Homework answers / question archive / G12 Physics: 2020-21 Assessed report (5% of GED grade) For this academic year, students will write a report that summarises one application of physics, related to a particular topic

G12 Physics: 2020-21 Assessed report (5% of GED grade) For this academic year, students will write a report that summarises one application of physics, related to a particular topic

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G12 Physics: 2020-21

Assessed report (5% of GED grade)

For this academic year, students will write a report that summarises one application of physics, related to a particular topic.

To write the report:

a) follow the guidance provided by: (a) briefing document; (b) extract from Ministry of Education assessment handbook (pages 29 & 50); (c) video to give overview of report & explain referencing;

b) choose an application that interests you, from a physics topic with which you are familiar;

c) read articles, reports, descriptions of this application from various sources (to get good information, we suggest 3-6 different sources);

d) write a well-structured original report that clearly explains the following points:

• What is the application, about which you are reporting - your explanation of the application should be both detailed and succinct!

• Why this application is of particular interest to you

• Why this application is important to the scientific community and/or the general public

• What you have learned, in the course of doing your research and writing the report

e) you MUST use referencing for any information that is not your own! Use the APA referencing format for declaring your sources (it is absolutely alright to use sources for your information; but opinions and evaluations of the information should be your own. Whenever giving the opinions of other people, make sure these are all correctly cited (referenced in text and in bibliography).

the report should be in word 500 words to 600words in three pages

check the file below for more info

the topic is attached in the picture belowG12 Physics: 2020-21 Assessed report (5% of GED grade) For this academic year, students will write a report that summarises one application of physics, related to a particular topic. To write the report: a) follow the guidance provided by: (a) briefing document; (b) extract from Ministry of Education assessment handbook (pages 29 & 50); (c) video to give overview of report & explain referencing; b) choose an application that interests you, from a physics topic with which you are familiar; c) read articles, reports, descriptions of this application from various sources (to get good information, we suggest 3-6 different sources); d) write a well-structured original report that clearly explains the following points: • What is the application, about which you are reporting - your explanation of the application should be both detailed and succinct! • Why this application is of particular interest to you • Why this application is important to the scientific community and/or the general public • What you have learned, in the course of doing your research and writing the report e) you MUST use referencing for any information that is not your own! Use the APA referencing format for declaring your sources (it is absolutely alright to use sources for your information; but opinions and evaluations of the information should be your own. Whenever giving the opinions of other people, make sure these are all correctly cited (referenced in text and in bibliography). Formatting: • • • Maximum 3 pages for report, and 1 further page for references. Use font-12 for text. Any diagrams should be labelled: Figure 1; Figure 2; etc., with a caption that briefly describes what the diagram shows. Diagrams should be referred to within the report text: e.g. “… (Figure 1)”. Marking: Marking will be done according to the criteria in the MoE document (1 mark for each criteria) Any questions about these notes or requirements, or clarifications needed - please do ask. Wishing you a successful - and interesting - experience of writing the assessed report! Structure and detail required for Report Document: Report Title: .............................................. Student’s name: ................................................... Class:........... Date:....................... Introduction This section should contain the personal interest of this topic, theoretical background and literature review. A brief introduction about the Physics topic along with its relevance with your syllabus ( unit, outcome number and statement) . You should list down all applications of the physics outcomes and then state what application you have chosen for this report. Main Body In this section, you need to explain the topic in detail. If your topic is about a process, then you should explain every single reaction in detail starting from the basic information. If this report requires any circuit diagram. Then you should draw a neat and clean circuit diagram. All derivations and calculations must be clearly presented. Your main body can be presented more effectively with the use of diagrams, tables, flow charts of the processes, reaction sequences and/or any other graphical representations. Conclusion Your conclusion should reflect your opinion about the topic. It should summarise the link of Physics knowledge to that application. Any limitations or extension of the research topic can be included in order to show your enthusiasm for the topic. Possible extension ⇒ You can talk about other chemistry topics related to this title. References Use APA style referencing, a helpful tool is attached to google doc Possible an application of resistors in electricl circuit. Regards h Add private comment 20.2 Kirchhoff's first and second day | It should be remembered that both electromotive force and potential difference hav direction. This must be considered when working out the equation for Kirchhoff's second law. For example, in the circuit of Figure 2013, two cells have been connected in opposition Here the total clectromotive force in the circuit is - Ey, and I Kirchhoff's second law -= V + VIR + IR Resistors in series Figure 2011 shows two resistors of resistances and Reconnected in series, and a single resistor of resistance Requivalent to them. The current in the resistors, and in their equivalent single resistor, is the same 1 1 Figure 2013 R R R equivalent to Figure 20.14 Resistors in series The total potential difference across the two resistors must be the same as that across the single resistor. If Vand V are the potential differences across each resistor V=V, V But since potential difference is given by multiplying the current by the resistance, IR IR, IR Dividing by the current / RR,R This equation can be extended so that the equivalent resistance R of several resistors connected in series is given by the expression R=R1+RE+R+... Thus The combined resistance of resistors in series is the sum of all the individual resistances Resistors in parallel New consider two resistors of resistance Rand R, connected in parallel, as shown in Figure 20.15. The current through each will be different, but they will each have the same potential dilference. The equivalent single resistor of resistance will have the same potential difference across it, but the current will be the total current through the separate resistors R equivalent to Figure 20.15 Resistors in parallel 161 9:24 Done 5 of 5 Retake lly Kind her first low and using resistance.d/current, so WRW WR Dividing by the potential difference 1/R=1/R + 1/R This equation can be extended that the equivalent rettance of everal resist inted in parallel stiven by 1 RA TI The reciprocal of the combined resistance of resistors in paralel is the sum of the reciproci of all the individual resistances Note that 1 For two identical resistors in parallel, the combined resistance is equal to half of the va of each one 2 For resistors in parate the combined resistance is always less than the value of the smallest individual resistance Example 30 30 30 322 Calculate the equivalent resistance of the arrangement of resistors in Figure 2016 The arrangement is equivalent to two resistors in parallel, so the combined resistance is given by 1-1/ 61/ 62/6: Don't forget to find the reciprocal of this value) Thus 30 Now it's your turn 6 Calculate the equivalent resistance of the arrangement of resistors in Figure 20.17 first find the resistance of the parallel combination 7 Calculate the effective resistance between the points A and B in the network in Figure 2018 2022 Figure 20.16 40 352 222 100 100 Figure 20.17 50 Figure 20.18 20.3 Potential dividers and potentiometers Thoroscomed in seties with a cell cach teve a potential difference may be wed to divide the end of the cell . This is illustrated in Pugure 2012 N The current in the theme, because they are in seriesThus Devinst the first equation by the secondes The D 4 EP AS Level 20 D.C. circuits @ centre panom Figure 20.1 Creu By the end of this topic, you will be able to: (c) derive, using Kirchhoff's laws, a formula for the 20.1 (a) recall and use appropriate circuit symbols combined resistance of two or more resistors in (b) draw and interpret circuit diagrams containing series sources, switches, resistors, ammeters and (d) solve problems using the formula for the voltmeters and/or any other type of component combined resistance of two or more resistors in referred to in the syllabus (c) define electromotive force (e.m.f) in terms of the series (e) derive, using Kirchhoff's laws, a formula for the energy transferred by a source in driving unit combined resistance of two or more resistors in charge round a complete circuit (1) distinguish between e.m.f. and potential parallel difference (p.d) in terms of energy considerations (1) solve problems using the formula for the (e) understand the effects of the internal resistance combined resistance of two or more resistors in of source of e.m.f. on the terminal potential difference (g) apply Kirchhoff's laws to solve simple circuit 20.2 (a) recall Kirchhoff's first law and appreciate the link problems to conservation of charge 20.3 (a) understand the principle of a potential divider (b) recall Kirchhoff's second law and appreciate the circuit as a source of variable p.d. link to conservation of energy (b) recall and solve problems using the principle of the potentiometer as a means of comparing potential differences parallel Starting points • Basic knowledge of appropriate circuit symbols. . Methods for drawing and interpreting circuit diagrams . Electric current is the rate of flow of charge. • A potential difference is required to provide energy to move charge through a device 20.1 Electrical circuits When reporting an electrical experiment, or describing a circuit, it is essential to know exactly how the components are connected. This could be done by taking a photograph, but this technique has disadvantages; the photograph in Figure 194, for example, is not clear and does not show all the components. You could sketch a block diagram, in which the components are indicated as rectangular boxes labelled cell ammeter', 'resistor', etc. The blocks would then be connected with lines to indicate the wiring. This is also unsatisfactory; it takes a lot of time to label all the boxes. It is much better to draw the circuit diagram using a set of symbols that is recognised by everyone and which do not need to be labelled. Figure 20.1 shows the symbols that you are likely to need in school and college work, and which you will meet in examination questions. (You will have met many of them already) it is important that you learn these so that you can recognise them straight away. The only labels you are likely to see on them will be the values of the components, for example 15V for a cell or 220 for a resistor. 156 20.1 Electrical circuits cell battery switch earth relay lamp fixed resistor 4 variable resistor light-dependent resistor thermistor diode capacitor ammeter voltmeter -the sin galvanometer centre-zero galvanometer transformer (iron core) transformer (air core) Junction of conductors light-emitting diode sin Figure 20.1 Circuit symbols the . in in Electromotive force and potential difference When charge passes through a power supply such as a battery, it gains electrical energy. The power supply is said to have an electromotive force, or e.m.f. for short. The electromotive force measures, in volts, the electrical energy gained by each coulomb of charge that passes through the power supply. Note that, in spite of its name, the e.m.f. is not a force. The energy gained by the charge comes from the chemical energy of the battery e.mf. -energy converted from other forms to electrical charge When charge passes through a resistor, its electrical energy is converted to heat energy in the resistor. The resistor has a potential difference (p.d.) across it. The potential difference measures, in volts, the electrical energy converted for each coulomb of charge that passes through it. vice p.d. -energy converted from electrical to other forms charge Example Two lamps are connected in series to a battery. State the energy transformation that occurs in (a) the battery (b) the lamps. (a) chemical to electrical (b) electrical to thermal (heat) and light. Now it's your turn 1 Each lamp in the example above has a resistance Rand the e.m.f. of the battery is E. The current in the circuit is I. State the rate of energy transformation in (a) the battery (b) a lamp. 157 20.2 N Effect of internal resistance on power from a battery The power delivered by a battery to a variable external load resistance can be investigated using the circuit of Figure 20.4. Readings of current I and potential difference Vie across the load are taken for different values of the variable load resistor. The product V/ gives the power dissipated in the load, and the quotient Vu/I gives the load resistance R Figure 205 shows the variation with load resistance Rof the power V/ dissipated. The graph indicates that there is a maximum power delivered by the battery at one value of the external resistance. This value is equal to the internal resistance r of the battery load Figure 20.4 Circuit for investigating power transfer to an external load 0 load resistance Figure 20.5 Graph of power delivered to external load against load resistance A battery delivers maximum power to a circuit when the load resistance of the circuit is equal to the internal resistance of the battery. Example A high-resistance voltmeter reads 13.0V when it is connected across the terminals of a battery. The voltmeter reading drops to 12.0V when the battery delivers a current of 3.0A to a lamp. State the e.m.f. of the battery. Calculate the potential difference across the internal resistance (the lost volts) when the battery is connected to the lamp. Calculate the internal resistance of the battery, The e.m.f. is 13.0V, since this is the voltmeter reading when the battery is delivering negligible current Using V, = E - VR. lost volts = V, =13.0 - 12.0 = 1.0V. Using V, = lr, r = 1.0/3.0 = 0.33 Now it's your turn 2 Three identical cells, each of e.m.f. 1.5V, are connected in series to a 150 lamp. The current in the circuit is 0.27A. Calculate the internal resistance of each cell. 3 A cell of e.m.f. 1.5V has an internal resistance of 0.50 0. (a) Calculate the maximum current it can deliver. Under what circumstances does it deliver this maximum current? (b) Calculate also the maximum power it can deliver to an external load. Under what circumstances does it deliver this maximum power? 10.15 A 20.2 Kirchhoff's first and second laws 0.15 A 0.15 A Figure 20,6 The current at each point in a Conservation of charge: Kirchhoff's first law A series circuit is one in which the components are connected one after another, forming one complete loop. You have probably connected an ammeter at different points in a series circuit to show that it reads the same current at each point (see Figure 20.6) Series Circuit is the same. 159 power supply load R Internal resistance When a car engine is started with the headlights switched on the headlights sometimes dim. This is because the car battery has resistance All power supplies have some resistance between their terminals, called internal electrical energy in the power supply itself. The power supply becomes warm when it resistance. This causes the charge circulating in the circuit to dissipate some delivers a current Figure 20.2 shows a power supply which has em. E and internal resistance delivers a current / when connected to an external resistor of resistance R (called the load). V is the potential difference across the load, and is the potential difference across the internal resistance. Using conservation of energy, E- Vu+ V The potential difference Vacross the load is thus given by V = E-V: Vi is called the terminal potential difference The terminal potential difference is the p.d between the terminals of a cell when a current is being delivered Figure 20.2 The terminal potential difference is always less than the electromotive force when the power supply delivers a current. This is because of the potential difference across the internal resistance. The potential difference across the internal resistance is sometimes called the lost volts. lost volts = e.m.f. - terminal p.d. In contrast, the electromotive force is the terminal potential difference when the cell is on open circuit (when no current is delivered). This e.m.f may be measured by connecting a very high resistance voltmeter across the terminals of the cell You can use the circuit in Figure 20.3 to show that the greater the current delivered by the power supply, the lower its terminal potential difference. As more lamps are connected in parallel to the power supply, the current increases and the lost volts, given by Figure 20.3 Effect of circuit current on terminal potential difference lost volts = current x internal resistance will increase. Thus the terminal potential difference decreases. To return to the example of starting a car with its headlights switched on, a large current (perhaps 100A) is supplied to the starter motor by the battery. There will then be a large potential difference across the internal resistance, that is, the lost voltage will be large. The terminal potential difference will drop and the lights will dim. In the terminology of Figure 20.2. VR = IR and V, = Ir, so E = VR+V, becomes E = IR + Ir, or E = I(+1) The maximum current that a power supply can deliver will be when its terminals are short-circuited by a wire of negligible resistance, so that R = 0. In this case, the potential difference across the internal resistance will equal the e.m.f. of the cell. The terminal p.d is then zero, Warning: do not try out this experiment, as the wire gets very hot; there is also a danger of the battery exploding. Quite often, in problems, the internal resistance of a supply is assumed to be negligible, so that the potential difference Vix across the load is equal to the emf of the power supply. power supply load Internal resistance When a car engine is started with the headlights switched on, the headlights sometimes dim. This is because the car battery has resistance. All power supplies have some resistance between their terminals, called internal resistance. This causes the charge circulating in the circuit to dissipate some electrical energy in the power supply itself. The power supply becomes warm when delivers a current Figure 20.2 shows a power supply which has e.mf. E and internal resistance r.lt delivers a current / when connected to an external resistor of resistance R (called the load). Vis the potential difference across the load, and is the potential difference across the internal resistance. Using conservation of energy, E = x + V The potential difference Vacross the load is thus given by V = E-V Vi is called the terminal potential difference, R Figure 20.2 The terminal potential difference is the p.d between the terminals of a cell when a current is being delivered The terminal potential difference is always less than the electromotive force when the power supply delivers a current. This is because of the potential difference across the internal resistance. The potential difference across the internal resistance is sometimes called the lost volts. lost volts = e.m.f. - terminal p.d. In contrast, the electromotive force is the terminal potential difference when the cell is on open circuit (when no current is delivered). This e.m.f. may be measured by connecting a very high resistance voltmeter across the terminals of the cell. You can use the circuit in Figure 20.3 to show that the greater the current delivered by the power supply, the lower its terminal potential difference. As more lamps are connected in parallel to the power supply, the current increases and the lost volts, given by Figure 20.3 Effect of circuit current on terminal potential difference lost volts = current x internal resistance will increase. Thus the terminal potential difference decreases. To return to the example of starting a car with its headlights switched on, a large current (perhaps 100 A) is supplied to the starter motor by the battery. There will then be a large potential difference across the internal resistance; that is the lost voltage will be large. The terminal potential difference will drop and the lights will dim. In the terminology of Figure 20.2, Vx = IR and V, = Ir, so E = VX + V, becomes E = IR + Ir, or E = (R+ r) The maximum current that a power supply can deliver will be when its terminals are short-circuited by a wire of negligible resistance, so that R = 0. In this case, the potential difference across the internal resistance will equal the em.f. of the cell. The terminal p.d is then zero. Warning: do not try out this experiment, as the wire gets very hot; there is also a danger of the battery exploding. Quite often, in problems, the internal resistance of a supply is assumed to be negligible, so that the potential difference Vy across the load is equal to the em.f. of the power supply 20 Datuits By Kirchhoff's first law, and using resistance pdcurrent, so I VIR VR = V/N + VIR Dividing by the potential difference V, 1/R=1/R + 1/R This equation can be extended so that the equivalent resistance of several resistors connected in parallel is given by 1 Figure 20. 1 RY R3 Thus The reciprocal of the combined resistance of resistors in parallel is the sum of the reciprocals of all the individual resistances Note that 1 For two identical resistors in parallel, the combined resistance is equal to half of the value of each one 2 For resistors in parallel, the combined resistance is always less than the value of the smallest individual resistance 3 Ω 3 Ω 3 Ω 302 Example Calculate the equivalent resistance of the arrangement of resistors in Figure 20.16. The arrangement is equivalent to two 60 resistors in parallel, so the combined resistance R is given by 1/R = 1/6 + 1/6 = 2/6. (Don't forget to find the reciprocal of this value.) Thus R=30 Now it's your turn 6 Calculate the equivalent resistance of the arrangement of resistors in Figure 20.17. Hint: first find the resistance of the parallel combination 7 Calculate the effective resistance between the points A and B in the network in Figure 20.18. Figure 20.16 40 2012 322 ? 222 Figure 20.17 10022 1092 Figure 5022 of a pote B Figure 20.18 R RS 20.3 Potential dividers and potentiometers Two resistors connected in series with a cell each have a potential difference. They may be used to divide the e.m.f. of the cell. This is illustrated in Figure 20.19. The current in each resistor is the same, because they are in series. Thus V; = IR and V3 = IR». Dividing the first equation by the second gives V,/V, = R./R2. The ratio of V mure 20.19 The potential divider 52 20 D.C. circuits 3 A ???? A parallel circuit is one where the current can take alternative routes in different loom In a parallel circuit, the current divides at a junction, but the current entering the function is the same as the current leaving it (see Figure 20.7). The fact that the current does not used up at a junction is because current is the rate of flow of charge, and charges cannot accumulate or get used up at a junction. The consequence of this conservation of electric charge is known as Kirchhoff's first law. This law is usually stated as follows The sum of the currents entering a junction in a circuit is always equal to the sum of the currents leaving it At the junction shown in Figure 20.8. Figure 20.7 The current divides in a parallel circuit Figur 13 Example For the circuit of Figure 20.9, state the readings of the ammeters A, A, and Az. 0.175A 1 13 1 = 1; + 1 + Figure 20.8 25 mA 75 mA Figure 20.9 Figure 20.10 1.5A A, would read 175 mA, as the current entering the power supply must be the same as the current leaving it A, would read 75 - 25 = 50 mA, as the total current entering a junction is the same as the total current leaving it. A, would read 175 - 75 = 100 mA. 2A 2.5 A Now it's your turn 4 The lamps in Figure 20.10 are identical. There is a current of 0.50 A through the battery, What is the current in each lamp? 5 Figure 20.11 shows one junction in a circuit. Calculate the ammeter reading. Figure 20.11 Conservation of energy: Kirchhoff's second law Charge flowing round a circuit gains electrical energy on passing through the battery and loses electrical energy on passing through the rest of the circuit. From the law of conservation of energy, we know that the total energy must remain the same. The consequence of this conservation of energy is known as Kirchhoff's second law. This law may be stated as follows. The sum of the electromotive forces in a closed circuit is equal to the sum of the potential differences. E Vi+ V Figure 2012 Figure 20.12 shows a circuit containing a battery, lamp and resistor in series. Applying Kirchhoff's second law, the electromotive force in the circuit is the e.mf. E of the battery. The sum of the potential differences is the p.d. Vi across the lamp plus the p.d. V, across the resistor . Thus, E = Vi + V. If the current in the circuit is I and the resistances of the lamp and resistor are R, and R, respectively, the p.d.s can be written as Vi = IR, and V, = IR, SO E = IR; + IR 160 It should be remembered that both electromotive force and potential difference have direction. This must be considered when working out the equation for Kirchhoff's second law. For example, in the circuit of Figure 20.13, two cells have been connected in opposition. Here the total electromotive force in the circuit is E-E, and by Kirchhoff's second law E - Ey = 1; + V = IR, + IR- Resistors in series Figure 2014 shows two resistors of resistances R, and R, connected in series, and a single resistor of resistance Requivalent to them. The current/ in the resistors, and in their equivalent single resistor, is the same 1 R R R Figure 20.13 equivalent to Figure 20.14 Resistors in series The total potential difference Vacross the two resistors must be the same as that across the single resistor. If V; and V are the potential differences across each resistor. V = V1 + V But since potential difference is given by multiplying the current by the resistance, IR = IR; + IR Dividing by the current ! R=R; + R2 This equation can be extended so that the equivalent resistance Rof several resistors connected in series is given by the expression R = R +R+R+. Thus The combined resistance of resistors in series is the sum of all the individual resistances. Resistors in parallel Now consider two resistors of resistance R, and R, connected in parallel, as shown in Figure 20.15. The current through each will be different, but they will each have the same potential difference. The equivalent single resistor of resistance R will have the same potential difference across it, but the current will be the total current through the separate resistors R R equivalent to R Figure 20.15 Resistors in parallel 161 20.3 Potential dividers and potentiometers 12 V C B the voltages across the two resistors is the same as the ratio of their resistances. If the potential difference across the combination were 12V and R, were equal to Ry, then each resistor would have 6V across it. If i were twice the magnitude of Ry, then Vi would be 8V and would be iv A potentiometer is a continuously variable potential divider. In Topic 19, a variable voltage supply was used to vary the voltage across different circuit components. A variable resistor, or theostat, may be used to produce a continuously variable voltage. Such a variable resistor is shown in Figure 20.20. The fixed ends AB are connected across the battery so that there is the full battery voltage across the whole resistor As with the potential divider, the ratio of the voltages across AC and CB will be the same as the ratio of the resistances of AC and CB. When the sliding contact C is at the end B. the output voltage will be 12V. When the sliding contact is at end A, then the output voltage will be zero. So, as the sliding contact is moved from A to B, the output voltage varies continuously from zero up to the battery voltage. In terms of the terminal pd. V of the cell, the output Vent of the potential divider is given by VR Vout (R1 + R2) Where R, is the resistance of AC and R, is the resistance of CB. Figure 20.20 Potentiometer circuit A variable resistor connected in this way is called a potentiometer. A type of potentiometer is shown in Figure 20.21. Note the three connections If a device with a variable resistance is connected in series with a fixed resistor, and the combination is connected to a cell or battery to make a potential divider, then we have the situation of a potential divider that is variable between certain limits. The device of variable resistance could be, for example, a light-dependent resistor or a thermistor. Changes in the illumination or the temperature cause a change in the resistance of one component of the potential divider, so that the potential difference across this component changes. The change in the potential difference can be used to operate control circuitry if, for example, the illumination becomes too low or too high, or the temperature falls outside certain limits. These two components are studied in detail as sensing devices in the A level course. A potentiometer can also be used as a means of comparing potential differences. The circuit of Figure 20.22 illustrates the principle. In this circuit the variable potentiometer resistor consists of a length of uniform resistance wire, stretched along a metre rule. Contact can be made to any point on this wire using a sliding contact Suppose that the cell A has a known e.m.f. Ex. This cell is switched into the circuit using the two-way switch. The sliding contact is then moved along the wire until the centre-zero galvanometer reads zero. The length ly of the wire from the common zero end to the sliding contact is noted. Cell B has an unknown e.m.f. Ey. This cell is then switched into the circuit and the balancing process repeated. Suppose that the position at which the galvanometer reads zero is then a distance ly from the common zero to the sliding contact. The ratio of the e.mfs is the ratio of the balance lengths; that is, Ey/E) = ly/1x, and Ey can be determined in terms of the known e.m.f. EA Figure 20.21 internal and external views of a potentiometer two-way switch EA A ER Figure 20.22 Potentiometer used to compare cell e.mfs