000 C.
b One coulomb is 6.25 × 1018 electrons.
How many electrons flow through the
heater in this time?
The number is 3.6 × 104 × 6.25 × 1018 =
2.25 × 1023 electrons.
Conduction in metals
In a metal, most of the electrons are attached to particular atoms. Only about one electron per
atom is free to move. When a piece of metal is connected to a battery, the battery sets up an
electric field in the wire. This makes an electric force on the electrons. The free electrons move
in the direction of this force. The more free electrons, the bigger the current.
• Good conductors have many free electrons.
• Good insulators have very few free electrons.
The average velocity of the electrons is called the drift velocity. The higher the drift
velocity, the bigger the current.
Direction of current flow
We say that current flows from positive to negative, from a place of higher voltage to
a place of lower voltage as shown in the diagram.
I
+
–
Sometimes this is called conventional current. This convention was adopted before it
was discovered that in metals the current is a flow of electrons. Electrons, being negative, flow
from negative to positive. When people talk of this electron flow, they use the term ‘electron
current’.
238 Beginning Physics Workbook
A rule for current
total current in = total current out
The examples below illustrate this rule.
Examples
1 There are 6 A arriving at point Z so there must be 6 A leaving. Thus I = 5 A.
2A
I
4A
1A
Z
2 When the 10 A current gets to the branch in the circuit, it must split up. We see
that 6 A goes through the top resistor. The rest of the current must go through the
bottom resistor. Thus I1 = 4 A.
CHAPTER 10
Currents are not created nor destroyed. This is because charge is not created or destroyed.
At any point in a circuit:
DIRECT CURRENT CIRCUITS
Note that we could not have the convention that the direction of the current was the
direction of charge flow. For example, when current flows in an electrolyte, positive charges
move one way while the negative charges move the other way. We would not want to say that
the current goes in both directions!
These currents then rejoin and return to the battery. Thus I2 = 10 A.
R1
10 A
R2
6A
I1
I2
We see that the current returning to the battery equals the current that left it.
This is always true. Every charge that leaves the battery returns. The charges do
however have less energy when they return.
3 Suppose the two lamps are identical. Then the 10 A current would split up equally.
Each lamp would have 5 A passing through it.
10 A
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VOLTAGE
Voltage is a measure of the energy per unit charge. It is sometimes called potential.
When a charge q has an electrical potential energy Ep at a point, then the voltage V at the
point is given by:
V=
Ep
q
Voltage has the symbol V, and the unit volt, V.
Examples
1 A car battery is rated at 12 V. This means the battery gives 12 J of electrical energy
to each coulomb of charge it delivers.
2 A lamp on a Christmas tree is found to have a voltage
difference of 23 V across it. This means that when charge of
one coulomb passes through the lamp, that charge will lose
23 J of electrical energy. This electrical energy will turn into
other forms of energy such as heat and light.
A rule for voltage differences
Voltage differences add up to zero around any circuit loop.
This is because energy is conserved. When charge travels round a circuit loop and returns to
where it started from, then it must return to that place with the same energy (and voltage) it
had before. This is like orienteers: when they come back to their starting point, they have the
same gravitational potential energy they had when they started.
Examples
1 Consider a coulomb of charge leaving
6V
V2
the positive terminal of the battery. It
C
B
has 10 J of energy.
R1
R2
D
• It travels to A and then to B without
loss of energy (assuming the wires
are perfect conductors).
+ –
• After passing through R1, it has lost 6 J
A
E
10 V
of energy. This energy appears as heat
in R1 (and some light if R1 is a lamp).
• At C, the charge now has 4 J of energy. It must return to the battery with no
energy. Therefore it must lose this energy on passing through R2. This means that
the voltage across R2 must be 4 V. V2 = 4 V.
• Now the charge arrives at D with negligible energy, travels to E and to the negative
terminal of the battery.
As the charge passes through the battery, it gains 10 J of energy (from the chemical
energy stored in the battery). It is now ready to go round the circuit again.
240 Beginning Physics Workbook
3 Here there are two routes a coulomb
of charge can follow from the positive
terminal to the negative terminal.
• It may travel through the top
resistor and then through the left
resistor. The voltages along this
path must add up to 5 V. Thus
V1 = 2 volts.
• Alternatively, the coulomb of
charge may travel through the
top resistor and then through the
right resistor. The voltages along
this path must also add up to 5 V.
Thus V2 = 3 volts.
12 V
+
V1
–
7V
V1
5V
DIRECT CURRENT CIRCUITS
V2
CHAPTER 10
2 Consider a coulomb of charge
leaving the battery. It leaves with
12 J of energy. It returns with 0 J.
• If it travels down the left branch
of the circuit it must lose all this
energy in the one resistor there.
Thus V1 = 12 volts.
• If, instead, the charge travels down
the right branch of the circuit it will
lose the 12 J partly in one resistor
and partly in the other. We see that
it loses 7 J in the second resistor.
Therefore it must lose the other
5 J in the first resistor.
Thus V2 = 5 volts.
+
–
3V
Note that the two resistors in
parallel have the same voltage across
them. This must always be the case.
A coulomb of charge must lose the
same amount of energy whichever
resistor it chooses to go through.
V2
COMPONENTS IN SERIES
The current that flows through R1 must equal
the current that flows through R2 (see top
diagram).
R1
To find the current through a component, we
connect an ammeter in series with it (as shown
in the bottom diagram). This is because we know
the current through the component will be the
same as the current through the ammeter.
A
R2
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Direct current circuits
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COMPONENTS IN PARALLEL
The voltage across R1 must equal the voltage across R2
(see the top diagram).
R1
R2
To find the voltage across a component, we connect
a voltmeter in parallel with it (as shown in the bottom
diagram). This is because we know the voltage across the
component equals the voltage across the voltmeter.
V
OHM’S LAW
For many conductors, we find that the voltage across them is proportional to the current
through them: V ∝ I. For example, when we triple the voltage across them we find that the
current through them also triples.
We introduce a constant of proportionality called the resistance, R, of the conductor.
V = IR
where:
V = voltage (V)
I = current (amperes, A)
R = resistance (ohms, Ω).
This relationship is called Ohm’s law.
We can tell whether a component obeys Ohm’s law by looking at the shape of its voltage vs
current graph. For Ohm’s law to be obeyed, the graph must be a straight line through the
origin (see Graph 1 below).
Graph 1 represents an
ohmic conductor.
Ohm’s law is obeyed.
The slope of the graph
gives the resistance.
V
I
Graphs 2 and 3 represent
V
non-ohmic conductors.
Ohm’s law is not obeyed.
V
Diode
Lamp
I
242 Beginning Physics Workbook
I
Example
A lamp carries a current of 500 mA when connected to a 3 V battery. Find its
resistance.
I = 500 mA = 500 × 10–3 A = 0.5 A
R=
V
3
=
=6Ω
I
0.5
DIRECT CURRENT CIRCUITS
Most of the resistors that you’ll use in
the lab adhere to Ohm’s law. They are
also known as ‘ohmic conductors’. If a
component’s resistance changes when
its temperature changes (see Graph 2),
then it is known as a ‘non-ohmic
conductor’.
Our electricity system uses a variety
of conductors and resistors.
CHAPTER 10
Graph 2 shows that the resistance of the device is increasing. This sort of graph is typical of
lamps. As we push more current through the lamp, it heats up more and this makes its resistance
higher too. If we could somehow keep the wire inside the lamp at a constant temperature, then
the resistance would also stay constant.
COMBINING RESISTANCES
Resistances in series
R1, R2, R3, ... in series have an equivalent resistance of Rtotal where:
Rtotal = R1 + R2 + R3 + ...
Example
R2
6Ω
R1
3Ω
≡
R
9Ω
Resistances in parallel
R1, R2, R3, ... in parallel have an equivalent resistance of Rtotal where:
1
1
1
1
=
+
+
+…
Rtotal
R1
R2
R3
For two resistances in parallel, use:
Rtotal =
R1 × R2
R1 + R2
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Direct current circuits
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Example
3Ω
R1
≡
R
2Ω
R2
6Ω
Example
Find the currents marked in the following circuit.
I3
7.5 Ω
3Ω
I1
I2
15 Ω
2V
Step one
We find the equivalent resistance of the circuit. For the two resistors in parallel,
15 × 7.5
we get
= 5.0 Ω
15 + 7.5
The two resistors are in series with the 3 Ω resistor.
Thus Rtotal = 3 Ω + 5 Ω
=8Ω
Step two
We find the current in the circuit:
I=
V
2
= = 0.25 A
R 8
Thus I1 = 0.25 A
Note that we use the circuit resistance and the circuit voltage to find the circuit
current.
Step three
We find the voltage across the 3 Ω resistor. The current through it is 0.25 A.
Therefore the voltage across it is:
V = 0.25 × 3 = 0.75 V
Step four
We find the voltage across the 7.5 Ω and 15 Ω resistors. The voltage across each is:
V = 2 – 0.75 = 1.25 V
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DIRECT CURRENT CIRCUITS
V 1.25
=
= 0.17 A
R
7.5
V 1.25
I2 =
=
= 0.083 A
R
15
Therefore
I3 =
Similarly
Note that I1 = I2 + I3. We expect this because we know that the current through the
3 Ω resistor splits up, some going through the 7.5 Ω resistor and the rest going through
the 15 Ω resistor.
Note also that I3 = 2 × I2. This is also to be expected. It is twice as easy for current
to get through the 7.5 Ω resistor as it is for current to get through the 15 Ω resistor. So,
twice as much current goes through the 7.5 Ω resistor.
THE VOLTAGE DIVIDER*
A voltage divider uses two resistors to produce an output voltage lower than the input
voltage. The diagram in the following example shows a typical arrangement.
CHAPTER 10
Step five
We find the current through the 7.5 Ω and 15 Ω resistors. The voltage across each
is 1.25 V.
Example
The input resistance for the voltage divider
shown is 12 V. The circuit resistance is 60 Ω.
Thus the circuit current is:
I=
20 Ω
V
12
=
= 0.2 A
R 60
The output voltage is the voltage across
the 40 Ω resistor. This is:
12 V
Input
voltage
V = IR = 0.2 × 40 = 8 V.
We can also calculate this more easily.
Since the current through the resistors is the
same, the voltage across them is proportional
to the resistance. Across 60 Ω we have 12 V.
40
= 8 V.
Thus across 40 Ω we have 12 ×
60
40 Ω
Output
voltage
Difficulties with voltage dividers
1 The output voltage decreases when current is drawn
from the output
Suppose in the above example some device is connected to the output. Then the current this
device draws must flow through the 20 Ω resistor but not through the 40 Ω resistor. Now, the
current through the 40 Ω resistor should equal the current through the 20 Ω resistor but it will
be smaller, smaller than it should be. Thus the voltage across the 40 Ω resistor will be smaller
than it should be too. This (reduced) voltage is the output voltage.
The more current the device draws, the worse the problem.
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2 Some of the current supplied is wasted
The above lowering of voltage can be made less by using lower resistance values in the voltage
divider. This makes the currents through the two resistors larger and the current drawn by
the device less significant. However, it also means most of the current is not going through
the device at all. In this sense, a lot of the current is wasted.
The potentiometer
The potentiometer is a variable resistance device that overcomes some of the above
difficulties with voltage dividers. To obtain a particular voltage from the divider, we can simply
vary the resistance ratio till the desired output voltage is obtained.
Example
Volume controls on radios, stereos, TVs and so on almost always use the arrangement
illustrated.
In the first diagram, we see that the output voltage is
almost as high as the input voltage. This corresponds to
a high volume setting.
Input
voltage
High
output
voltage
Input
voltage
In the second diagram, the output voltage is low and so
the volume is low.
Low
output
voltage
POWER
When a current flows through a resistor, there is a
potential difference across the resistor. This means
that the charge goes into the resistor with more
energy than it comes out with. The lost energy is
usually turned into heat.
The amount of energy given to the resistor each
second is the power.
power =
change in Ep
time taken
qV
=
t
q
=V ×
t
Ep = qV
I=
A collection of different light bulbs.
q
t
P = VI
where:
P = power, measured in watts (W), or joules per second (J s–1)
V = voltage (volts, V)
I = current (amperes, A).
246 Beginning Physics Workbook
DIRECT CURRENT CIRCUITS
Example
A heater draws a current of 5 A from the mains (230 V) for two hours. How much
energy has it received?
We first calculate the power.
power = VI
= 230 × 5
= 1150 W
From this we can calculate the energy.
energy = power x time
= 1150 × 2 × 60 × 60
= 8.28 × 106 J
= 8 MJ (1 sig fig)
Note that the time of two hours
had to be changed into seconds.
CHAPTER 10
Practical work
A) VOLTAGE RELATIONSHIPS
•
•
•
Set up the circuits shown in the diagrams below.
Measure the voltages labelled. You could use a moving coil voltmeter or a multimeter
(in parallel). Record these values beside each diagram in the spaces provided.
Determine the relationship between the supply voltage (Vs) and the voltages measured
around the circuit.
1
Vs
V1
V2
Vs =
V.
V1 =
V.
V2 =
V.
V3 =
V.
Vs =
V.
V1 =
V.
V2 =
V.
V3 =
V.
V3
2
Vs
V2
V1
V3
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Direct current circuits
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BEGINNING PHYSICS WORKBOOK
3
Vs
V1
Vs =
V.
V1 =
V.
V2 =
V.
V3 =
V.
V2
V3
4
Vs
V1
V2
Vs =
V.
V1 =
V.
V2 =
V.
V3 =
V.
V3
Complete the following conclusion:
The s
v
each l
of the circuit.
is equal to the total potential difference across
B) CURRENT RELATIONSHIPS
•
•
•
Set up the following circuits.
Measure the currents labelled. You could use a moving coil ammeter or a multimeter
(in series). Record these values beside each diagram in the spaces provided.
Determine the relationship between the currents.
1
I1
I3
I2
248 Beginning Physics Workbook
I1 =
A.
I2 =
A.
I3 =
A.
I1
I4
I2
I1 =
A.
I2 =
A.
I3 =
A.
I4 =
A.
DIRECT CURRENT CIRCUITS
2
I3
Complete the following conclusions:
The current around a single loop is
.
In a circuit with a parallel branch, however, the current
The total current
.
the total current
.
CHAPTER 10
C) OHMIC AND NON-OHMIC
CONDUCTORS
Measure the current through and voltage across:
1 an ohmic conductor (e.g. a ceramic resistor)
2 two non-ohmic conductors, i.e. a filament bulb and a diode.
Set up separate circuits for each.
• Start with the lowest supply voltage and build up to the
maximum. Allow the temperature of the conductor to
increase significantly.
• Graph V against I for the components.
A
V
V (V)
Ohmic conductor:
I (A)
V (V)
I (A)
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Direct current circuits
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BEGINNING PHYSICS WORKBOOK
V (V)
Non-ohmic conductor 1
Component:
I (A)
V (V)
I (A)
V (V)
Non-ohmic conductor 2
Component:
I (A)
V (V)
I (A)
D) THE RESISTANCE OF A LONG,
THIN WIRE
The resistance of a component can be
measured using a multimeter. This is
because the multimeter (when set to
measure resistance) supplies a known
current and measures the voltage across
the component.
Multimeter
Ω
Some wires of different thicknesses.
Component
250 Beginning Physics Workbook
0
10
20
30
40
50
60
70
80
90
cm
100
l (cm)
CHAPTER 10
R (Ω)
DIRECT CURRENT CIRCUITS
1 Hold a length of fine wire against a metre ruler. Place one lead of a multimeter on the wire
at 0 m. Place the second lead progressively at every 10 cm interval. Measure the resistance
of the wire as l, the length, is varied. Sketch a graph of R, resistance, against l, length.
R (Ω)
l (m)
2 To measure the relationship between resistance and area, treat one strand of the wire as
having one standard unit of area.
Measure the resistance of a 30 cm length of:
a single wire
b doubled wire
c tripled wire
etc.
Try to get at least five different
R (Ω)
‘cross-sectional areas’.
Sketch a graph of resistance,
R (in Ω) against cross-sectional
area, A (in ‘strands’).
A (strands)
R (Ω)
A (strands)
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BEGINNING PHYSICS WORKBOOK
3 Comment on how your results compare with the theory that:
ρl
Resistance, R =
A
Resistivity is a constant value for
where
a particular metal. For example,
l = the length of wire (m)
the resistivity of copper (at room
A = the cross-sectional area (m2)
temperature) is 1.67 × 10–8 Ω m.
ρ = the ‘resistivity’ of the type of metal (Ω m).
Comments:
Extension
Use your results, along with an accurate reading of the cross-sectional area, to find the
resistivity of the metal used in your experiments.
Questions for discussion
1 a
Use the idea of drift velocity to explain why increasing the battery voltage increases the
current in the metal.
b Use the idea of free electrons to explain how some insulators ‘break down’ and start
conducting if the applied voltage becomes too large.
252 Beginning Physics Workbook
DIRECT CURRENT CIRCUITS
2 Vincent wants to measure the current through a resistor
and the voltage across the resistor at the same time. He
decides there are two ways he could set up the circuit. The
two circuits are shown in the diagrams.
a In each circuit, one meter measures what it should and
one meter does not. For each circuit, say which meter
does not measure what it should. Explain what the
meter is actually measuring.
R1
A
V
R1
A
V
CHAPTER 10
b Which circuit would be better if the resistor had high resistance?
c Which circuit would be better if the resistor had low resistance?
3 Household mains supply voltage is 230 V,
yet transmission voltages in power lines can
be 1000 times as high or more.
a What are the advantages in having very
high voltages for transmission?
b What are the advantages of having 230 V for mains?
4 Is series or parallel the best system for household wiring? Does your choice make for a very
large or a very small resistance when all outlets are operating?
5 A typical household fuse is 10 A. The fuse from the street to the house might be 40 A.
Why the difference?
Chapter 10
Direct current circuits
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BEGINNING PHYSICS WORKBOOK
Problems and exercises
1 A 10 Ω resistor is connected to a 1.5 V cell as shown.
a What is the direction of conventional current through
the resistor?
1.5 V
10 Ω
b Calculate the size of the current that flows through the
10 Ω resistor.
c Calculate the power dissipated in the resistor.
2 Find the unknown voltage in each of the circuits below.
a
10 V
2V
b
V=?
6V
2V
V=?
2V
254 Beginning Physics Workbook
5V
DIRECT CURRENT CIRCUITS
c
8V
V=?
4V
CHAPTER 10
4V
3 Find the unknown current in each of the circuits below.
a
2A
I=?
b
3A
2A
I=?
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Direct current circuits
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BEGINNING PHYSICS WORKBOOK
c
I=?
2A
2A
4 A solenoid is supplied with 240 V and draws a current of 8.0 A. Calculate the resistance of
the solenoid.
5 When Moana’s bedside lamp is connected to the 230 V
mains, it has a resistance of l000 Ω.
a Find the current in the lamp.
b What do you think would happen to the resistance
if the mains voltage dropped to 150 V? Explain.
6 For the circuit shown, find
a the current through each resistor and the voltage across each resistor.
3V
3Ω
6Ω
18 Ω
256 Beginning Physics Workbook
the battery voltage.
DIRECT CURRENT CIRCUITS
b
7 The potentiometer in a volume control is set as shown.
The input voltage is 0.5 V.
a Find the output voltage.
10 kΩ
15 kΩ
CHAPTER 10
b Show how the potentiometer would be set if the output
voltage were 0.15 V.
8 Show how three 1 Ω resistors can be connected to make a total resistance of:
a 3Ω
b 1.5 Ω
c 0.67 Ω
d 0.33 Ω.
9 Draw a diagram to show how you could connect three 10 Ω resistors to make a total
resistance of 6.67 Ω.
Chapter 10
Direct current circuits
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BEGINNING PHYSICS WORKBOOK
10 Find the total resistance of each of the following networks.
a
2Ω
b
2Ω
2Ω
3Ω
5Ω
4Ω
c
4Ω
8Ω
6Ω
d
3Ω
5Ω
3Ω
5Ω
3Ω
11
a Find the total resistance when a 6 Ω, a 9 Ω and a 12 Ω resistor are connected in:
i series
ii parallel.
258 Beginning Physics Workbook
12
5Ω
2Ω
5Ω
R
CHAPTER 10
The total resistance of this circuit is the same as that of the resistor R. Find the value of resistor
R in the circuit.
DIRECT CURRENT CIRCUITS
b Show how these same three resistors can be connected to form a total resistance of
13 Ω.
13 For each of the following circuits, find the unknown readings.
a
A ?
V
12 V
10 Ω
b
A ?
V
12 V
10 Ω
10 Ω
V
?
Chapter 10
Direct current circuits
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BEGINNING PHYSICS WORKBOOK
c
V
A ?
?
10 Ω
10 Ω
V
12 V
d
A ?
V
12 V
10 Ω
10 Ω
10 Ω
V
?
14
6.0 V
6.0 Ω
12 Ω
A
9.0 Ω
For the circuit above, find:
a the total resistance
260 Beginning Physics Workbook
DIRECT CURRENT CIRCUITS
b the current delivered by the battery
c the voltage across the 12 Ω resistor
CHAPTER 10
d the voltage across the parallel branch
e the reading on the ammeter.
15 A component is labelled as having a power rating of 180 W.
a Calculate the current it will draw if supplied with 20 V.
b What is the resistance of the component?
16 A heater with a resistance of 200 Ω is connected to a 230 V power supply.
a Calculate the size of the current drawn by the heater.
b What is the power output of the heater?
c How much energy does the heater use if it is left running for
2 hours?
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BEGINNING PHYSICS WORKBOOK
For the following circuit:
17
8V
5Ω
15 Ω
A
10 Ω
a calculate the total resistance of the circuit
b find the size of the current through the 5 Ω resistor
c find the voltage drop across each resistor
d find the total energy dissipated by this circuit in 10 s.
18 Three resistors are connected to a 12 V cell, as shown below.
12 V
A
R
10 Ω
4Ω
a Calculate the reading on the ammeter if
the size of the unknown resistor, R, is:
i 4Ω
262 Beginning Physics Workbook
Hint: you will first need to find the
total resistance of the circuit.
iii
2Ω
DIRECT CURRENT CIRCUITS
ii
8 Ω.
ii
iii
CHAPTER 10
b Calculate the total power output of the cell if the unknown resistor is:
i 4Ω
2Ω
8 Ω.
19 In the circuit shown, the ammeter shows a reading
of 4 A.
a Find the total power output of the 12 V battery.
12 V
A
R
20 Ω
4Ω
b Calculate the value of R, the unknown resistor.
5Ω
Chapter 10
Direct current circuits
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BEGINNING PHYSICS WORKBOOK
c Find the voltage drop across the parallel section of the circuit.
d Calculate the current flowing through each of the four resistors.
e Calculate the rate at which electrical energy is dissipated in each of the resistors.
f
Compare your answers to questions a and e.
264 Beginning Physics Workbook