After completing this chapter, students should be able to:
1. state that current is a rate of flow of charge and that it is measured in amperes 2. distinguish between conventional current and electron flow 3. recall and apply the relationship charge = current x time to new situations or to solve related
problems 4. define electromotive force (e.m.f.) as the work done by a source in driving a unit charge around a
complete circuit 5. state that the e.m.f. of a source and the potential difference (p.d.) across a circuit component is
measured in volts 6. define the p.d. across a component in a circuit as the work done to drive a unit charge through the
component 7. state the definition that resistance = p.d. -;- current 8. apply the relationship R = V -;- I to new situations or to solve related problems 9. describe an experiment to determine the resistance of a metallic conductor using a voltmeter and an
ammeter, and make the necessary calculations 10. recall and apply the formulae for the effective resistance of a number of resistors in series and in
parallel to new situations or to solve related problems 11. recall and apply the relationship of the proportionality between resistance and the length and cross
sectional area of a wire to new situations or to solve related problems
16.1 Conventional Current and Electron Flow page 251 1. Emphasise that a current is a "loop flow" of charge which is the mechanism for energy transfer.
2. At this stage define the ampere as the coulomb/second rather that the more accurate version.
3. Explain that a charged battery stores a certain amount of charge. For example 20 Ah can supply 20 A for 1 hand 10 A for 2 h etc. The charge it stores is given by Q = 20 Ah = 20 A x 1 h = 20 A x 3600 s = 72 000 C
4 . Explain that an ammeter must be connected in series to the circuit and describe the use of an ammeter with different ranges.
Answer to Think Time question page 251
When charges move, it gives rise to a current
Answers to Section Review questions page 253
1. An electric current is the rate of flow of charge.
2. Current = charge = 4 C = 0.8 A time 5s
3. Total amount of charge = 1600 rnA x l h = 1600 X 10-3 A x 3600 s = 5760 C
16.2 Electromotive Force page 254 1. Explain that e.mJ. is the property of a source of electrical energy, include: cells, generators .
2. Explain that e.mJ. is measured by the work done by a source in driving a unit charge around a complete circuit.
3. Analogy for the understanding of e.mJ.: Coal trucks travel in a continuous loop from the coal mine to the power station and travel back empty. There is a fixed amount of coal per lorry. The truck is the charge and the coal it carries is the energy.
4. Calculate the total e.m.f. where several sources are arranged in series and discuss how this is used in the design of batteries.
5. Discuss the advantage of making a battery from several equal voltage sources of e.m.f. arranged in parallel. The cells will run longer and able to supply higher current.
Answers to Section Review questions page 255
1. The electromotive force of an electrical source is the work done by the source in driving a unit charge around a complete circuit.
2. The electromotive force can be measured by connecting a voltmeter directly across the terminals of the cell when the cell is not connected to any other components.
16.3 Potential Difference page 255 1. Potential difference concerns with where the energy ends up . It is difference from e.mJ. E.m.f.
concerns with where the energy comes from.
2. Potential difference across a component in a circuit is the work done in the component when one unit charge passed through the component.
3. Explain that a voltmeter must be connected across the component and describe the use of a voltmeter with different ranges.
Answer to Think Time question page 255
The bulb will not light up because the cell is the source of energy.
Answers to Section Review questions page 257
1. The potential difference across a component in a circuit is the work done to drive a unit charge through the component.
.. work done 20 J 2. Potent:J.al difference = = - - = 10 V
charge 2C 3. Kinetic energy = work done = QV
= 1.6 X 10-19 C x 500 V = 8.0 X 10- 17 J
16.4 Resistance page 257 1. Emphasise to students that resistance is the ratio of potential difference across the component to the
current flowing through it. The gradient of potential difference against current graph does not give resistance except in the case when potential difference is directly proportional to current (when Ohm's law is obeyed).
2. To understand factors affecting resistance: (a) a thick wire is just many thin ones laid side by side, so current is easier to flow and resistance
decreases. (b) a long wire is just many short ones laid end to end, so current is more difficult to flow and
resistance increases. (c) a hot wire has its molecules moving with greater amplitude in the wire which in tum make it
difficult for the electrons to flow, so resistance increases.
3. When filament bulbs blow, why is it happened when they are switched on?
[Answer: When the bulbs are switched on, the resistance is lowest then, so the current is having highest value.]
4. Discuss the advantage of using wire of high resistivity. The high resistivity enables us to use a short length of wire to give us the required resistance.
5. Emphasise to students that when the component obeys Ohm's law, its resistance is constant.
Answer to Think Time question page 258
The resistance decreases.
Answer to Think Time question page 260
When diameter is doubled, the area is four times, so the resistance is !i . 4
Answer to Think Time question page 263
Two identical resistors in parallel, effective resistance = !i 2
Three identical resistors in parallel, effective resistance = R 3
n identical resistors in parallel, effective resistance = !i n
Answers to Section Review questions page 265
1. R=~=lOV = 5n I 2A
2. (a) R ex I
3.
When the length is doubled, the resistance is doubled. SoR =40 Q
1 (b) Rex -
A
(c)
When the cross-sectional area is doubled, the resistance is halved. SoR= 10Q
In the second case, large current flows through the filament, increasing its temperature and resistance.
Physics in Society: Solid State Physics page 266 Answers to Q
1. Transistors enabled the development of integrated circuits on silicon chips and this leads subsequently to the development of powerful computer systems and initiated the computer revolution.
2. No heat will be generated when a current flow through a material with zero resistance. There is no necessity to cool the computer system. This leads to greater efficiency.
3. Much of the automated systems and machines that we take for granted today will not be available.
4. Yes. In some sectors the computer system replaces manual labour leading to unemployment. However, the technological revolution can also create new job opportunities in society.
Answers to Misconception Analysis page 268 1. True 2. True 3. False. The unit should be joules per coulomb. 4. False. E.mJ. is defined as the work done by the source in driving a unit charge through the whole
circuit. 5. 6. 7. 8.
True True False. False.
The potential difference across both resistors is the same. The resistance is halved.
Answers to Multiple Choice Questions page 268, 269 1. D 2. D Q = It = 3 A x 60 s = 180 C 3. A
4. B E 10 J
v= Q = 2 C = 5.0 V
5. A V 2.5 V
R = [= 5.0 A = 0.5 n
6. Rx Lx Ay 1 2 1
C -= - x -=- x -= -Ry Ly Ax 2 1 1
7. 1 1 1
D "8 = R + R ; therefore R = 16 n
Connected in series, effective resistance = (16 n +16 n) = 32 n 8. A 10 n connected in parallel with a wire gives an effective resistance of 0 n.
The two 20 .Q resistors connected in parallel gives an effective resistance of 10 n.
Answers to Structured Questions page 269, 270 l. (a) series - Wand Y parallel- X and Z
(c) connection Y: E = (1.5 + 1.5) V = 3.0 V connection Z: E = 1.5 V
Science in Focus: Physics '0 ' Level
2. . Q 12 C
(a) (1) 1= - = - = 4 A t 3 s
(b) (i) Q = It = 4 A x 8 s = 32 C
3. 1= V = 6V = 2A (a) R 3 Q
(c) V = IR = 2 A x 6 Q = 12 V
4. (a) 80
(ii) 1= 360 C = 3 A 2x60s
(ii) Q = 4 A x (2 x 60) s = 480 C
V 3.0V (b)R = - =--= 6Q
I 0.5A
80
Chapter 16 Current of Electricity
-~-40
(b) 180
20 60 20 80
90 (c)
20
5. R ex l and 1
Roc -A
So R = pl or R2 = l2AI = l2(d l )2 A RI l lA2 II (d2)2
R 3(1.2)2 (a) 1.8 Q = 2(1.2)2
R = 2.7 Q
b R _ 2(1.2)2 ( ) 1.8 Q - 2(2.4)2
R = 0.45 Q
R 5(1.2)2 (c) 1.8 Q = 2(0.6)2
R = 18 Q
6. (a) 1= ~ = 12V = 2.4 rnA R 5ill
--- --c=J----c=J- --- ---c=:J--
d 2
where A = 11: -4
(b) Q = It = 2.4 x 10 - 3 A x 0 0 x 60) s = 1.44 C
(c) E = QV = 1.44Cx 12V = 17.3J
Answers to Critical Thinking Questions page 270 1. This is possible because the number of electrons involved in typical electric circuit is extremely large.
2. It indicates the amount of charge. 800 mAh means the cell can supply 800 rnA for 1 h, 400 rnA for 2 h and 80 rnA for 10 h etc. The charge it stores is given by
Q = 800 mAh = 800 rnA x 1 h = 800 X 10-3 A x 3600 s = 2880 C
3. The number of cars corresponds to the charge Q and the number of cars moving per second corresponds to the current.
Extension page 270 1. Some examples:
• Late to school because MRT service affected. • Traffic lights not working results in massive traffic jams during the rush hour. • No light, fan or air conditioner in the class room. • Not safe to go out at night because the whole housing estate will be dark.
• The lift is not working, so have to climb up and down the staircase in high rise buildings.
• No entertainment at home because television not working.
• No access to the Internet because the computer is not working. • You can still use battery operated transistor radio because television and radio stations mostly
remained on the air with the help of backup generators.
• No entertainment outside because shops and shopping centres all closed down (no lights, no air
con) and no electricity for cooking.
• Have to throwaway food like fish and meat in the refrigerator because they are no longer kept at a low temperature.
• Can only pay by cash when buying things. Cash registers and ATM machines not working.
• When the black out started, fixed line phone network will be overload by people calling home. Hand phone will experience service disruptions as transmission towers were overloaded with the
sudden increase in volume of calls.
• Water supply system loses pressure because electric pump not working.
2. The resistivity of a conductor decreases with temperature but there is always a lower limit.
Superconductors are materials that have zero resistivity below a critical temperature. The following website has a good list of references and external links:
http://en. wikipedia. orglwiki/Supercond uctor
Uses of superconductors: Magnetic Levitation Trains, Magnetic Resonance Imaging, particles
accelerators used in high-energy particle research, transformers, fault limiters and power cables Refer to Websites: http://www.superconductors.org!Uses.htm
