Content Page Summary of Relevant Content 1-5 Problem Solving Techniques 6-10 Worksheet 1: Multiple Choice Questions (10) 11-14 Worksheet 2: Multiple Choice Questions (10) 15-17 Structured Questions 18-19 nswers Multiple Choice Questions 20 Model nswers Structured Questions 20-22 Located at Funda Community College Soweto P.O.Box 39067 2016 Booysens Tel: 011-938 1666/7 Fax: 011-938 3603 Email: sec@global.co.za
SCIENCE EDUCTION CENTRE! SOWETO/DIEPKLOOF P.O.BOX 39067 BOOYSENS 2016!!! " 011 9381666/7 # 011 9383603 email: sec@global.co.za lthough electricity contains many parallels with motion, it involves a fundamentally different force. Electricity is the study of the motion and effect of charges just as mechanics is the study of the motion and effect of masses. The subject of electricity is usually split into two areas: Static electricity (is still) and current electricity (flows). Insulators and Conductors Materials usually fall into one of two groups: conductors (all metals are good conductors) or insulators (e.g., wood, plastic). There is a small, but very important, third group, known as semiconductors; these are used in electronics. conductor of electricity is a material that allows electricity to flow through it. n insulator is a barrier to electricity and will not allow a current to flow through it. Because of this, an insulator may act as a store for electricity as it has no way of escaping or leaking away. Static Electricity (or: Electrostatics) There are two types of charge: positive (+ve) and negative (-ve). n uncharged piece of material has an equal number of each, so there appears to be no charge at all. Friction between materials causes a transfer of charge from one material to another. When negative charges are separated, one material will be positively charged and the other will be negatively charged. It is only the negative charges that move. Static charges will: attract other materials that are uncharged; attract other materials that have the opposite charge; repel other materials with the same charge. The only true test for an object being electrically charged is repulsion. Then you are sure that both objects carry the same type (+ve or -ve) of charge. Static electricity can usefully be employed in: chimneys (to attract the smoke particles in order to reduce environmental pollution); electrostatic photocopiers; painting of car bodies. Static electricity can be harmful. Tall buildings have lightning conductors running from the highest point right down into the ground. These are to take away any charge that may hit them if they are struck by lightning during a thunderstorm. The current of electricity goes harmlessly to earth. Current Electricity ELECTRICITY Whenever charges move we speak of current electricity. When you touch an object that is charged, you will provide a pathway for that electricity. n electric current (symbol I) will flow through you to earth and you will receive an electric shock. n electric current needs a complete circuit to flow in. If there is a break anywhere in a circuit the current flow will cease. Current is measured in amperes: The unit of current flow is the ampere, but when written it may be shortened to amps or, e.g. 5 amps, 2. Current values are measured with an ammeter. mmeters suitable for use with d.c. will usually only work one way round. The positive (usually red) connection of the ammeter is always connected to the positive side of the source. common mistake is to assume that current is lost as it moves around a circuit. This is not true. Whatever current flows into a device must flow out again. Similarly, any current leaving one terminal of a source must return to the other. Electric current is a flow of charge. There are two types of current: alternating current (a.c., will de discussed later) and direct current (d.c.). Direct current sources supply electricity that flows in one direction only. It is accepted that d.c. currents flow in the direction positive (+ve) to negative (-ve). This is known as the conventional 2
direction of electric current. n arrow indicates the direction of current flow. Electrons flow in the opposite direction: from (-ve) to (+ve). Electricity and Electrons When the scientists were first working with simple electricity experiments - starting about in the 17 th century - such as static electricity, they had no idea what it was. If two objects created static electricity between them, the simplest explanation was that one object had an excess and the other was lacking in something, though they had no idea what in. material with excess was called positive. material with a shortage was called negative. Naturally there would be a flow from positive to negative. Unfortunately in making this guess of a flow from positive to negative, the early scientists made a mistake. They had no idea about electrons and that they were the cause of it. Only in 1911 was the electron established as a particle responsible for the flow of electricity. Charge Charge (and charge transfer) is a way of describing an 'amount of electricity'. The unit of charge (symbol Q) is the coulomb (symbol C). capacitor is a device that stores charge. Charge = Current (in amperes) x Time (in seconds) Potential Difference (or: Voltage) The driving force that makes an electric current flow around a circuit is known as potential difference (p.d.). pd is always measured between two points in a circuit. Potential difference is measured in volts (symbol V) with a voltmeter. Voltmeters are connected to a circuit last of all, with one connection either side of the component under investigation. Voltmeters are usually oneway devices and must be connected correctly. P.d.s in series add together in the same way as resistors: two 1.5V cells in series add together to provide a p.d. of 3V. Sharing p.d. around a circuit In a series circuit, the p.d. value decreases around the circuit. In a parallel section, the p.d. value is the same for all components in that section. The sum total of all the individual p.d.s in a circuit will equal the p.d. of the battery. Ohm's Law There are three measurements that can be made for any electrical circuit: current (ampere), p.d. (volt) and resistance (ohm). Ohm s law states that: P.d. [V] = Current [] x Resistance [Ω] V = I x R This can be demonstrated by taking pairs of readings (current / pd) for a resistor in a circuit controlled by a rheostat and calculating the resistance in each case (Fig. 1). (Resistance tends to increase with temperature.) + rheostat V p.d. (volts) current (ampere) resistance (V/I) current () p.d. (V) experiment table graph Fig. 1: Investigating Ohm s law 3
Series and Parallel Circuits ll the components in a series circuit (resistors, bulbs, etc.) follow on one after another in a single loop or chain. parallel circuit has all its components laid out and connected side by side, rather like parallel railway lines. It is very unusual to have a completely parallel circuit. More often a circuit is a mixture of parallel and series sections. One advantage of parallel circuits is that if one part of the parallel section fails to work, it has no effect on the others. Current is the same at all points round a series circuit. In a parallel circuit, the current in the parallel branches is equal to the current in the main circuit. 1 = 1 + 2 2 Fig. 2: a parallel circuit Resistance The term resistance is used to describe any object/device/material that opposes the flow of electricity. Many exam questions will refer to a resistance in a circuit (see later 'worked examples'). large resistance will allow less current to flow than a small resistance. Resistance is measured in ohms, symbol Ω. Resistors: Individual resistors are of two types: fixed or variable. Fixed resistors are colour-coded and are available in several values. Variable resistors (rheostats) allow you to vary the resistance and so control the amount of current flowing. volume control on a radio is one example of the use of a variable resistor. Resistors allow you to limit/control the current flowing in a circuit. Resistors in series: When resistors are laid out in series in a circuit the total resistance increases. They add together in the following way: R total = R 1 + R 2 (wherer total is the total resistance of the circuit; R 1, R 2, etc., are the individual resistors.) Resistors in parallel: set of resistors in parallel will allow more current to flow through them than if the same resistors had been arranged in series. The total resistance decreases in a parallel circuit. 1/R total = 1/R 1 + 1/R 2 (More on resistors in a later section.) Effects of Electric Currents There are three main effects caused by an electric current flowing through a device: magnetic effect; heating effect; chemical effect. Electromagnetism: Whenever electricity flows through a conductor, a magnetic field is created around that conductor. This is known as electromagnetism. Heating: The passage of electricity through a material has a heating effect on the material. Some materials heat up more than others. For this reason, fuses are included in many circuits. They are designed to heat up and melt, or break, before the other materials in the circuit overheat. Chemical: When electricity passes through some materials it causes a chemical reaction to occur. Electrolysis/electroplating are examples of this effect. The reverse is also true; a mixture of certain chemicals will produce an electric current when they react together. battery uses this effect. 4
lternating Current lternating currents, such as household mains, alternate (undergo a complete change of direction) with a frequency of 50Hz (symbol 50Hz a.c. or 50Hz ). Fig. 3 n a.c. generator In Fig. 3 coil BCD rotates in a magnetic field. When the coil turns, this induces a voltage between its ends. s the coil rotates to the following positions: (i) (ii) (iii) (iv) (v) D D D D D This produces an alternating voltage: V ii i iii v time iv The National Grid System: Why.C.? Transmitting electrical energy is most efficiently done by using very high voltages and keeping the current values as small as possible. This can only be achieved with the use of transformers. Without them too much energy would be wasted as heat. Transformers only work with a.c. Electrical energy from the power station, at 22000 V, is first transformed up to approximately 275000 V. This may be further transformed to 400000 V. It is then fed to the National Grid system. The electrical energy is received by the local substation which transforms the grid voltage down to 11 000V. nother series of transformers finally reduce this to 240V a.c. for household use. Transformer transformer has two separate coils of wire on a soft-iron laminated core. The alternating current in the primary coil induces a current in the secondary coil. If the number of turns in the primary coil 5
is less than the number m the sec it is known as a step-up transformer. The reverse situation is known as a step-down transformer. The rule for calculating voltages in a transformer is: V p = N p where V p is the primary voltage, and N p is the number of turns on the primary; V s is the secondary voltage and N s is the number of turns on the secondary coil. Household Supply Mains electricity in South frica is supplied at 240V and 50Hz a.c. (50 Hz ). From the main junction box various circuits are made available. Inside the junction or distribution box are several individual fuses, each leading to one of the circuits. Wire colour-coding: There are two sets of colour codes for dealing with household electricity. The mains cables that supply electricity within the house, i.e., ring mains, etc., are colour-coded: red - live; black - neutral; bare copper - earth. Live, neutral and earth: The live cable - usually red - in the household mains is made to push and then pull the electricity around the circuit. The neutral wire - usually black - only completes the circuit each time. In effect the live cable is the source of electricity all the time. For this reason fuses are always placed on the live side of a circuit. Household appliances have a different colourcode: brown/live; blue/neutral; green/yellow - earth. Fuses: Whenever current flows through a material, then the material begins to warm up. The amount of heat produced depends on the current flowing. higher current will result in greater heating. To stop wires from overheating, fuses are included in electrical circuits. Fuses melt when the current flow through them reaches a certain maximum value so they are labelled in amperes. The correct fuse rating for any particular application may be found from either of the following: by knowing the current value; by calculating the current value from the power and voltage values. If the current value is not known it should be calculated from the equation: Current = Power/Voltage Electrical Energy and Electrical Power Increase any one of the following quantities and the total energy used will increase: p.d. (V volts); current (I amps); time (t seconds) Energy: E = I x V x t (joules) Power: P =I x V (watts) (1000W = 1 kilowatt or 1 kw) Example Find the power output of a power pack that drives a 12V electric motor by supplying a 3 ampere current. How much energy will be consumed in 10 s? To find the power output: P = I x V =3 x 12 =36 watts To find the energy output in 10s: E = I x V x t = 3 x 12 x 10 = 360 joules V s N s Electricity: Basic Concepts/Useful nalogies Electricity Potential Difference, measured in VOLTS Current, measured in MPERES Resistance, measured in OHMS Cell/Battery Switch Voltmeter mmeter Water electric pressure electric flow electric friction electric pump electric tap electric pressure gauge electric flow meter 6
ELECTRICITY ND ELECTRICL CIRCUITS HOW TO SOLVE PROBLEMS When tackling problems in electricity it is important to realise that resistors are usually the most important components in an electrical circuit (there are normally more resistors in a circuit than any other component and they are found on their own and as part of other components). KEY IDE: RESISTOR CUSES ELECTRICL FRICTION and does to an electric current what in mechanics friction does to moving bodies. RESISTOR: 1. opposes the movement (flow) of current. 2. lowers the electric pressure (voltage) of a flowing current. 3. removes energy from a circuit (energy usually generated by the power source - the battery) by changing it to heat (just as friction does in mechanics). 4. only opposes a current if that current flows through it. IMPORTNT: If there is no flow then a resistor has no effect and might as well not be in the circuit. RESISTORS IN SERIES: I R 1 R 2 R 3 I 1. R TOTL =R 1 +R 2 +R 3 + 2. ll the current flows through each resistor. 3. The more resistors in series the greater the total resistance. 4. R TOTL is always greater than the biggest resistor. RESISTORS IN PRLLEL: I 1 R 1 I I 2 I R 2 I 3 R 3 1.) 1/R total = 1/R 1 + 1/R 2 + 1/R 3 2.) The current that flows through R 1 does not flow through R 2 and R 3. 3.) The more resistors in parallel the smaller the total resistance. 4.) R TOTL is always smaller than the smallest resistor. OTHER IMPORTNT FCTS BOUT RESISTORS IN PRLLEL: (i) If we have 4Ω 4Ω then R TOTL is 1/2 of one of them: R TOTL = 7
(ii) If we have: 9Ω 9Ω then R TOTL is 1/3 of one of them: R TOTL = 3Ω 9Ω (iii) If a current flows through resistors in parallel, the smaller the resistance the bigger the current that flows through it, i.e. I. 2 Ω 2 2 3 Ω 3/5 of 2 will flow through the resistor and 2/5 will flow through the 3Ω resistor. II. 2 Ω 3 3 4 Ω If 1 flows through the 4Ω resistor, then 2 will flow through the resistor. (Rule: 1/2 the resistance, twice the current from Ohm's Law: V = I x R.) Since V is the same for two resistors in parallel, if R halves, then I doubles. FCTS BOUT RESISTNCE: 1.) Electric stoves, kettles, etc. use the fact that anything that has resistance generates heat when a current passes through it. Certain metals have high resistance and high melting points. These are ideal as elements in stoves, etc. 2.) Most metals, when they get hot increase their resistance. light globe draws more current as you turn it on (the filament is cold) compared to once it has been on for a few seconds (the filament is now hot). This means that as you turn on a light bulb it draws more current than it will draw once it has been on for a few seconds. 3.) ll components used in a circuit have resistance. Wires, ammeters and on switches have such a low resistance, however, that we assume them to be zero. ll cells have resistance inside them (internal resistance) which we are either given in a question, asked to find or told to ignore. 8
HOW TO FIND THE TOTL RESISTNCE OF COMPLEX CIRCUIT? 1.) Check that the battery - let's assume the source of electricity provides a direct current - does not have internal resistance (r i ). If it does, draw it into the circuit next to the battery in series with it (see diagram below). 2) Find any two or more resistors next to each other in series (the current that flows through any one of them must all flow through the other(s)). Use the formula R T = r 1 + r 2 + and find R T (the total resistance). Replace these resistors with one resistor of value R T. When there are no resistors left in series next to each other, find two or more in parallel and use the formula: 1/R total = 1/R 1 + 1/R 2 + to find their total resistance and redraw the circuit replacing those in parallel with one of value R T. Continue until you have only one resistor left in the circuit which has a value equalling the total of all those that were present at the beginning (including r i ). LET US DO N EXMPLE: Find the total resistance of the following circuit. The battery has an internal resistance of 1Ω. 6Ω 1 6Ω NSWER: First we draw the internal resistance of the battery next to the battery (it does not matter whether to the left or right of the battery). That will give us diagram 2. You could combine the 1Ω with the which are in series and get 3Ω but it is often very convenient to leave r i until the end. 1Ω (r i ) 4Ω diagram 1 It is wise to keep the circuit in its original layout - diagram 3 - until you have each section reduced to one resistor before bringing them all together - diagram 4 -. 6Ω 1 6Ω 4Ω diagram 2 9
so we start with the resistors in parallel: 1/R t = 1/R 1 + 1/R 2 + 1/R 3 = 1/12 + 1/6 + 1/4 = 1/12 + 2/12 + 3/12 = 6/12 R t = 12/6 = Now we replace the three resistors in series by one resistor of and we get diagram 3... 1Ω diagram 3 ll resistors left in the circuit are in series. R total = r 1 + r 2 + r 3 + r 4 = 1 + 2 + 2 + 6 = 11Ω R total nd our simplified circuit does look like diagram 4: 11Ω diagram 4 INTERNL RESISTNCE OF CELL cell consists of two parts. 1. the pump part (that increases the voltage). 2. the resistance part (the friction inside the cell). We buy cells for 1. but can t avoid 2. When drawing a cell, which has internal resistance (R 1 or R i ), draw the two parts separately. one cell 1. pump part 2. internal resistance part (increases (decreases voltage if voltage) current is flowing) 10
CELLS IN PRLLEL * Gives low voltage (same as one cell) * Give high current I TOTL = I 1 + I 2 +I 3 + * Gives low R. CELLS IN SERIES (the usual way) * Gives high voltage: V TOTL = V 1 + V 2 + V 3 * Gives low current: I TOTL = I of one cell (assuming they are all the same). * Gives high R i OHM'S LW The resistance (R) of a conductor is directly proportional to the potential difference (V) across the conductor and inversely proportional to the current (I) flowing through it. Note 1: n ohmic conductor is a conductor that obeys ohm's law. R = V/I Note 2: The graph V against I is a straight line for an ohmic conductor (a curve for a non-ohmic conductor). Note 3: The slope (gradient) of a V against I graph gives resistance; the area under the graph gives power. # Q = I x t where: OTHER RELEVNT FORMULE Q = charge passing a point in a circuit (in Coulomb) I = current flowing past that point (in ampere) t = time (in seconds) # Power = V x I = I 2 x R = V 2 /R # Heat produced (energy dissipated) in joules. = power x t = V x I x t = I 2 x R x t = V 2 / R x t 11
SCIENCE EDUCTION CENTRE! SOWETO/DIEPKLOOF P.O.BOX 39067 BOOYSENS 2016!!! " 011 9381666/7 # 011 9383603 email: sec@global.co.za Physical Science Topic: ELECTRICITY Grade: 12 Worksheet 1: Multiple Choice Questions Time: 30 Minutes Instructions: Make a cross over the letter, B, C or D to show the correct answer. 1. If a current of 3 is divided by the following circuit, the current flowing through the 4 Ω resistor is 3 8Ω 3 B 2 C 1,5 D 1 4Ω 2. The diagram below shows a portion of a circuit into which a current I is flowing. 1 8Ω 2 3 10Ω I Which ammeter shows the highest reading? 1 B 2 C 3 D ll three ammeters give the same reading 3. The diagram to the right represents a part of a circuit containing an ohmic resistor, a voltmeter and an ammeter. If the reading on the ammeter increases the reading on voltmeter V B C D increases in the same ratio. increases, but not in the same ratio. remains unchanged. decreases in the same ratio. V 12
4. battery is connected to two identical light bulbs in parallel as well as another identical bulb in series. n ammeter and a voltmeter are also connected as shown. V One of the light bulbs connected in parallel is unscrewed. How will the ammeter and the voltmeter readings change? Voltmeter reading mmeter reading Increases Increases B Increases Decreases C Increases Unchanged D Decreases Decreases 5. learner connects a circuit as is shown in the diagram. He uses a source of electricity with an emf of 12 V. 12 V V R Which one of the following best gives the ammeter and voltmeter readings which the learner is most likely to get with this circuit? mmeter Voltmeter Reads zero reads zero B Reads zero reads 12 V C very large reading reads zero D very large reading reads 12 V 13
6. Three identical resistors of 4 Ω are connected to give a combined resistance of 6 Ω. Which of the following circuit diagrams illustrates how this was done? B 4 Ω 4 Ω 4 Ω 4 Ω 4 Ω 4 Ω C D 4 Ω 4 Ω 4 Ω 4 Ω 4 Ω 4 Ω 7. In the circuit to the right B 1, B 2 and B 3 are identical light bulbs. The internal resistance of the battery can be ignored. Which statement is true regarding the relative brightness of the bulbs? B 2 B 3 B 1 The three bulbs glow with the same brightness. B B 2 and B 3 glow with the same brightness but brighter than B 1. C B 2 and B 3 glow with the same brightness but less brightly than B 1. D B 1 glows brighter than B 2 while B 2 in turn glows brighter than B 3. 8. Two identical light bulbs, X and Y, are connected in series as shown in the sketch. How will the brightness of the bulbs change if switch S is closed? X S Y X Y Brighter Brighter B Dimmer Dimmer C Brighter Not lit up D Not lit up Brighter 14
9. student connects three identical resistors as shown in the sketch below. 12 V R R R V 1 V 2 The potential difference across the battery is 12 V. What are the readings on V 1 and V 2 respectively? V 1 V 2 4 8 B 6 6 C 8 4 D 9 3 10. 9 V battery is composed of six 1,5 V cells which are connected in series. Each cell has an internal resistance of 0,2 Ω. What is the highest current that can be obtained from such a battery? 7,5 B 1,5 C 1,2 D 0,3 15
SCIENCE EDUCTION CENTRE! SOWETO/DIEPKLOOF P.O.BOX 39067 BOOYSENS 2016!!! " 011 9381666/7 # 011 9383603 email: sec@global.co.za Physical Science ELECTRICITY / ELECTROSTTICS Grade: 12 Worksheet 2: Multiple Choice Questions Time: 30 Minutes Instructions: Make a cross over the letter, B, C or D to show the correct answer. 1. Light bulbs designed for operation at 240V and of power ratings 60W and 100W can be connected in various ways to a standard 240V power supply. Which one of the following combinations of bulbs will give the most light? 240V B 240V 100W 60W 100W 100W C 240V D 240V 100W 60W 100W 100W 2. The diagram below shows a portion of a circuit into which a current I is flowing. 1 8Ω 2 3 10Ω I Which ammeter shows the highest reading? 1 B 2 C 3 D ll three ammeters give the same reading 3. Two point charges +q 1 and +q 2 are kept separated at a distance r. The force of repulsion between them is F. If each of the charges is doubled, the force of repulsion between the charges will be: 16F B 8F C 4F D F 16
4. There is an electrostatic force of attraction between any two of three metal spheres P, Q and R. possible combination for the charges on P, Q and R is P positive; Q negative; R positive B P positive; Q negative; R neutral C P positive; Q positive; R positive D P neutral; Q positive; R neutral 5. The electrical force per unit charge experienced by a small positive test charge in an electric field is called the potential difference B the electric field strength C the electrical potential energy D one volt 6. The three resistors in the circuit shown are identical. When switch S is closed, the reading on voltmeter V 1 will increase and that on V 2 will decrease B increase and that on V 2 will also increase C remain the same while that on V 2 will increase D remain the same while that on V 2 will decrease. S R R R V 1 V 2 7. Three identical resistors each of 2 Ω are connected to give a combined resistance of 3 Ω. Which of the following circuit diagrams illustrates how this was done? B C D 17
8. The rate of flow of charge through a conductor is a measure of B C D power resistance potential difference current 9. battery is connected to two identical bulbs in parallel as well as another identical bulb in series. n ammeter and a voltmeter are also connected as shown in the diagram below. V One of the light bulbs connected in parallel is unscrewed. How will the ammeter and the voltmeter reading change? Voltmeter reading mmeter reading increases increases B increases decreases C increases unchanged D decreases decreases 10. learner connects an electric circuit as shown in the diagram below. She uses a battery with an emf of 6V. 6 V R V Which one of the following best gives the ammeter and voltmeter readings which the learner is most likely to get with this circuit? ammeter voltmeter reads zero reads zero B reads zero reads 6 V C very large reading reads zero D very large reading reads 6 V 18
STRUCTURED QUESTIONS QUESTION 1: girl sets up an electrical circuit from equipment she finds in the science storeroom. She finds ten torch bulbs, each with a resistance of 2 Ω. She also finds two torch cells, a multimeter and a switch. She connects the components as shown in the circuit diagram below. 1,5 V 1,5 V 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω 1.1 What is the value of the emf of the battery? (2) 1.2 Show, by calculation, that the effective resistance of the resistors in parallel is 4Ω. (6) 1.3 Calculate the total resistance of the circuit. (2) 1.4 The girl connects the multimeter in series with the battery and sets the multimeter to read ampere. Calculate the reading on the meter. (4) 1.5 Then she connects the multimeter across the resistors in parallel and sets the multimeter to read volts. Calculate the reading on the meter. (4) Question 2: In the circuit below the 12 V battery has negligible internal resistance and L 1, L 2 and L 3 are three identical light bulbs. 12 V L 2 S 2 L 1 S 1 L 3 S 3 2.1 If only S 1 is closed (S 2 and S 3 are open) which bulbs(s), if any, would light up? (2) 2.2 When both S 1 and S 2 are closed with S 3 open, how do the brightness of the bulbs L 1, L 2 and L 3 compare with each other? (3) 2.3 When all three switches are closed, how does the brightness of the bulbs L 1, L 2 and L 3 compare with each other? (4) 2.4 Calculate the total resistance of the circuit if each bulb has a resistance of 2 Ω, and all the switches are closed. (5) 2.5 What is the current through L 1 when all switches are closed? (4) 19
Question 3 The 24 V battery in the given circuit has negligible internal resistance. 24 V 1 S 1 V 6Ω 4Ω 2 S 2 1 3.1 What is the reading on the voltmeter V when both switches, S 1 and S 2, are open? (3) 3.2 What is the total resistance of the circuit if both switches, S 1 and S 2, are closed? (5) 3.3 What is the reading on the ammeter 1 and 2, when both switches, S 1 and S 2, are closed? (5) Question 4 (HG) battery with an emf of 16 V and internal resistance r i is connected in a circuit as shown in the diagram below. 16 V / r I V R R S 12 Ω When the switch S is open a current of 3,2 is recorded on the ammeter and the reading on voltmeter is 12,8V. 4.1 Determine the value of resistor R. (6) 4.2 Calculate the internal resistance r i. (4) 4.3 Calculate the reading on the ammeter when the switch S is closed. (5) 4.4 Would the reading on the voltmeter increase, decrease or remain the same when S is closed? Explain your answer without doing any further calculations. (3) 4.5 Determine the amount of heat released in 10 minutes by either of the resistors R when the switch S is open. (5) 20
NSWER SECTION Multiple Choice Questions: Worksheet I Worksheet II 1 B 1 B 2 D 2 D 3 3 C 4 B 4 B 5 B 5 B 6 B 6 7 C 7 B 8 D 8 D 9 C 9 B 10 10 B STRUCTURED QUESTIONS (Model nswers) Question 1 1.1 What is the value of the emf of the battery? (2) There are two cells, each with an emf of 1,5V. Hence: emf of battery = 3 V!! 1.2 Show, by calculation, that the effective resistance of the resistors in parallel is 4Ω. (6) top branch: 2+2+2= 6Ω! bottom branch: 2+2+2+2+2+2= 1! lways start calculations with the formula: 1/R total = 1/R 1 + 1/R 2! then you substitute: 1/R t = 1/6 + 1/12 = 2/12 + 1/12 = 3/12!! R t = 12/3 = 4Ω! 1.3 Calculate the total resistance of the circuit. (2) total resistance = + 4Ω = 6Ω!! 1.4 The girl connects the multimeter in series with the battery and sets the multimeter to read ampere. Calculate the reading on the meter. (4) 1. formula: I = V/R! 2. substitution: I = 3/6!! 3. answer with unit: I = 0,5! 1.5 Then she connects the multimeter across the resistors in parallel and sets the multimeter to read volts. Calculate the reading on the meter. (4) V = I x R! V = 0.5 x 4!! V = 2 V! Question 2 2.1 When S 2 and S 3 are both open no current can flow since the circuit is not closed. Hence: no bulb will light up. 21
2.2 When S 2 is closed the current will flow through L 1 and L 2 but not L 3. Therefore L 3 will not light up. Since the same current flows through the bulbs L 1 and L 2, they will have the same brightness. 2.3 When all switches are closed all the current will pass through L 1 and then it will divide. Half of the current flows through L 2 and the other half through L 3. s a result L 2 and L 3 will have the same brightness while L 1 will shine brighter than the bulbs connected in parallel. 2.4 In a parallel circuit the total resistance is given by: 1/R t = 1/R 1 + 1/R 2 = 1/2 + 1/2 = 1/1 R t = 1 Ω L 2 L 1 therefore: R total = R 1 + R 2 = 1Ω + R total = 3Ω L 3 2.5 ll the current flows through L 1 : I = V/R = 12V/3Ω I = 4 (note: the current in L 2 and L 3 will be 2.) Question 3 3.1 What is the reading on the voltmeter V when both switches, S 1 and S 2, are open? (3) When S 1 and S 2 are open no current will flow in the circuit since the circuit is not closed. The resistors don't have any effect when there is no current flow! (Don't assume the 4Ω resistor does have any effect on the voltmeter reading!!!) ccordingly the reading on the voltmeter will show the emf of the battery. Reading on voltmeter (emf) = 24 V 3.2 What is the total resistance of the circuit if both switches, S 1 and S 2, are closed? (5) Make a sketch to find the total resistance of the circuit: 6Ω Starting with the resistors in parallel: 1 1/R t = 1/R 1 + 1/R 2 = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 R t = 12/3 = 4Ω Now we have two resistors in series: R T = R 1 + R 2 R T = 4Ω + 4Ω = 8Ω 4Ω 4Ω 4Ω 3.3 What is the reading on the ammeters 1 and 2, when both switches are closed? (5) ll current in the circuit flows through 1. pplying Ohm's Law: V = I x R I = V/R = 24V/8Ω = 3 mpere Reading of 1 = 3 Reading on 2 will give the current flowing through the 1 resistors. The p.d. across the parallel connection will be 12 V (total resistance is 8Ω, therefore the drop in p.d across 4Ω will be 12V). Hence: I = V/R = 12/12 = 1 reading on 2 22
or: 2/3 of total current will flow through the 6Ω resistor and 1/3 through the 1 resistor. 1 amp flows through 2 3 2 1 2 6Ω 1 3 Question 4 4.1 Determine the value of resistor R. (6) Circuit (with S R total = V/I! open): = 12,8/3,2! = 4 Ω! 3,2 12,8V R R 1/R T = 1/R 1 + 1/R 2! 1/4 = 1/R + 1/R = 2/R! R/2 = 4/1 R = 8 Ω! 4.2 Calculate the internal resistance r i. (4) Two methods can be applied: ) r i = V (lost volts) / I total or B) Emf = I (R + r i )! = (16-12,8)/3,2 16 = 3,2 (4 + r i )!! = 3,2/3,2 16/3,2 = 4 + r i r i = 1 Ω 5-4 = r i r i = 1 Ω! 4.3 Calculate the reading on the ammeter when the switch S is closed. (5) With S closed the total resistance becomes: 1/R t = 1/R 1 + 1/R 2 = 1/4 + 1/12 = 3/12 + 1/12 = 4/12! R t = 3Ω! I = Emf/(R + r I ) = 16/(3 + 1)!! I = 4! (don't forget the internal resistance of the battery!!!) 4.4 Would the reading on the voltmeter increase, decrease or remain the same when S is closed? Explain your answer without doing any further calculations. (3) Decrease! Resistance decreases. Current I increases Voltage V decreases!! (or: V l increases V decreases!!) 4.5 Determine the amount of heat released in 10 minutes by either of the resistors R when the switch S is open. (5) W = I 2 R t or: W = V I t or: W = V 2 t/r! = (1,6) 2 x 8 x 600 = 12,8 x 1,6 x 600 = (12,8) 2 x 600/8!!! W = 12288 J W = 12288 J W = 12288 J! 23