4 Electricity and MagnetismChapter 2 Electric Circuit
2Electric Circuit
New Senior Secondary Physics at Work1OxfordUniversity Press 2010
4 Electricity and MagnetismChapter 2 Electric Circuit
Practice2.1 (p.54)
1(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 terminal of the battery.
2D
3(a)The current is doubled.
(b)The current is halved.
4Q= It = (1800 103) 60 60 = 6480 C
6480 C of charge can be driven to flow through the battery.
5(a)The reading is 0.49 A. The full-scale reading is 1A.
(b)The reading is 67 mA. The full-scale reading is 100 mA.
6(a)(b)
(c)
7Current === 0.2 A
8Q = It= 20 103 0.1 = 2 103 C
2 103 C of charge passes through the cow.
Since 1C of charge is equal to 6 1018 electrons,
number of electrons passing through the cow
= 2 103 6 1018= 1.2 1016
Practice2.2 (p.62)
1B
2A
3C
4Potential difference === 0.1 V
5In 1 minute, by V =,
Q === 120 C
The amount of charge flows through the 3-V battery in 1 minute is 120 C.
6The 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.
7By V =,
Q === 10 C
Current === 100 A
8Q = 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.
9In 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
Therefore, arrangement in Fig c gives a higher total e.m.f.
Practice2.3 (p. 74)
1A
2A
3C
4By V = IR,
I === 0.25 A
The current through the resistor is 0.25 A.
5R === 73.3
The overall resistance of the computer is
73.3 .
6V= IR = 0.35 17 = 5.95 V
The MP3 player requires 5.95 V to operate.
Thereforeit needsAA dry cells.
7V = IR = 10 10–3 1 103 = 10 V
The voltage across the resistor is 10 V.
8(a)The current is halved.
(b)The current is halved.
9(a)(b)
Ohm’s law is not applicable beyond the point ‘’.
This is because the voltage across the wire is no longer directly proportional to the current passing through it.
(c)(i)R === 5
The resistance of the wire is 5 .
(ii)R=== 6
The resistance of the wire is 6 .
10(a)Theminimum voltage is 0.8 V.
(b)No, it does not obey Ohm’s law.
11(a) === 0.001 m
The resistivity of the wire is 0.001 m.
(b)Since Rl, we have=.
R2 == 2 5 = 10
The new resistance is 10 .
Practice2.4 (p. 83)
1A
When the switch is open, current passes through Y, so the equivalent resistance of the two bulbs increases. Hence the current through the circuit decreases.
2C
By V = IR,
voltage across 3- resistorV3- = 0.5 3
= 1.5 V
voltage across 1- resistor V1- = 0.5 1
= 0.5 V
Voltage across 2- resistor = 2 V
Current passing 2- resistor === 1 A
Total current in the circuit = 1.5 A
Voltage across R = 6 2 = 4 V
R === 2.67
3C
4I3=I8 ==6 A
I4 = I5 = I6 =I7==3 A
I1 = I2 = I9 =I10=6+3=9A
5(a)
R = 0.5
The equivalent resistance is 0.5 .
(b)=
R = 0.33
The equivalent resistance is 0.33 .
(c)R == 1.33
The equivalent resistance is 1.33 .
(d)R == 1.67
The equivalent resistance is 1.67.
(e)R == 2.5
The equivalent resistance is 2.5 .
6Since the bulbs are connected in series, the current passing them is the same.Besides, by
V = IR, their voltage ratio is RX : RY : RZ.
(a)(i)Current ratio = 1 : 1 : 1
(ii)Voltage ratio = 1 : 2 : 3
(b)(i)Current ratio = 1 : 1 : 1
(ii)Voltage ratio = 3 : 4 : 12
(c)(i)Current ratio = 1 : 1 : 1
(ii)Voltage ratio = RX :RY:RZ
7Since the bulbs are connected in parallel, the voltage across them is the same. Besides,by, the ratio of the current passing them is .
(a)(i)Current ratio = 6 : 3 : 2
(ii)Voltage ratio = 1 : 1 : 1
(b)(i)Current ratio = 4 : 3 : 1
(ii)Voltage ratio = 1 : 1 : 1
(c)(i)Current ratio =
(ii)Voltage ratio = 1 : 1 : 1
8(a)Total resistance in the circuit
= 2 + 10 = 12
I === 0.5 A
Current of 0.5 A passes the 2-light bulb.
(b)Since there is a shortcircuit,
total resistance in the circuit = 2
I === 3 A
Current of 3 A passes the 2-light bulb.
9(a)R = 2 + 2 + 2 = 6
The equivalent resistance is 6.
(b)
R= 1.5
The equivalent resistance is 1.5.
(c)R =3 ++ 3 = 9
The equivalent resistance is 9.
10(a)Current flowing through 5- resistor
=0.6 A
Voltage across PQ
= voltage across 5- resistor
= I5-R5- = 0.6 5 = 3 V
(b)Current flowing through 10- resistor
=== 0.3 A
The reading of ammeter A2 is 0.3 A.
(c)Current flowing through 2- resistor
= 0.6 + 0.3 = 0.9 A
Voltage across 2- resistor
= I2-R2- = 0.9 2 = 1.8 V
Voltage of the battery
= VPQ + V2- = 3 + 1.8 = 4.8 V
11(a)Equivalent resistance of the circuit
= 3 += 6
Current through 3- resistor
= current drawn from the battery
=== 1 A
Voltage across 3- resistor
I3-R3- = 1 3 = 3 V
Voltage across 4- / 12- resistor
= 6 3 = 3 V
Current through 4- resistor
=== 0.75 A
Current through 12- resistor
=== 0.25 A
(b)As shown in the calculation in (a), the voltage across each resistor is 3 V.
(c)Equivalent resistance of 4- and 12- resistors == 3
Voltage across all resistors are the same.
Voltage across 12- resistor
= voltage across 3- resistor
= 0.4 3 = 1.2 V
Reading of ammeter A2 == 0.1 A
12(a)(i)The reading of the voltmeter remains unchanged.
(ii)The reading of the ammeter A1 increases.
(iii)The reading of the ammeter A2 remains unchanged.
(b)The voltage across PQ remains unchanged.
13(a)The statement is incorrect. The current through all three bulbs connected in series is the same.
(b)The statement is correct. The voltage across each bulb connected in series is in the same ratio as their resistance.
(c)The statement is incorrect. If the equivalent resistance of three identical bulbs connected in series is 3 , the resistance of each bulb is 1 (1 + 1 + 1 = 3 ).
14(a)The statement is incorrect. The voltageacross every bulb connected in parallel is the same.
(b)The statement is correct. The current through each bulb connected in parallel is in the same ratio as the reciprocal of their resistance.
(c)The statement is incorrect. If the equivalent resistance of three identical bulbs connected in parallel is 3 , the resistance of each bulb is 9:
Practice2.5 (p. 95)
1D
As the ammeter which the resistance is zero is connected to the two resistors in parallel, the equivalent resistance across the two resistors and the ammeter is zero. The total resistance in the circuit is 1 .
Therefore, the total current === 3 A
The reading of the ammeter is 3 A.
Reading of the voltmeter
= voltage across the 1- resistor
= IR = 3 1 = 3 V
2D
3(a)No. They should be equal as they are in series.
(b)Yes. V0is thetotal voltage across R and the ammeter.
(c)No. As V0is greater than the voltage across R, and I0 is equal to the current through R, the ratiois greater than the actual resistance of R.
(d)The arrangement is suitable for measuring high resistance as the undesired effect of the resistance of the ammeter is much smaller.
4(a)Voltage across the internal resistance
=98.4=0.6V
Internal resistance of the battery
== 0.188
(b)Voltage across R = 4.0 2.0 = 8.0 V
Voltage across the internal resistance
=9 8.0 = 1.0 V
Internal resistance of the battery
== 0.25
(c)Current through R
== 1.87 A
Voltage across the internal resistance
=6 5.6 =0.4 V
Internal resistance of the battery
== 0.214
5(a)
A / B3.998 104 A / 2.394103 A
Equal to / Larger than
0% / 500%
2V / 1.995V
Larger than / Equal to
0.04% / 0%
5002 / 833
Larger than / Smaller than
0.04% / 500%
For circuitA:
(i)Measured current
== 3.998 104A
(ii)Measured current is exact since the ammeter and resistor are connected in series.
(iii)Percentage error of current = 0
(iv)Measured voltage
= e.m.f of the battery = 2 V
(v)The voltmeter measures the total voltage across the ammeter and the resistor, so the measured voltage is larger than the true value.
(vi)Actual voltage
= 2
= 1.9992 V
Percentage error of voltage
== 0.04%
(vii)Calculated resistance
= 5000 + 2 = 5002
(viii)The calculated value is larger than the true value (5000 ).
(ix)Percentage error of resistance
= 100% = 0.04%
For circuitB:
(i)Equivalent resistance of the circuit
= 2 += 835.3
Measured current
== 2.394 103 A
(ii)The ammeter measures the total current through the resistor and the voltmeter, so the measured current is larger than the true value.
(iii)Measured voltage across the resistor
== 1.995 V
Actual current across the resistor
== 3.990 104 A
Percentage error of current
=
= 500%
(iv)Measured voltage = 1.995 V (see part (iii))
(v)The measured voltage across the resistor is exact since the voltmeter is connected in parallel with the resistor.
(vi)Percentage error of voltage = 0
(vii)Calculated resistance
== 833
(viii)The calculated value is smaller than the true value (5000 ).
(ix)Percentage error of resistance
== 500%
(b)By comparing the percentage error of the resistance in circuit A and circuit B, it is found that circuit A is a better way to measure large resistance.
Revision exercise2
Multiple-choice (p. 100)
1B
2C
3B
4C
5B
6C
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)
Conventional (p. 102)
1(a)Resistivity is the strength of an ohmic conductor to oppose the flow of electric current at a certain temperature. (1A)
(b)Resistance of the copper wire
=(1M)
=
= 0.0203 (1A)
2(a)(i)The resistance of the eureka wire remains unchanged.(1A)
(ii)The resistance of the eureka wire increases.(1A)
(b)The resistance increases and proportional relation is not obeyed when the current through the wire is high enough to heat up the wire sufficiently. (1A)
3(a)More charge passes the wire every second.(1A)
By V = IR, when the voltage increases, the current flowing through the wire, i.e. the amount of charge passing through the wire per unit time, increases. (1A)
(b)Less charge passes the wire every second.(1A)
Since the resistance of the wire increases with its length, (1A)
if the length of the wire increases, the current 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 every second.(1A)
Since the resistance of the wire increases with decreasing diameter, (1A)
if the diameter of the wire decreases, the current flowing through the wire, i.e. the amount of charge passing through the wire per unit time, decreases (V = IR).
4(a)Resistor Y has a lower resistance.(1A)
(b)If X and Y are connected in series, the equivalent resistance will be higher than the resistance of X or that of Y alone. (1A)
The V–I graph of the combined resistor will lie in region K.(1A)
(c)If X and Y are connected in parallel, the equivalent resistance will be lower than the resistance of X or that of Y alone. (1A)
The V–I graph of the combined resistor will lie in region M.(1A)
5(a)
(1A)
(b)
(1A)
(c)
(1A)
(d)
(1A)
6(a)Equivalent resistance between AB
=(1M)
= 4 (1A)
(b)Equivalent resistance of the whole resistor network
= 5 + 4 + 16(1M)
= 25 (1A)
(c)Voltage across parallel branches is the same.
Voltage across the 8-resistor between AB
=
e.m.f. of the battery(1M)
=
= 2 V(1A)
(d)I =(1M)
== 0.25 A(1A)
The current passing through 8-resistor between AB is 0.25 A.
7When S is open, current of 1 A passes the 4- resistor and R2. The voltage across R2 is 8 V.
R2 =(1M)
= = 8 (1A)
When S is closed, total current drawn from the battery is 1.5 A. The voltage across R2 is 6 V.
Equivalent resistance of R1 and R2
=== 4 (1M)
Consider the equivalent resistance of R1 and R2
= 4
= 4
R1 = 8 (1A)
8(a)If a metallic balloon is in touch with a live cable and an earthed object, a short circuit is formed. (1A)
The voltage at the power cable is very high, so a very large current would flow from the cable to the object through the metallic balloon. (1A)
This overheats the cable and causes electricity failure.(1A)
(b)Plastic is an insulating material.(1A)
The plastic shoes prevent current from flowing through the electriciansto the earth even if they touch high-voltage power cables accidentally. (1A)
9(a)A bird standing on a power transmission cable will not get an electric shock.(1A)
This is because the potential differencebetween the points where the bird stands is small.(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 high-voltage power transmission cable, the potential difference between the cable and the earth, where the person stands, is huge. (1A)
By V = IR, the current passing the body of the person is very large and may kill that person.(1A)
If the wire of the kite touches two power transmission cables at the same time, it can cause short-circuit and result in disastrous effects. (1A)
10(a)
(Correct connection of ammeter.)(1A)
(Correct connection of voltmeter.)(1A)
Measure the current I through the wire by the ammeter,(1A)
when a known voltage V is applied across the wire.(1A)
The resistance is calculated using the formula R = V/I.(1A)
(b)The resistance of a wire of uniform cross-sectional area and its length are in direct proportion. (1A)
(c)(i)Wire Phas higher resistance. (1A)
(ii)Wire Q is thicker.(1A)
For a wire, its resistanceR, where l is its length and A is its cross-sectional area.(1A)
Since Q always has a smallerresistance than P of the same length, Q is thicker.(1A)
11(a)By R =,(1M)
circuit (i): R == 500 (1A)
circuit (ii): R == 1000 (1A)
(b)Circuit (ii) gives a more accurate result.(1A)
The voltage measured in circuit (i) is 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 a low accuracy for the value of R. (1A)
The current measured in circuit (ii) is correct but the voltage measured is incorrect, since it measures the voltage across both the resistor and the ammeter. (1A)
The resistance of the resistor is much larger than the internal resistance of the ammeter, so the measured voltage is only slightly larger than the actual one, leading to a high accuracy for the value of R. (1A)
(c)Circuit (i) should be used.(1A)
This is because the resistance of the unknown resistor is much smaller than that of the voltmeter and only negligible amount of current would pass the voltmeter. (1A)
Then the readings of the ammeter in circuit (i) would be close to the actual current passing the unknown resistor. (1A)
If circuit (ii) is used instead, since the resistance of the unknown resistor is comparable to that of the ammeter, the voltage across them would also be comparable and the voltage measured would be much larger than the actual voltage across the unknown resistor. (1A)
12(a)An ammeter is connected in series with a circuit component.(1A)
The total resistance of the circuit is the sum of the resistance of the ammeter and the equivalent resistance of other circuit components. (1A)
If the ammeter has high resistance, the total resistance of the circuit will be significantly increased and the current flowing in the circuit will be significantly reduced. (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 voltageacross the component will be significantly reduced. (1A)
This greatly affects the circuit.
13Let x be the number of bulbs; Rb be the resistance of each bulb.
The equivalent resistance of the circuit when the bulbs are connected in series = xRb(1A)
The equivalent resistance of the circuit whenf the bulbs are connected in parallel =(1A)
By V = IR,
for the connection in series:
10 = 0.01 xRb
Rb= ……(1)
for the connection in parallel:
10 = 1
Rb=10x ……(2)
(1M)
Substituting (1) into (2), we have:
= 10x
x = 10
The number of bulbs is 10.(1A)
14(a)
(Correct connection of ammeter: ‘+’ and ‘’ terminals.)(2A)
(Correct connection of voltmeter: ‘+’ and ‘’ terminals.)(2A)
(Correct connection with 5 wires.)(1A)
(b)R =, where R, V and I are resistance of the resistor, voltmeter reading and ammeter reading respectively. (1A)
(c)The resistance of an ammeter should be very small while that of a voltmeter should be very large. (1A)
(d)The experimental value will be lower than the actual value.(1A)
If the resistance of X is comparable to that of the voltmeter, the current measured by the ammeter is much larger than the actual current passing X as a large fraction of the current measured passes the voltmeter. (1A)
Besides, the voltage measured is equal to the voltage across X.(1A)
By R =,(1A)
the experimental value is lower than the actual resistance of X.
(e)
(Correct connection of ammeter.)(1A)
(Correct connection of voltmeter.)(1A)
15(a)(i)Since the lamps are connected in series, the circuit will be broken if one of the lamps breaks. (1A)
(ii)The light dims.(1A)
(iii)When more lamps are connected, the total resistance of the circuit increases and the voltage across each lamp decreases. (1A)
This makes the lamps dimmer.
(b)(i)If a filament breaks, the current can pass through the resistor connected in parallel to the filament and the circuit is still complete. (1A)
(ii)The other lamps will be much dimmer.(1A)
Since the lamps are connected in series and the resistance of R is much larger than that of a filament, the voltage across each filament is much smaller than that across the resistor of the broken lamp. Therefore, the lamps will be very dim. (1A)
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)
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. (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. (1A)
(b)They are opposite.(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. (1A)
New Senior Secondary Physics at Work1OxfordUniversity Press 2010