D
Electricity and Magnetism
Chapter 15 Electric Circuit
15 Electric Circuit
Practice 15.1 (p. 36)
1
(c)
(a) No, the connection of the ammeter is not
proper. The ammeter is wrongly connected
in parallel with the bulb.
(b) No, the connection of the ammeter is not
proper. The negative terminal of the
7
ammeter is wrongly connected to the
By Q = It,
720 = I  60  60
positive terminal of the battery. Also, the
I = 0.2 A
positive terminal of the ammeter is
The size of the current is 0.2 A.
wrongly connected to the negative
8
terminal of the battery.
2
D
3
(a) The current is doubled.
Q = 20  103  0.1 = 2  103 C
2  103 C of charge passes through a cow.
Since 1C of charge is a flow of 6  1018
(b) The current is halved.
4
electrons per second,
By Q = It,
number of electrons passing through the cow
Q = 1.8  60  60
= 2  103  6  1018 = 1.2  1016
= 6480 C
6480 C of charge can be driven to flow
Practice 15.2 (p. 42)
through the battery.
5
(a) The reading is 0.49 A. The full-scale
reading is 1 A.
(b) The reading is 67 mA. The full-scale
reading is 100 mA.
6
By Q = It,
1
B
2
A
3
C
4
By E = QV,
36 = 360  V
(a) & (b)
V = 0.1 V
The voltage across the bulb is 0.1 V.
5
In 1 minute, by E = QV,
360 = Q  3
Q = 120 C
The amount of charge flows through a 3-V
battery in 1 minute is 120 C.
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6
Electricity and Magnetism
Chapter 15 Electric Circuit
The sentence is not correct. It should be
5
220 = 3  R
corrected as: ‘The voltage of a cell is 2 V’
R = 73.3 
means that 2 J of energy is provided to every
7
1 C of charge.
The overall resistance of the computer is
(a) Since the cells are connected in series,
73.3 .
total voltage = 1.5 + 1.5 = 3 V
6
total voltage = 1.5 V
= 5.95 V
By E = QV,
10 = Q  10
9
The MP3 player requires 5.95 V to operate.
5.95
It thus needs
 4 AA dry cells.
1.5
8
Q = 10 C
7
By Q = It,
(a) The current is halved.
(b) The current is halved.
10 = I  0.1
8
I = 100 A
By V = IR,
V = 10  10–3  1  103
The size of the current of a lightning flash is
= 10 V
100 A.
9
By V = IR,
V = 0.35  17
(b) Since the cells are connected in parallel,
8
By V = IR,
The voltage across the resistor is 10 V.
By Q = It,
9
Q = 0.5  4  60  60
(a) By V = IR,
220 = 11  R
= 7200 C
R = 20 
By E = QV,
The resistance of the heating element is
E = 7200  5
20  when operated at the rated voltage.
= 36 000 J
(b) By V = IR,
The amount of energy transferred to the
110 = I  20
battery is 36 000 J.
I = 5.5 A
The current flowing through the heating
Practice 15.3 (p. 52)
element is 5.5 A. The kettle cannot
1
A
2
A
3
C
4
By V = IR,
function normally in 110-V mains.
5 = I  20
I = 0.25 A
The current through a 20- resistor is 0.25 A.
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D
Electricity and Magnetism
Chapter 15 Electric Circuit
Practice 15.4 (p. 61)
10 (a) & (b)
1
A
2
C
By V = IR,
voltage across 3- resistor V3- = 0.5  3
= 1.5 V
voltage across 1- resistor V1- = 0.5  1
= 0.5 V
 Voltage across 2- resistor = 2 V
V 2
Current passing 2- resistor =  = 1 A
R 2
Total current in the circuit = 1.5 A
 Voltage across R = 6  2 = 4 V
4
V
R= =
= 2.67 
I 1 .5
Ohm's law is not applicable beyond the
point ''.
It is because the voltage across the wire is
no longer directly proportional to the
current passing through it.
(c) By V = IR,
V
1
(i) R = =
=5
I 0.2
C
4
D
5
(a)
1 1 1
 
R 1 1
R = 0.5 
The equivalent resistance is 0.5 .
1 1 1 1
(b)
  
R 1 1 1
The resistance of the wire is 5 .
V
6
(ii) R = =
=6
I 1 .0
R = 0.33 
The resistance of the wire is 6 .
11
3
The equivalent resistance is 0.33 .
1 1 1 1 1
(c)
   
R 1 1 1 1
Since the volume of the wire remains
unchanged as 10  2 = 20 mm3 after being
R = 0.25 
stretched, the new cross-sectional area will be
2
The equivalent resistance is 0.25 .
halved as 1 mm if the length has been
doubled.
By R  1/A, halving the cross-sectional area
results in the resistance being four times as the
original. Thus the new resistance will be 4R.
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6
Electricity and Magnetism
Chapter 15 Electric Circuit
Let the resistance of each resistor be 1 .
9
= 2 + 10 = 12 
V
By I  ,
R
6
I
 0.5 A
12
For (1): The equivalent resistance R
1
 1 1 1
= 1      = 1.33 
 1 1 1
For (2): The equivalent resistance R
1
1 1 
= 1     = 1.67 
1 2 
Current of 0.5 A passes the 2- light bulb.
(b) Since there is a short circuit,
For (3): The equivalent resistance R
 1 1
= 1   
 1 1
total resistance in the circuit = 2 
V
By I  ,
R
6
I  3A
2
1
 1 = 2.5 
The ascending order is (1), (2), (3).
7
Since the bulbs are connected in series, current
Current of 3 A passing the 2- light bulb.
passing them is the same. And, by
10 (a) Equivalent resistance = 2 + 2 + 2 = 6 
V = IR, their voltage ratio is RX : RY : RZ.
(a) (i)
1
1 1
(b) Equivalent resistance =    = 1.5 
6 2
Current ratio = 1 : 1 : 1
(ii) Voltage ratio = 1 : 2 : 3
(b) (i)
(c) Equivalent resistance
Current ratio = 1 : 1 : 1
1
1
1

= 3 + 
 +3=9
6
2

2

2


11 (a) Current flowing through 5- resistor I A
(ii) Voltage ratio = 3 : 4 : 12
(c) (i)
Current ratio = 1 : 1 : 1
(ii) Voltage ratio = RX : RY : RZ
8
(a) Total resistance in the circuit
1
Since the bulbs are connected in parallel,
= 0.6 A
voltage across them is the same. And,
V
by I  , the ratio of the current passing them
R
1
1
1
:
:
is
.
R X RY RZ
By V = IR, the voltage across PQ
= I A R5- Ω = 0.6  5 = 3 V
(a) (i)
1
(b) Current flowing through 10- resistor
VPQ
3


 0.3 A
R10- Ω 10
Current ratio = 6 : 3 : 2
The reading of ammeter A2 is 0.3 A.
(ii) Voltage ratio = 1 : 1 : 1
(b) (i)
(c) Current flowing through 2- resistor
Current ratio = 4 : 3 : 1
= 0.9 A
(ii) Voltage ratio = 1 : 1 : 1
1
1
1
:
:
(c) (i) Current ratio =
R X RY RZ
Voltage across 2- resistor
= IR = 0.9  2 = 1.8 V
Voltage of the battery
(ii) Voltage ratio = 1 : 1 : 1
New Physics at Work (Second Edition)
= VPQ + V2- = 3 + 1.8 = 4.8 V
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D
Electricity and Magnetism
Chapter 15 Electric Circuit
12 (a) Equivalent resistance of the circuit R
1 1 
= 3 +  
 4 12 
13 (a) (i)
1
The reading of the voltmeter remains
unchanged.
(ii) The reading of the ammeter A1
=6
increases.
V
By I  ,
R
(iii) The reading of the ammeter A2
remains unchanged.
current drawn from the battery
(b) The voltage across PQ remains
= current through 3- resistor
6
= =1A
6
unchanged.
14 (a) The statement is incorrect. The current
 Voltage across 3- resistor = 3 V
through all three bulbs connected in series
 Voltage across 4- / 12- resistor
is the same.
(b) The statement is correct.
=63=3V
(c) The statement is incorrect. If the equivalent
Similarly,
resistance of three identical bulbs
current through 4- resistor
V 3
= = = 0.75 A
R 4
connected in series is 3 , the resistance of
each bulb is 1  (1 + 1 + 1 = 3 ).
current through 12- resistor
3
V
= =
= 0.25 A
R 12
15 (a) The statement is incorrect. The voltage
across every bulb connected in parallel is
the same.
Current through 3-, 4- and 12-
(b) The statement is correct.
resistors are 1 A, 0.75 A and 0.25 A
(c) The statement is incorrect. If the equivalent
respectively.
resistance of three identical bulbs
(b) As shown in the calculation in (a), the
connected in parallel is 3 , the resistance
 1 1 1  1

of each bulb is 9       3   .
 9 9 9 

voltage across each resistor is 3 V.
(c) Since the current flowing through 3-
1
resistor is always
A = 4 times as the
0.25
current flowing through 12- resistor, the
Revision exercise 15
readings of ammeter A1 will also be four
Multiple-choice (p. 66)
times as the readings of ammeter A2.
Section A
The reading of ammeter A2
0.4
=
= 0.1 A
4
1
B
2
C
3
(HKCEE 2004 Paper II Q31)
4
(HKCEE 2005 Paper II Q39)
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D
Electricity and Magnetism
Chapter 15 Electric Circuit
Section B
2
(a) Resistor Y has a lower resistance.
(1A)
5
B
(b) If X and Y are connected in series, the
6
B
equivalent resistance will be higher than
7
(HKCEE 2005 Paper II Q40)
the resistance of X or that of Y alone.(1A)
8
(HKCEE 2005 Paper II Q41)
The V–I graph of the combined resistor
9
C
will lie in region K.
(1A)
(c) If X and Y are connected in parallel, the
Conventional (p. 68)
equivalent resistance will be lower than
Section A
the resistance of X or that of Y alone.(1A)
1
The V–I graph of the combined resistor
(a) More charge passes the wire each second.
(1A)
will lie in region M.
By V = IR, when the voltage increases, the
3
(a) (i)
current I flowing through the wire, i.e. the
per unit time, increases.
The resistance of the eureka wire
remains unchanged.
amount of charge passing through the wire
(1A)
(1A)
(ii) The resistance of the eureka wire
(1A)
increases.
(b) Less charge passes the wire each second.
(1A)
(b) The proportional relation is not obeyed
(1A)
when the current through the wire is high
Since the resistance of the wire increases
enough to heat up the wire sufficiently
with its length,
such that its resistance increases.
(1A)
if the length of the wire increases, the
4
(1A)
(a)
current I flowing through the wire, i.e. the
amount of charge passing through the wire
per unit time, decreases (V = IR).
(c) Less charge passes the wire each second.
(1A)
(1A)
(b)
Since the resistance of the wire increases
with decreasing diameter,
(1A)
if the diameter of the wire decreases, the
(1A)
current I flowing through the wire, i.e. the
(c)
amount of charge passing through the wire
per unit time, decreases (V = IR).
(1A)
(d)
(1A)
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5
Electricity and Magnetism
Chapter 15 Electric Circuit
(a) Equivalent resistance between AB
1
 1

 
 3  4 1 8 
7
1
4
high.
(1M + 1A)
(1A)
If a metallic balloon is in touch with a live
(b) Equivalent resistance of the whole resistor
network  5  4  16  25  (1M + 1A)
cable and an earthed object, a very large
(c) Voltage in parallel branches is the same.
object through the metallic balloon. A
current would flow from the cable to the
Voltage across the 8- resistor between
short circuit is therefore formed.
AB
This overheats the cable and causes
equivalent resistance between AB

equivalent resistance of the network
 voltage of the battery
electricity failure.
(1A)
(For effective communication.)
(1C)
(b) Plastic is an insulating material.
(1A)
4

 12 .5  2 V
25
(1M+1A)
(d) By V = IR,
(1A)
The plastic shoes prevent current from
flowing through the electricians even if
(1M)
they touch high-voltage power cables
2=I8
I = 0.25 A
accidentally.
(1A)
8
The current passing through 8- resistor
(1A)
(a) A bird standing on a power transmission
cable will not get an electric shock. (1A)
between AB is 0.25 A.
6
(a) The voltage at the power cable is very
It is because the voltage across the points
When S is open, current of 1 A passes 4-
where the bird stands is small.
resistor and R2. The voltage across R2 is 8 V.
V
By I  ,
(1M)
R
8
R2 = = 8 
(1A)
1
(1A)
By V = IR, the current passing the body of
the bird is very small and the bird will not
get an electric shock.
(1A)
(b) When a kite is entangled with a
When S is closed, total current drawn from the
high-voltage power transmission cable, the
battery is 1.5 A. The voltage across R2 is 6 V.
V
By I  , equivalent resistance of R1 and R2
R
6
=
=4
(1A)
1 .5
voltage between the cable and the Earth,
And, the equivalent resistance
person.
1
where the person stands, is huge.
By V = IR, the current passing the body of
the person is very large and may kill that
1
 1
 1 1
1 
 =    (= 4 )
=  

 R 8
 R1 R2 
 1

 R1 = 8 
(1A)
If the wire of the kite touches two power
transmission cables at the same time, it
can cause short-circuit and result in
(1A)
disastrous effects.
The values of R1 and R2 are both 8 .
New Physics at Work (Second Edition)
(1A)
12
(1A)
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D
9
Electricity and Magnetism
Chapter 15 Electric Circuit
(b) It is because the ammeter in circuit (i)
(a)
gives the sum of the current passing the
unknown resistor and the voltmeter, (1A)
while the ammeter in circuit (ii) measures
the current passing the unknown resistor
only.
(1A)
Circuit (ii) gives a more accurate result.
(1A)
(Correct connection of ammeter.)
(1A)
(Correct connection of voltmeter.)
(1A)
The reasons are as follows:
Since the voltmeter readings measured in
Measure the current I through the wire by
the ammeter,
circuits (i) and (ii) are equal to the voltage
(1A)
across the battery, R is much larger than
when a known voltage V is applied across
the wire.
1  (the resistance of the ammeter). (1A)
(1A)
Hence, the voltage in circuit (ii) is roughly
The resistance is calculated using the
formula R = V/I.
equal to that across the unknown resistor.
(1A)
(1A)
(b) The resistance of a wire of uniform
Also, in circuit (i), the ammeter reading is
cross-sectional area and its length are in
direct proportion.
(c) (i)
larger than the actual current passing
(1A)
Wire P has higher resistance.
through the unknown resistor.
(1A)
(ii) Wire Q is thicker.
Hence, circuit (ii) gives a more accurate
(1A)
For a wire, its resistance R 
result.
l
,
A
(c) Circuit (i) should be used.
(1A)
It is because the resistance of the unknown
where l is its length and A is its
cross-sectional area.
(1A)
resistor is much smaller than that of the
(1A)
Since Q always has a smaller
voltmeter and only negligible amount of
resistance for all lengths of the wire,
current would pass the voltmeter.
Q is thicker.
Then the readings of the ammeter and the
(1A)
(1A)
voltmeter in circuit (i) would be close to
the actual current passing and the actual
Section B
10 (a) By R =
V
,
I
circuit (i): R =
circuit (ii): R =
voltage across the unknown resistor. (1A)
(1M)
12
24  10  3
12
12  10  3
= 500 
If circuit (ii) is used instead, since the
resistance of the unknown resistor is
(1A)
comparable to that of the ammeter, the
= 1000  (1A)
voltage measured would be much larger
than the actual voltage across the unknown
resistor.
New Physics at Work (Second Edition)
13
(1A)
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D
Electricity and Magnetism
Chapter 15 Electric Circuit
11 (a) Consider a simple circuit in which a cell is
12
Let x be the number of identical bulbs; Rb be
connected to an unknown resistor T.
the resistance of each bulb.
To measure the current passing T, an
The equivalent resistance of the circuit if the
ammeter is connected in series with T.
bulbs are connected in series = xRb
(1A)
(1A)
The equivalent resistance of the circuit if the
It is known that the total resistance of the
bulbs are connected in parallel = Rb/x
circuit is the sum of the resistance of the
By V = IR,
ammeter and T.
for the connection in series:
(1A)
(1A)
If the resistance of the ammeter is not very
10 = 0.01  xRb
small, after connecting the ammeter, the
Rb = 1000/x ............................................ (1)
total resistance of the circuit will be larger
V
and, by I  , the current flowing in the
R
for the connection in parallel:
10 = 1  Rb/x
Rb = 10x ................................................. (2)
circuit will be smaller than the expected
value.
(1M)
(1A)
Substituting (1) into (2), we have:
1000
 10 x
x
Hence, the resistance of an ammeter
should be very small.
x 2  100
(b) Consider a simple circuit in which two
x  10
resistors, T and U, are connected in series
The number of bulbs is 10.
to a cell.
(1A)
13 (a)
To measure the voltage across T, a
voltmeter is connected in parallel to it.
(1A)
It is known that the equivalent resistance
of T and the voltmeter is
R
T
1
 Rvoltmeter1

1
, where RT and
Rvoltmeter are the resistance of T and
voltmeter respectively.
(1A)
(Correct connection of ammeter: '+' and ''
If the resistance of the voltmeter is not
terminals.)
very large, after connecting the voltmeter,
(Correct connection of voltmeter: '+' and
the equivalent resistance will be smaller
than the resistance of T and the voltage
across T will be different from the
expected value.
(2A)
'' terminals.)
(2A)
(Correct connection with 5 wires.)
(1A)
(1A)
Hence, the resistance of a voltmeter
should be very large.
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D
Electricity and Magnetism
(b) R =
Chapter 15 Electric Circuit
V
, where R, V and I are resistance of
I
the resistor, voltmeter reading and
ammeter reading respectively.
14
(HKCEE 2005 Paper I Q9)
15
(HKCEE 2001 Paper I Q10)
16 (a) (i)
(1A)
series, the circuit will be broken if
(c) The resistance of an ammeter should be
one of the lamps breaks.
very small while that of a voltmeter should
be very large.
(ii) The light dims.
(1A)
(1A)
(1A)
(iii) When more lamps are connected, the
(d) The experimental value will be lower than
the actual value.
Since the lamps are connected in
total resistance of the circuit
(1A)
increases and the voltage across each
If the resistance of X is comparable to that
lamp decreases.
of the voltmeter, the current measured by
(1A)
This makes the lamps dimmer.
the ammeter is larger than the actual
(b) (i)
current passing X as a fraction of the
If a filament breaks, the current can
pass through the resistor connected in
current measured passes the voltmeter.
parallel to the filament and the circuit
(1A)
is still complete.
And, since the ammeter has much smaller
(1A)
(ii) The other lamps will be much
resistance than the voltmeter and X, the
dimmer.
equivalent resistance of X and the
(1A)
Since the lamps are connected in
voltmeter is still much larger than the
series and the resistance of R is much
resistance of the ammeter and the voltage
larger than that of a filament, the
measured is still close to the voltage of the
battery.
(1A)
By V = IR,
(1A)
voltage across each filament is much
smaller than that across the resistor of
the broken lamp. Hence, the lamps
the resistance calculated is lower than the
will be very dim.
(1A)
actual resistance of X.
(e)
(Correct connection of ammeter.)
(1A)
(Correct connection of voltmeter.)
(1A)
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D
Electricity and Magnetism
Chapter 15 Electric Circuit
Physics in articles (p. 73)
(a) He is incorrect.
(1A)
Even if a piece of metal is not connected to a
battery, free electrons inside the metal move
rapidly.
(1A)
Since free electrons collide with positive ions
inside the metal, they change their moving
directions and their overall displacement, not
distance travelled, is zero.
(b) They are opposite.
(1A)
(1A)
(c) When a current passes through a piece of
metal, electrons are accelerated by the electric
field and gain kinetic energy.
(1A)
Then electrons transfer the kinetic energy
gained to ions in collisions. This increases the
internal energy of the metal and produces the
heating effect.
New Physics at Work (Second Edition)
(1A)
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