Download 4 Electricity and Magnetism Chapter 2 Electric Circuit 2 Electric

4
2
Electricity and Magnetism
Chapter 2 Electric Circuit
Electric Circuit
Practice 2.1 (p. 54)
1
(c)
(a) The connection of the ammeter is not
proper. The ammeter is not connected in a
chain with the bulb and the battery.
(b) The connection of the ammeter is not
proper. The negative terminal of the
ammeter is wrongly connected to the
positive terminal of the battery, and the
positive terminal of the ammeter is
wrongly connected to the negative
Current =
8
Q = It = 20  103  0.1 = 2  103 C
2  103 C of charge passes through the cow.
terminal of the battery.
2
D
3
(a) The current is doubled.
Since 1 C of charge is equal to 6  1018
electrons,
number of electrons passing through the cow
(b) The current is halved.
4
= 2  103  6  1018 = 1.2  1016
Q = It = (1800  103)  60  60 = 6480 C
6480 C of charge can be driven to flow
Practice 2.2 (p. 62)
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
Q
720
=
= 0.2 A
t 60  60
7
1
B
2
A
3
C
4
Potential difference =
5
In 1 minute, by V =
(a) & (b)
Q=
E 36
=
= 0.1 V
Q 360
E
,
Q
E 360
=
= 120 C
V
3
The amount of charge flows through the 3-V
battery in 1 minute is 120 C.
6
The sentence is not correct. It should be
corrected as: ‘The e.m.f. of a cell is 2 V’
means that 2 J of energy is provided to every
1 C of charge.
New Senior Secondary Physics at Work
1
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4
7
Electricity and Magnetism
By V =
Chapter 2 Electric Circuit
E
,
Q
9
(a) & (b)
E 10 9
=
= 10 C
V 10 8
Q 10
Current = =
= 100 A
t 0 .1
Q=
8
Q = It = 0.5  4  60  60 = 7200 C
E = QV = 7200  5 = 36 000 J
The amount of energy transferred to the
battery is 36 000 J.
9
In Fig c, since the cells are connected in series,
total voltage = 1.5  3 = 4.5 V
In Fig d, since the cells are connected in
parallel, total voltage = 1.5 V
Ohm’s law is not applicable beyond the
Therefore, arrangement in Fig c gives a higher
point ‘’.
total e.m.f.
This is because the voltage across the wire
is no longer directly proportional to the
Practice 2.3 (p. 74)
1
A
2
A
3
C
4
By V = IR,
5
V
I= =
= 0.25 A
R 20
5
current passing through it.
1
V
(c) (i) R = =
=5
I 0 .2
The resistance of the wire is 5 .
V
6
(ii) R = =
=6
I 1 .0
The resistance of the wire is 6 .
The current through the resistor is 0.25 A.
V 220
R= =
= 73.3 
3
I
10 (a) The minimum voltage is 0.8 V.
(b) No, it does not obey Ohm’s law.
RA (5)( 2  10 6 )
11 (a)  =
=
= 0.001  m
l
10  10 3
The overall resistance of the computer is
73.3 .
6
The resistivity of the wire is 0.001  m.
R
l
(b) Since R  l, we have 2 = 2 .
R1 l1
V = IR = 0.35  17 = 5.95 V
The MP3 player requires 5.95 V to operate.
5.95
 4 AA dry cells.
Therefore it needs
1 .5
7
R2 =
V = IR = 10  10–3  1  103 = 10 V
The new resistance is 10 .
The voltage across the resistor is 10 V.
8
(a) The current is halved.
Practice 2.4 (p. 83)
(b) The current is halved.
New Senior Secondary Physics at Work
l2
 R1 = 2  5 = 10 
l1
1
2
A
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4
Electricity and Magnetism
Chapter 2 Electric Circuit
When the switch is open, current passes
(e) R = 1 
through Y, so the equivalent resistance of the
two bulbs increases. Hence the current
6
C
(a) (i)
= 1.5 V
(b) (i)
= 0.5 V
(c) (i)
7
 Voltage across R = 6  2 = 4 V
4
V
R= =
= 2.67 
I 1 .5
Since the bulbs are connected in parallel, the
voltage across them is the same. Besides,
V
by I  , the ratio of the current passing them
R
1
1
1
is
.
:
:
R X RY RZ
6
=6A
1
(a) (i)
6
I4 = I5 = I6 = I7 = = 3 A
2
Current ratio = 6 : 3 : 2
(ii) Voltage ratio = 1 : 1 : 1
(b) (i)
I1 = I2 = I 9 = I10 = 6 + 3 = 9 A
1 1 1
 
(a)
R 1 1
Current ratio = 4 : 3 : 1
(ii) Voltage ratio = 1 : 1 : 1
1
1
1
:
:
(c) (i) Current ratio =
R X RY RZ
R = 0.5 
(ii) Voltage ratio = 1 : 1 : 1
The equivalent resistance is 0.5 .
1 1 1 1
(b)
=  
R 1 1 1
8
(a) Total resistance in the circuit
= 2 + 10 = 12 
6
V
I= =
= 0.5 A
R 12
R = 0.33 
The equivalent resistance is 0.33 .
1
(c) R = 1 
= 1.33 
1 1 1
 
1 1 1
Current of 0.5 A passes the 2- light bulb.
(b) Since there is a short circuit,
total resistance in the circuit = 2 
V 6
I= = =3A
R 2
The equivalent resistance is 1.33 .
1
(d) R = 1 
= 1.67 
1
1

11 1
Current of 3 A passes the 2- light bulb.
9
The equivalent resistance is 1.67 .
New Senior Secondary Physics at Work
Current ratio = 1 : 1 : 1
(ii) Voltage ratio = RX : RY : RZ
Total current in the circuit = 1.5 A
5
Current ratio = 1 : 1 : 1
(ii) Voltage ratio = 3 : 4 : 12
 Voltage across 2- resistor = 2 V
V 2
Current passing 2- resistor = = = 1 A
R 2
I3 = I8 =
Current ratio = 1 : 1 : 1
(ii) Voltage ratio = 1 : 2 : 3
voltage across 1- resistor V1- = 0.5  1
4
Since the bulbs are connected in series, the
V = IR, their voltage ratio is RX : RY : RZ.
voltage across 3- resistor V3- = 0.5  3
C
 1 = 2.5 
current passing them is the same. Besides, by
By V = IR,
3
1 1

1 1
The equivalent resistance is 2.5 .
through the circuit decreases.
2
1
(a) R = 2 + 2 + 2 = 6 
The equivalent resistance is 6 .
3
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4
Electricity and Magnetism
(b)
Chapter 2 Electric Circuit
1 1 1
 
R 6 2
Current through 12- resistor
V
3
= 12 =
= 0.25 A
R12 12
R = 1.5 
The equivalent resistance is 1.5 .
(b) As shown in the calculation in (a), the
1
1
1

(c) R = 3 +  
 +3=9
6 22 2
voltage across each resistor is 3 V.
(c) Equivalent resistance of 4- and 12-
1
The equivalent resistance is 9 .
1 1 
resistors =    = 3 
 4 12 
10 (a) Current flowing through 5- resistor
= 0.6 A
 Voltage across all resistors are the
Voltage across PQ
same.
= voltage across 5- resistor
Voltage across 12- resistor
= I5-R5- = 0.6  5 = 3 V
= voltage across 3- resistor
= 0.4  3 = 1.2 V
(b) Current flowing through 10- resistor
VPQ
3
=
=
= 0.3 A
R10-Ω 10
Reading of ammeter A2 =
12 (a) (i)
The reading of ammeter A2 is 0.3 A.
1 .2
= 0.1 A
12
The reading of the voltmeter remains
unchanged.
(c) Current flowing through 2- resistor
(ii) The reading of the ammeter A1
= 0.6 + 0.3 = 0.9 A
increases.
Voltage across 2- resistor
= I2-R2- = 0.9  2 = 1.8 V
(iii) The reading of the ammeter A2
remains unchanged.
Voltage of the battery
(b) The voltage across PQ remains
= VPQ + V2- = 3 + 1.8 = 4.8 V
unchanged.
11 (a) Equivalent resistance of the circuit
1
1 1 
= 3 +   = 6 
 4 12 
13 (a) The statement is incorrect. The current
through all three bulbs connected in series
is the same.
Current through 3- resistor
= current drawn from the battery
V 6
= = =1A
R 6
(b) The statement is correct. The voltage
Voltage across 3- resistor
(c) The statement is incorrect. If the
across each bulb connected in series is in
the same ratio as their resistance.
I3-R3- = 1  3 = 3 V
equivalent resistance of three identical
Voltage across 4- / 12- resistor
bulbs connected in series is 3 , the
=63=3V
resistance of each bulb is 1  (1 + 1 + 1 =
Current through 4- resistor
V
3
= 4 = = 0.75 A
R4 4
3 ).
14 (a) The statement is incorrect. The voltage
across every bulb connected in parallel is
the same.
New Senior Secondary Physics at Work
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4
Electricity and Magnetism
Chapter 2 Electric Circuit
(b) The statement is correct. The current
4
(a) Voltage across the internal resistance
through each bulb connected in parallel is
= 9  8.4 = 0.6 V
in the same ratio as the reciprocal of their
Internal resistance of the battery
0 .6
=
= 0.188 
3 .2
resistance.
(c) The statement is incorrect. If the
equivalent resistance of three identical
(b) Voltage across R = 4.0  2.0 = 8.0 V
bulbs connected in parallel is 3 , the
Voltage across the internal resistance
= 9  8.0 = 1.0 V
resistance of each bulb is 9 :
1 1 1
   
9 9 9
1
Internal resistance of the battery
1 .0
=
= 0.25 
4 .0
3
(c) Current through R
5 .6
=
= 1.87 A
3 .0
Practice 2.5 (p. 95)
1
D
As the ammeter which the resistance is zero is
Voltage across the internal resistance
connected to the two resistors in parallel, the
= 6  5.6 = 0.4 V
equivalent resistance across the two resistors
Internal resistance of the battery
0.4
=
= 0.214 
1.87
and the ammeter is zero. The total resistance in
the circuit is 1 .
Therefore, the total current =
V 3
= =3A
R 1
5
(a)
A
B
The reading of the ammeter is 3 A.
3.998  10 A
2.394  103 A
Reading of the voltmeter
Equal to
Larger than
= voltage across the 1- resistor
0%
500%
2V
1.995 V
Larger than
Equal to
0.04%
0%
5002 
833 
Larger than
Smaller than
0.04%
500%
4
= IR = 3  1 = 3 V
2
D
3
(a) No. They should be equal as they are in
series.
(b) Yes. V0 is the total voltage across R and
the ammeter.
(c) No. As V0 is greater than the voltage
For circuit A:
across R, and I0 is equal to the current
V
through R, the ratio 0 is greater than the
I0
(i)
actual resistance of R.
Measured current
2
=
= 3.998  104 A
5000  2
(ii) Measured current is exact since the
(d) The arrangement is suitable for measuring
ammeter and resistor are connected
high resistance as the undesired effect of
in series.
the resistance of the ammeter is much
(iii) Percentage error of current = 0
smaller.
New Senior Secondary Physics at Work
5
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4
Electricity and Magnetism
Chapter 2 Electric Circuit
(iv) Measured voltage
Percentage error of current
= e.m.f of the battery = 2 V
=
2.394 10 3  3.990 10 4
3.990 10  4
(v) The voltmeter measures the total
= 500%
voltage across the ammeter and the
(iv) Measured voltage = 1.995 V (see part
resistor, so the measured voltage is
(iii))
larger than the true value.
(v) The measured voltage across the
(vi) Actual voltage
5000
=2
5000  2
resistor is exact since the voltmeter is
connected in parallel with the
= 1.9992 V
resistor.
Percentage error of voltage
2  1.9992
=
 100 % = 0.04%
1.9992
(vi) Percentage error of voltage = 0
(vii) Calculated resistance
1 
 1
=


5000
1000


(vii) Calculated resistance
= 5000 + 2 = 5002 
(ix) Percentage error of resistance
5000  833
=
 100 % = 500%
833
(ix) Percentage error of resistance
5002  5000
 100% = 0.04%
5000
(b) By comparing the percentage error of the
For circuit B:
resistance in circuit A and circuit B, it is
Equivalent resistance of the circuit
1 
 1

= 2 +

 1000 5000 
= 833 
the true value (5000 ).
the true value (5000 ).
(i)
1
(viii)The calculated value is smaller than
(viii) The calculated value is larger than
=
 100 %
1
found that circuit A is a better way to
= 835.3 
measure large resistance.
Measured current
2
=
= 2.394  103 A
835 .3
Revision exercise 2
Multiple-choice (p. 100)
1
B
current through the resistor and the
2
C
voltmeter, so the measured current is
3
B
larger than the true value.
4
C
5
B
6
C
7
(HKCEE 2005 Paper II Q39)
8
(HKCEE 2005 Paper II Q40)
9
(HKCEE 2005 Paper II Q41)
10
(HKCEE 2007 Paper II Q20)
11
(HKCEE 2007 Paper II Q23)
(ii) The ammeter measures the total
(iii) Measured voltage across the resistor
 833 .3 
= 2
 = 1.995 V
 835 .3 
Actual current across the resistor
1.995
=
= 3.990  104 A
5000
New Senior Secondary Physics at Work
6
 Oxford University Press 2010
4
Electricity and Magnetism
Chapter 2 Electric Circuit
Conventional (p. 102)
if the diameter of the wire decreases, the
1
current flowing through the wire, i.e. the
(a) Resistivity is the strength of an ohmic
conductor to oppose the flow of electric
amount of charge passing through the wire
current at a certain temperature.
per unit time, decreases (V = IR).
(b) Resistance of the copper wire
l
=
A
1.7 10 8  0.15
=
π  (0.2 10 3 ) 2
= 0.0203 
2
(a) (i)
(1A)
4
equivalent resistance will be higher than
the resistance of X or that of Y alone.(1A)
The V–I graph of the combined resistor
(1A)
will lie in region K.
(1A)
(c) If X and Y are connected in parallel, the
(1A)
equivalent resistance will be lower than
(ii) The resistance of the eureka wire
increases.
(1A)
(b) If X and Y are connected in series, the
(1M)
The resistance of the eureka wire
remains unchanged.
(a) Resistor Y has a lower resistance.
the resistance of X or that of Y alone.(1A)
(1A)
The V–I graph of the combined resistor
(b) The resistance increases and proportional
will lie in region M.
relation is not obeyed when the current
5
(1A)
(a)
through the wire is high enough to heat up
the wire sufficiently.
3
(1A)
(a) More charge passes the wire every second.
(1A)
(1A)
By V = IR, when the voltage increases, the
current flowing through the wire, i.e. the
(b)
amount of charge passing through the wire
per unit time, increases.
(1A)
(b) Less charge passes the wire every second.
(1A)
(1A)
(c)
Since the resistance of the wire increases
with its length,
(1A)
if the length of the wire increases, the
(1A)
current flowing through the wire, i.e. the
(d)
amount of charge passing through the wire
per unit time, decreases (V = IR).
(1A)
(c) Less charge passes the wire every second.
6
(1A)
(a) Equivalent resistance between AB
Since the resistance of the wire increases
1
 1
 
=
 3  4 1 8 
with decreasing diameter,
=4
New Senior Secondary Physics at Work
(1A)
7
1
(1M)
(1A)
 Oxford University Press 2010
4
Electricity and Magnetism
Chapter 2 Electric Circuit
(b) Equivalent resistance of the whole resistor
8
network
cable and an earthed object, a short circuit
= 5 + 4 + 16
(1M)
is formed.
= 25 
(1A)
The voltage at the power cable is very
(c) Voltage across parallel branches is the
(1A)
high, so a very large current would flow
same.
from the cable to the object through the
Voltage across the 8- resistor between
metallic balloon.
AB
This overheats the cable and causes
equivalent resistance between AB
=
equivalent resistance of the network
electricity failure.
 e.m.f. of the battery
4
=  12 .5
25
=2V
V
(d) I =
R
2
= = 0.25 A
8
7
(a) If a metallic balloon is in touch with a live
(b) Plastic is an insulating material.
(1M)
(1A)
(1A)
(1A)
The plastic shoes prevent current from
flowing through the electricians to the
earth even if they touch high-voltage
(1A)
power cables accidentally.
(1M)
9
(1A)
(a) A bird standing on a power transmission
cable will not get an electric shock. (1A)
(1A)
This is because the potential difference
The current passing through 8- resistor
between the points where the bird stands
between AB is 0.25 A.
is small.
(1A)
When S is open, current of 1 A passes the 4-
By V = IR, the current passing the body of
resistor and R2. The voltage across R2 is 8 V.
V
R2 =
(1M)
I
8
=
=8
(1A)
1
the bird is very small and the bird will not
get an electric shock.
(b) When a kite is entangled with a
high-voltage power transmission cable,
the potential difference between the cable
When S is closed, total current drawn from the
and the earth, where the person stands, is
battery is 1.5 A. The voltage across R2 is 6 V.
Equivalent resistance of R1 and R2
6
V
= =
=4
I 1 .5
(1A)
huge.
(1A)
By V = IR, the current passing the body of
(1M)
the person is very large and may kill that
Consider the equivalent resistance of R1 and
person.
R2
If the wire of the kite touches two power
1
transmission cables at the same time, it
 1
1 
 

R R  =4
2
 1
can cause short-circuit and result in
disastrous effects.
1
 1 1
   =4
 R 8
 1

 R1 = 8 
New Senior Secondary Physics at Work
(1A)
(1A)
(1A)
8
 Oxford University Press 2010
4
Electricity and Magnetism
Chapter 2 Electric Circuit
The voltage measured in circuit (i) is
10 (a)
correct but the current measured is
incorrect, since it measures the current
through both the resistor and the voltmeter.
(1A)
The resistance of the resistor and the
internal resistance of the voltmeter are
comparable, so the measured current is
much larger than the actual one, leading to
(Correct connection of ammeter.)
(1A)
a low accuracy for the value of R.
(Correct connection of voltmeter.)
(1A)
The current measured in circuit (ii) is
(1A)
Measure the current I through the wire by
correct but the voltage measured is
the ammeter,
incorrect, since it measures the voltage
(1A)
when a known voltage V is applied across
the wire.
across both the resistor and the ammeter.
(1A)
(1A)
The resistance is calculated using the
The resistance of the resistor is much
formula R = V/I.
larger than the internal resistance of the
(1A)
(b) The resistance of a wire of uniform
ammeter, so the measured voltage is only
cross-sectional area and its length are in
slightly larger than the actual one, leading
direct proportion.
to a high accuracy for the value of R.(1A)
(c) (i)
(1A)
Wire P has higher resistance.
(1A)
(c) Circuit (i) should be used.
(ii) Wire Q is thicker.
(1A)
l
For a wire, its resistance R  ,
A
This is because the resistance of the
unknown resistor is much smaller than
that of the voltmeter and only negligible
where l is its length and A is its
cross-sectional area.
amount of current would pass the
(1A)
voltmeter.
Since Q always has a smaller
circuit (i): R =
circuit (ii): R =
circuit (i) would be close to the actual
(1A)
current passing the unknown resistor.(1A)
(1M)
12
24  10
12
3
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
voltage across them would also be
= 1000  (1A)
comparable and the voltage measured
(b) Circuit (ii) gives a more accurate result.
would be much larger than the actual
(1A)
New Senior Secondary Physics at Work
(1A)
Then the readings of the ammeter in
resistance than P of the same length,
Q is thicker.
V
11 (a) By R = ,
I
(1A)
voltage across the unknown resistor. (1A)
9
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4
Electricity and Magnetism
Chapter 2 Electric Circuit
12 (a) An ammeter is connected in series with a
circuit component.
for the connection in parallel:
R
10 = 1  b
x
(1A)
The total resistance of the circuit is the
Rb = 10x
sum of the resistance of the ammeter and
……(2)
(1M)
the equivalent resistance of other circuit
components.
Substituting (1) into (2), we have:
1000
= 10x
x
(1A)
If the ammeter has high resistance, the
total resistance of the circuit will be
x = 10
significantly increased and the current
The number of bulbs is 10.
flowing in the circuit will be significantly
reduced.
(1A)
14 (a)
(1A)
This greatly affects the circuit.
(b) A voltmeter is connected in parallel with a
circuit component.
(1A)
The equivalent resistance of the
component and the voltmeter is smaller
than that of any of them alone.
(1A)
If the voltmeter has low resistance, the
equivalent resistance will be much smaller
than the resistance of the circuit
component and the voltage across the
component will be significantly reduced.
(Correct connection of ammeter: ‘+’ and
(1A)
‘’ terminals.)
This greatly affects the circuit.
13
(Correct connection of voltmeter: ‘+’ and
Let x be the number of bulbs; Rb be the
‘’ terminals.)
resistance of each bulb.
(2A)
(Correct connection with 5 wires.) (1A)
V
(b) R = , where R, V and I are resistance of
I
The equivalent resistance of the circuit when
the bulbs are connected in series = xRb (1A)
The equivalent resistance of the circuit whenf
R
the bulbs are connected in parallel = b (1A)
x
the resistor, voltmeter reading and
ammeter reading respectively.
(1A)
(c) The resistance of an ammeter should be
By V = IR,
very small while that of a voltmeter
for the connection in series:
should be very large.
10 = 0.01  xRb
1000
Rb =
……(1)
x
New Senior Secondary Physics at Work
(2A)
10
(1A)
 Oxford University Press 2010
4
Electricity and Magnetism
Chapter 2 Electric Circuit
(d) The experimental value will be lower than
the actual value.
(b) (i)
If a filament breaks, the current can
(1A)
pass through the resistor connected in
If the resistance of X is comparable to that
parallel to the filament and the circuit
of the voltmeter, the current measured by
is still complete.
the ammeter is much larger than the actual
(1A)
(ii) The other lamps will be much
current passing X as a large fraction of the
dimmer.
current measured passes the voltmeter.
Since the lamps are connected in
(1A)
(1A)
series and the resistance of R is much
Besides, the voltage measured is equal to
larger than that of a filament, the
the voltage across X.
V
By R = ,
I
(1A)
voltage across each filament is much
(1A)
smaller than that across the resistor
of the broken lamp. Therefore, the
the experimental value is lower than the
lamps will be very dim.
actual resistance of X.
(e)
16
(HKALE 2001 Paper I Q8)
17
(HKCEE 2005 Paper I Q9)
18
(HKALE 2005 Paper II Essay Q3)
19
(HKCEE 2006 Paper I Q11)
(1A)
Physics in articles (p. 107)
(a) He is incorrect.
(1A)
Even if a piece of metal is not connected to a
battery, free electrons inside the metal move
rapidly.
(Correct connection of ammeter.)
(1A)
(Correct connection of voltmeter.)
(1A)
15 (a) (i)
Since free electrons collide with positive ions
inside the metal, they change their moving
directions and their overall displacement, not
Since the lamps are connected in
distance travelled, is zero.
series, the circuit will be broken if
one of the lamps breaks.
(ii) The light dims.
(b) They are opposite.
(1A)
(1A)
metal, electrons are accelerated by the electric
field and gain kinetic energy.
total resistance of the circuit
(1A)
Then electrons transfer the kinetic energy
increases and the voltage across each
gained to ions in collisions. This increases the
(1A)
internal energy of the metal and produces the
This makes the lamps dimmer.
heating effect.
New Senior Secondary Physics at Work
(1A)
(c) When a current passes through a piece of
(1A)
(iii) When more lamps are connected, the
lamp decreases.
(1A)
11
(1A)
 Oxford University Press 2010