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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity CHAPTER 7: ELECTRICITY7.1 CHARGE AND ELECTRIC CURRENTVan de Graaf1. What is a Van de Graaff generator? Fill in each of the boxes the name of the part shown. A device that ……………….. and ………………….. at high voltage on its dome + + + + + dome + + + + + -1-
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity2. How are electrical charges produced by a Van de Graaff generator? And what type of charges is usually produced on the dome of the generator? When the motor of the Van de Graaff generator is switched on, it drives the ……………………….. This causes the rubber belt to against the …….……… and hence becomes …..……… The charge is then carried by the moving belt up to the …………… ………. where it is collected. A large amount of ……………. is built up on the dome ……………………. charges are usually produced on the dome of the generator. + + +3. What will happen if the charged dome of + + + + the Van de Graaff is connected to the earth via a micrometer? Explain. There is a …………………….. of the pointer of the microammeter. This indicates an electric current ……………………. -2-
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JPN Pahang Physics Module Form 5 Student’s Copy Chapter 7: Electricity 4. Predict what will happen if a discharging metal sphere to the charged dome. + + + + + ++ + When the discharging metal sphere is brought near the charged dome, …………………………… occurs. An electric current …………… 5. Predict what will happen if hair of a student is brought near to the charged dome. Give reasons for your answer. The metal dome …………. the hair and the hair stand ……………….. This is because of each strand of hair receives ……………….. charges and …………………….. each other. 6. The flow of electrical charges produces …………………. -3-
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityElectric Current 1. Electric current consists of a flow of …………...... 2. The more charges that flow through a cross section within a given time, the ……………… is the current. 3. Electric current is defined as the rate of flow of …………………………. Each second, 15 coulombs of charge cross the plane. The current is I = 15 amperes. One ampere is one coulomb per second. 4. In symbols, it is given as: I= where I = …………………….… Q = …………………….… t = …………..................... (i) The SI unit of charge is (Ampere / Coulomb / Volt) (ii) The SI unit of time is (minute / second / hour) (iii)The SI unit of current is (Ampere / Coulomb / Volt) is equivalent to (Cs // C-1s // Cs-1) I t (iv) By rearranging the above formula, Q = ( It / t / I ) 4. If one coulomb of charge flows past in one second, then the current is …………………. ampere. 5. 15 amperes means in ………………second, …………….. coulomb of charge through a cross section of a conductor. 6. In a metal wire, the charges are carried by…………………. 7. Each electron carries a charge of ……………………….. 8. 1 C of charge is…………………………….. -4-
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityElectric Fielda) An electric field is a ………………. in which an……………… experiences a…………..b) An electric field can be represented by a number of lines indicate both the……………. and ……………….. of the fieldc) The principles involved in drawing electric field lines are : (i) electric field lines always extend from a ……………… - charged object to a ………………..-charged object (ii) electric field lines never ………………….. each other, (ii) electric field lines are ……………….. in a ………………….. electric field.Demo: To study the electric field and the effects of an electric field.Apparatus & materialsExtra high tension (E.H.T) power supply (0 – 5 kV), petri dish, electrodes with differentshapes (pointed electrode and plane electrode), two metal plates, talcum powder, cooking oil,polystyrene ball coated with conducting paint, thread and candle.Method DEMOA) 1. Set up the apparatus as shown in the above figure 2. Switch on the E.H.T. power supply and adjust the voltage to 4 kV 3. Observed the pattern formed by the talcum powder for different types of electrodes. 4. Draw the pattern of the electric field lines. -5-
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity→ Draw the pattern of the electric field lines. ELECTRIC FIELD AROUND A POSITIVE CHARGE ELECTRIC FIELD AROUND A NEGATIVE CHARGE ELECTRIC FIELD AROUND A POSITIVE AND NEGATIVE CHARGE ELECTRIC FIELD AROUND TWO NEGATIVE CHARGES -6-
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity ELECTRIC FIELD AROUND TWO POSITIVE CHARGES ELECTRIC FIELD AROUND A NEGATIVE CHARGE AND A POSITIVELY CHARGED PLATE ELECTRIC FIELD AROUND A POSITIVE CHARGE AND A NEGATIVELY CHARGED PLATE ELECTRIC FIELD BETWEEN TWO CHARGED PARALLEL PLATES -7-
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity EFFECT OF AN ELECTRIC FIELD ON A POLYSTYRENE BALL Observation: The polystyrene ball oscillated between the two plates, touching one plate after another. Explanation: When the polystyrene ball touches the negatively charged plate, the ball receives negative charges from the plate 1. Place the polystyrene ball between the and experiences a repulsive force. two metal plates. The ball will then move to the positively 2. Switch on the E.H.T and displace the charged plate. polystyrene ball slightly so that it When the ball touches the plate, the ball touches one of the metal plates loses some of its negative charges to the plate and becomes positively charged. It then experiences a repulsive force. This process continues. -8-
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity EFFECT OF AN ELECTRIC FIELD ON A CANDLE FLAMEC) Observation: The candle flame splits into two portions in opposite direction. The portion that is attracted to the negative plate is very much larger than the portion of the flame that is attracted to the positive plate. Explanation: 1) Switch of the E.H.T and replace the The heat of the flame ionizes the air polystyrene ball with a lighted candle. molecules to become positive and 2) Sketch the flame observed when the negative charges. E.H.T. is switched on. The positive charges are attracted to the negative plate while the negative charges are attracted to the positive plate. The flame is dispersed in two opposite directions but more to the negative plate. The positive charges are heavier than the negative charges. This causes the uneven dispersion of the flame. Conclusion 1. Electric field is a ……………………………………………………………………….. 2. Like charges ………………. each other but opposite charges …………… each other. 3. Electric field lines are ……………………in an electric field. The direction of the field lines is from …………………….. to ………………………… -9-
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity Exercise 7.1 1. 5 C of charge flows through a wire in 10 s. What is the current in the wire? 2. A charge of 300 C flow through a bulb in every 2 minutes. What is the electric current in the bulb? 3. The current in a lamp is 0.2 A. Calculate the amount of electric charge that passes through the lamp in 1 hour.4. If a current of 0.8 A flows in a wire, how many electrons pass through the wire in one minute? (Given: The charge on an electron is 1.6 x 10-19 C) An electric current of 200 mA flows through a resistor for 3 seconds, what is the (a) electric charge (b) the number of electrons which flow through the resistor? - 10 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityIdeas of Potential Difference (a) (b) X water Y P Q ⇒ Pressure at point P is ……………… Gravitational potential energy at X is …………… than the pressure at point Q than the gravitational potential energy at Y. Water will flow from ……to …… The apple will fall from …… to …… when the when the valve is opened. apple is released. This due to the ……………….. in the This due to the …………………….. in the pressure of water gravitational potential energy. (c) Similarly, Point A is connected to…………………. terminal Point B is connected to …………………. terminal Electric potential at A is ……………. than the electric potential at B. Bulb Electric current flows from A to B, passing the bulb in the A B circuit and ……………….. the bulb. This is due to the electric ………………. between the two terminals. As the charges flow from A to B, work is done when electrical energy is transformed to ………….and …………… energy. The potential difference, V between two points in a circuit is defined as ……………………………………………………... ………………………………………………………………… ………………………………………………………………… The potential difference,V between the two points will be given by: Work W where W is …………………………. V = Quantityofch arg e = Q Q is …………………………. - 11 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityDevice and symbol Cells ammeter voltmeter Switch Constantan wire // connecting wire eureka wire bulb resistance rheostatMeasuring Current and Potential Difference/Voltage Measurement of electricity Measurement of potential difference/voltage (a) Electrical circuit (a) Electrical circuit (b) Circuit diagram (b) Circuit diagram Turn to next page→ - 12 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity1. Name the device used to measure electrical 1. Name the device used to measure current. potential difference.2. (a) What is the SI unit for current? 2. (a) What is the SI unit for potential difference? (b) What is the symbol for the unit of current? (b) What is the symbol for the unit of potential difference?3. How is an ammeter connected in an electrical circuit? 3. How is an voltmeter connected in an electrical circuit?4. The positive terminal of an ammeter is connected to which terminal of the dry 4. The positive terminal of a voltmeter is cell? connected to which terminal of the dry cell?5. What will happen if the positive terminal of the ammeter is connected to the negative terminal of the dry cell? - 13 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityExperiment: To investigate the relationship between current and potential differencefor an ohmic conductor. (a) (b)Figure (a) and figure (b) show two electrical circuits. Why do the ammeters show differentreadings? Why do the bulbs light up with different intensity?Referring to the figure (a) and (b),(i) Make one suitable inference.(ii) State one appropriate hypothesis that could be investigated.(iii) Design an experiment to investigate the hypothesis.(a) Inference The current flowing through the bulb is influenced by the potential difference across it.(b) Hypothesis To determine the relationship between current and potential difference for a(c) Aim constantan wire. (i) manipulated variable :(d) Variables (ii) responding variable : (iii) fixed variable : Apparatus / materials - 14 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityMethod : 1. Set up the apparatus as shown in the figure. 2. Turn on the switch and adjust the rheostat so that the ammeter reads the current, I= 0.2 A. 3. Read and record the potential difference, V across the wire. 4. Repeat steps 2 and 3 for I = 0.3 A, 0.4 A, 0.5 A, 0.6 A and 0.7 A.Tabulation of :data Current,I/A Volt, V/V 0.2 1.0 0.3 1.5 0.4 2.0 0.5 2.5 0.6 3.0 0.7 3.5Analysis of data : Draw a graph of V against I .Discussion : 1. From the graph plotted. (a) What is the shape of the V-I graph? The graph of V against I is a straight line that passes through origin (b) What is the relationship between V and I? This shows that the potential difference, V is directly proportional to the current, I. (c) Does the gradient change as the current increases? V The gradient ≡ the ratio of is a constant as current increases. I 2. The resistance, R, of the wire used in the experiment is equal to the gradient of the V-I graph. Determine the value of R. 3. What is the function of the rheostat in the circuit? It is to control the current flow in the circuitConclusion : The potential difference, V across a conductor increases when the current, I passing through it increases as long as the conductor is kept at constant temperature. - 15 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity Ohm’s Law (a) Ohm’s law states that the electric current, I flowing through a conductor is directly proportional to the potential difference across the ends of the conductor, if temperature and other physical conditions remain constant (b) By Ohm’s law: V ∝ I = constant × I V or I = constant (c) The constant is known as ………………………………. of the conductor. (d) The resistance, R is a term that describes …………………………………………….. ………………………………………………………………………………………….. It is also defined as the ratio……………………………………………………………. ………………………………………………………………………………………….. That is R= and V= (e) The unit of resistance is ………………………………… (f) An ……………………….. is one which obeys Ohm’s law, while a conductor which does not obey Ohm’s law is known as a ……………………….conductor - 17 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityFactors Affecting Resistance 1. The resistance of a conductor is a measure of the ability of the conductor to (resist / allow) the flow of an electric current through it. 2. From the formula V = IR, the current I is (directly / inversely) proportional to the resistance, R. 3. When the value of the resistance, R is large, the current, I flowing in the conductor is (small / large) 4. What are the factors affecting the resistance of a conductor? a) ……………………………………………………………. b) ……………………………………………………………. c) ……………………………………………………………. d) ……………………………………………………………. 5. Write down the relevant hypothesis for the factors affecting the resistance in the table below. Factors Diagram Hypothesis Graph The …………… the conductor, the …………….. its resistance Length of the conductor, l Resistance is …………………. proportional to the length of a conductor The ……………….….. the cross - The cross-sectional sectional area, the …….………… conductor, A the its resistance area of the Resistance is ……………...…….. proportional to the cross-sectional area of a conductor Different conductors with the same physical conditions have The type of the material of the ……………………. resistance conductor The …………………. The The temperature of temperature of a conductor, the the conductor …………………... the resistance - 18 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity 6. From, the following can be stated: Resistance of a conductor, R ∝ length Resistance of a conductor, R ∝ 1 cross-sectional area Hence, resistance of a conductor, R ∝ length cross-sectional area Or R∝ l or R= ρ l where ρ = resistivity of the A A substance - 19 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityExercise 7.21. Tick (√) the correct answers True False (a) Unit of potential difference is J C-1 (b) J C-1 ≡ volt, V The potential difference between two points is 1 volt if 1 joule (c) of work is required to move a charge of 1 coulomb from one point to another. 2 volt is two joules of work done to move 2 coulomb of charge (d) from one to another in an electric field. (e) Potential difference ≡ Voltage I t2. i) Electric charge, Q = ( It / / ) t I V Q ii) Work done, W = (QV / / ) Q V iii) Base on your answer in 2(i) and (ii) derive the work done, W in terms of I, V and t. W = QV = ItV3. If a charge of 5.0 C flows through a wire and the amount of electrical energy converted into heat is 2.5 J. Calculate the potential differences across the ends of the wire. W = QV 2.5 = 5.0 (V) V = 0.5 V4. A light bulb is switched on for a period of time. In that period of time, 5 C of charges passed through it and 25 J of electrical energy is converted to light and heat energy. What is the potential difference across the bulb? W = QV 20 = 6 (V) V = 3.33 V - 20 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity5. The potential difference of 10 V is used to operate an electric motor. How much work is done in moving 3 C of electric charge through the motor? W = QV = 3 (10) = 30 J6. When the potential difference across a bulb is Bulb 20 V, the current flow is 3 A. How much work 3A done to transform electrical energy to light and heat energy in 50 s? 20 V W = VIt = 20 (3) (50) = 3000 J7. What is the potential difference across a light bulb of resistance 5 Ω when the current that passes through it is 0.5 A? V = IR = 0.5 (5) = 2.5 V8. A potential difference of 3.0 V applied across a resistor of resistance R drives a current of 2.0 A through it. Calculate R. V = IR 3.0 = 2.0 (R) R = 1.5 Ω9. What is the value of the resistor in the figure, if the dry cells supply 2.0 V and the ammeter reading is 0.5 A? V = IR 2.0 = 0.5 (R) R = 4Ω - 21 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity10. If the bulb in the figure has a resistance of 6 Ω, what is the reading shown on the ammeter, if the dry cells supply 3 V? V = IR 3.0 = 6 (R) R = 0.5 Ω11. If a current of 0.5 A flows through the resistor of 3 Ω in the figure, calculate the voltage supplied by the dry cells? V = IR = 0.5 (3) R = 1.5 Ω12. The graph shows the result of an experiment to V/V determine the resistance of a wire. The resistance of the wire is 1.2 From V-I graph, resistance = gradient = = 2.4Ω I/A 0 513. An experiment was conducted to measure the current, I flowing through a constantan wire when the potential difference V across it was varied. The graph shows the results of the experiment. What is the resistance of the resistor? From V-I graph, resistance = gradient = = 2.0 x 10-3Ω - 22 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity14. Referring to the diagram on the right, calculate (a) The current flowing through the resistor. I 5Ω V = IR 12 V 12 = I (5) I = 2.4 A (b) The amount of electric charge that passes through the resistor in 30 s Q = It = 2.4 (30) = 72 C (c) The amount of work done to transform the electric energy to the heat energy in 30 s. W = QV or W = VIt = 72 (12) = 12(2.4)(30) = 864 C = 864 C15. Figure shows a torchlight that uses two 1.5 V dry cells. The two dry cells are able to provide a + 1.5 V - + 1.5 V - current of 0.3 A when the bulb is at its normal brightness. What is the resistance of the filament? V = IR 3.0 = 0.3(R) I = 10Ω16. The diagram shows four metal rods of P, Q, R and S made of the same substance. a) Which of the rod has the most resistance? P b) Which of the rod has the least resistance? S - 23 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity 17. The graph shows the relationship between the V/V potential difference, V and current, I flowing X through two conductors, X and Y. 8 Y a) Calculate the resistance of conductor X. From V-I graph, resistance = gradient 2 = 0 I/A 0 2 = 4Ω b) Calculate the resistance of conductor Y. From V-I graph, resistance = gradient = = 1Ω c) If the cross sectional area of X is 5.0 x 10-6 m2, and the length of X is 1.2 m, calculate its resistivity.18. The graph shows a graph of I against V for three I/A P conductors, P, Q and R. i) Compare the resistance of conductor P, Q and R. Q Q R ii) Explain your answer in (a) V/V From V-I graph, resistance = gradient The greater the gradient, the greater the resistance Gradient of P > Gradient of Q > Gradient of R - 24 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity 19. Figure shows a wire P of length, l with a cross- sectional area, A and a resistance, R. Another wire, Q is a conductor of the same material with a length of 3l and twice the cross-sectional area of P. What is resistance of Q in terms of R? Conductor P R = Conductor Q R’ = (notes: P and R have the same resistivity, ρ) = = R 20. PQ, is a piece of uniform wire of length 1 m with a resistance of 10Ω. Q is connected to an ammeter, a 2 Ω resistor and a 3 V battery. What is the reading on the ammeter when the jockey is at X? Resistance in the wire R is directly proportional to l 100 cm = 10 Ω Hence, 20 cm = (10) R = 2Ω Total resistance 2Ω + 2Ω = 4Ω Current, I = = = 0.75 A - 25 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity 21. Figure shows the circuit used to investigate the relationship between potential difference, V and current, I for a piece of constantan wire. The graph of V against I from the experiment is as shown in the figure below. (a) What quantities are kept constant in this experiment? Length // cross-sectional area // type of material // temperature of the wire (b) State the changes in the gradient of the graph, if i) the constantan wire is heated R ↑, gradient ↑ // the resistance increases, hence the gradient increases ii) a constantan wire of a smaller cross-sectional area is used R ↑, gradient ↑ // the resistance increases, hence the gradient increases iii) a shorter constantan wire is used R ↓, gradient ↓ // the resistance decreases, hence the gradient decreases - 26 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity7.3 SERIES AND PARALLEL CIRCUITSCurrent Flow and Potential Difference in Series and Parallel Circuit SERIES CIRCUIT PARALLEL CIRCUIT I V1 the current flows through each bulb/resistor is 1 the potential difference is the same across each the same. bulb/resistor I = I1 = I2 = I3 V = V1 = V2 = V32 the potential difference across each bulb / 2 the current passing through each bulb / resistor is resistor depends directly on its …………………. inversely proportional to the resistance of the The potential difference supplied by the dry cells resistor. The current in the circuit equals to the is shared by all the bulbs / resistors. sum of the currents passing through the bulbs / resistors in its parallel branches. V = V1 + V2 + V3 where V is the potential difference across the I = I1 + I2 + I3 where I is the total current battery from the battery3 If Ohm’s law is applied separately to each bulb / 3 If Ohm’s law is applied separately to each bulb / resistor, we get : resistor, we get : V = V1 + V2 + V3 I = I1 + I2 + I3 IR = IR1 + IR2 + IR3 V V V V R = R1 + R2 + R3 If each term in the equation is divided by I, we If each term in the equation is divided by V, we get the effective resistance get the effective resistance 1 1 1 1 R = R1 + R2 + + R3 R = R1 + R2 + R3 - 27 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityIdentify series circuit or parallel circuit (a) (b) (c) (d)Ammeter reading ≡ CurrentVoltmeter reading ≡ Potential difference ≡ Voltage - 28 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricitySolve problems using V = IRExercise 7.31. The two bulbs in the figure have a resistance of 2Ω and 3Ω respectively. If the voltage of the dry cell is 2.5 V, calculate (a) the effective resistance, R of the circuit Effective R = 2 + 3 = 5 Ω (b) the main current, I in the circuit (c) the potential difference across each bulb. V = IR 2Ω: V = IR = (0.5)(2) = 1V 2.5 =I(5) 3Ω: V = IR = (0.5)(3) = 1.5 V = 0.5 A2. There are two resistors in the circuit shown. Resistor R1 has a resistance of 1Ω. If a 3V voltage causes a current of 0.5A to flow through the circuit, calculate the resistance of R2. V = IR 3=0.5(1+R2) R2 = 5Ω - 30 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity3. The electrical current flowing through each branch, I1 and I2, is 5 A. Both bulbs have the same resistance, which is 2Ω. Calculate the voltage supplied. Parallelcircuit;V =V1=V2 = IR1 or = IR2 = 5(2) = 10 V4. The voltage supplied to the parallel is 3 V. R1 and R2 have a resistance of 5Ω and 20Ω. Calculate (a) the potential difference across each resistor 3 V (parallel circuit) (b) the effective resistance, R of the circuit 1/R = 1/5 + 1/20 =1/4 R=4Ω (c) the main current, I in the circuit (d) the current passing through each resistor V = IR 5Ω: V = IR 20 Ω: V = IR 3 =I(4) 3 =I(5) 3 =I(20) = 0.75 A I = 0.6 A I = 0.15 A5. In the circuit shown, what is the reading on the ammeter when switch, S (a) is open? (b) is closed? Effective R = 6 Ω Effective R = 4 Ω V = IR V = IR 12 =I(6) 12 =I(4) I=2A I=3A6. Determine the voltmeter reading. Determine the ammeter reading. (a) (a) (b) - 31 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity7. Calculate (d) (i) The potential difference across 8Ω (a) The effective resistance, R resistor. R = 12 Ω V = IR (b) The main current, I = 2(8) = 16 V I=2A (ii) The potential difference across 2.5Ω (c) The current passing through 8Ω and 2.5Ω resistor. resistors. V = IR I=2A = 2(2.5) = 5 V (e) The current passing through 6 Ω resistor. V = V8 + V2.5 +Vparallel 24 = 16 + 5 + Vparallel Vparallel = 3V V = IR 3 = I(6) I = 0.5 A - 32 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity8. The electrical components in our household appliances are connected in a combination of series and parallel circuits. The above figure shows a hair dryer which has components connected in series and parallel. Describe how the circuit works. The hair dryer has three switches A, B and C When switch A is switched on, the dryer will only blow air at ordinary room temperature When switches A and B are both switched on, the dryer will blow hot air. As a safety feature to prevent overheating, the heating element will not be switched on if the fan is not switched on The hair dryer has an energy saving feature. Switch C will switch on the dryer only when it is held by the hand of user The body of the hair dryer must be safe to hold and does not get hot easily - 33 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity7.4 ELECTROMOTIVE FORCE AND INTERNAL RESISTANCEElectromotive force Figure (a) Figure (b) Voltmeter reading, Voltmeter reading, e.m.f. potential difference, V < e.m.f., E,r R Current flowing No current flow1. An electrical circuit is set up as shown in figure (a). A high resistance voltmeter is connected across a dry cell which labeled 1.5 V.a) Figure (a) is (an open circuit / a closed circuit)b) There is (current flowing / no current flowing) in the circuit. The bulb (does not light up / lights up)c) The voltmeter reading shows the (amount of current flow across the dry cell / potential difference across the dry cell)d) The voltmeter reading is (0 V / 1.5 V / Less than 1.5 V)e) The potential difference across the cell in open circuit is (0 V / 1.5 V / Less than 1.5 V). Hence, the electromotive force, e.m.f., E is (0 V / 1.5 V / Less than 1.5 V)f) It means, (0 J / less than 1.5 J / 1.5 J / 3.0 J) of electrical energy is required to move 1 C charge across the cell or around a complete circuit. - 34 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity2. The switch is then closed as shown in figure (b).a) Figure (b) is (an open circuit / a closed circuit)b) There is (current flowing / no current flowing) in the circuit. The bulb (does not light up / lights up)c) The voltmeter reading is the (potential difference across the dry cell / potential difference across the bulb / electromotive force).d) The reading of the voltmeter when the switch is closed is (lower than/ the same as / higher than) when the switch is open.e) If the voltmeter reading in figure (b) is 1.3 V, it means, the electrical energy dissipated by 1C of charge after passing through the bulb is (0.2 J / 1.3 J / 1.5 J)f) The potential difference drops by (0.2 V/ 1.3 V / 1.5 V). It means, the potential difference lost across the internal resistance, r of the dry cell is (0.2 V/ 1.3 V / 1.5 V).g) State the relationship between e.m.f , E , potential difference across the bulb, VR and drop in potential difference due to internal resistance, Vr. Electromotive force, e.m.f., E = Potential Difference + Drop in Potential Difference across resistor, R due to internal resistance,r = VR + Vr where VR = IR and Vr = Ir = IR + Ir = I (R + r) - 35 -
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JPN Pahang Physics Module Form 5 Student’s Copy Chapter 7: Electricity 3. i. Why is the potential difference across the resistor not the same as the e.m.f. of the battery? The potential drops as much as 0.4 V across the internal resistance ii. Determine the value of the internal resistance. Since E = V + Ir 1.5 = 1.1 + 0.5 r r = 0.8 Ω Therefore, the value of the internal resistance is 0.8 Ωiii. Determine the value of the external resistor. Since V = IR 1.1 = 0.5 R R = 2.2 Ω Therefore, the value of the external resistance is 2.2 Ω - 36 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityActivity : To determine the values of the electromotive force (e.m.f.) and the internal resistance, r of the cell Voltmeter V Internal resistance + Dry cell - Ammeter Switch Rheostat To determine the values of the electromotive force (e.m.f.) andAim the internal resistance, r of the cellApparatus / Dry cells holder, ammeter (0 – 1 A), voltmeter(0 – 5 V), rheostat (0 – 15 Ω), connectingmaterials wires, switch, and 2 pieces of 1.5 V dry cell.Method : a) Set up the circuit as shown in the figure. b) Turn on the switch, and adjust the rheostat to give a small reading of the ammeter, I, 0.2 A. c) Read and record the readings of ammeter and voltmeter respectively d) Adjust the rheostat to produce four more sets of readings, I = 0.3 A, 0.4 A, 0.5 A and 0.6 A.Tabulation of :data Current,I/A Volt, V/V 0.2 2.6 0.3 2.5 0.4 2.4 0.5 2.2 0.6 2.0 0.7 1.9 - 37 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityAnalysis of data :Based on theabove data,draw a graph ofV against IDiscussion : 1. From the graph plotted, state the relationship between the potential difference, V across the cell and the current flow, I? The potential difference, V across the cell decreases as the current flow increases. 2. A cell has an internal resistance, r. This is the resistance against the movement of the charge due to the electrolyte in the cell. With the help of the figure, explain the result obtained in this experiment. When the current flowing through the circuit increases, the quantity of charge flowing per unit time increased. Hence, more energy was lost in moving a larger amount of charge across the electrolyte. Because of this, there was a bigger drop in potential difference measured by the voltmeter. 3. By using the equation E = V + Ir (a) write down V in terms of E, I and r. V = -rI + E (b) explain how can you determine the values of E and r from the graph plotted in this experiment. E = the vertical intercept of the V – I graph R = the gradient of the V – I graph (c) determine the values of E and r from the graph. By extrapolating the graph until it cuts the vertical axis, E = 2.9 V r = - gradient = 1.4 Ω - 38 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityExercise 7.41 A voltmeter connected directly across a battery gives a reading of 1.5 V. The voltmeter reading drops to 1.35 V when a bulb is connected to the battery and the ammeter reading is 0.3 A. Find the internal resistance of the battery. E = 3.0 V, V = 1.35 V, I = 0.3 A Substitute in : E = V + Ir 1.5 = 1.35 + 0.3(r) r = 0.5 Ω2. A circuit contains a cell of e.m.f 3.0 V and internal resistance, r. If the external resistor has a value of 10.0 Ω and the potential difference across it is 2.5 V, find the value of the current, I in the circuit and the internal resistance, r. E = 3.0 V, R = 10 Ω, V = 2.5 V Calculate current : V = IR Calculate internal resistance : E = I(R + r) r = 2.0 Ω3 A simple circuit consisting of a 2 V dry cell with an internal resistance of 0.5Ω. When the switch is closed, the ammeter reading is 0.4 A. Calculate (a) the voltmeter reading in open circuit The voltmeter reading = e.m.f. = 2 V (b) the resistance, R (c) the voltmeter reading in closed circuit E = I(R + r) V = IR 2 = 0.4(R + 0.5) = 0.4 (4.5) R = 4.5 Ω = 1.8 V4 Find the voltmeter reading and the resistance, R of the resistor. E = V + Ir e.m.f. 12 = V + 0.5 (1.2) V = 11.4 V V = IR 11.4 = 0.5 (R) R = 22.8 Ω - 39 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity5 A cell of e.m.f., E and internal resistor, r is connected to /V a rheostat. The ammeter reading, I and the voltmeter reading, V are recorded for different resistance, R of the 6 rheostat. The graph of V against I is as shown. 2 From the graph, determine /A 2 a) the electromotive force, e.m.f., E b) the internal resistor, r of the cell E = V + Ir r = - gradient Rearrange :V = E - Ir = - (6 - 2) Equivalent : y = mx + c 2 Hence, from V – I graph : E = c = intercept of V-axis =2Ω =6V6 V/V The graph V against I shown was obtained from an experiment. 1.5 a) Sketch a circuit diagram for the experiment 0.2 1/A 5 b) From the graph, determine i) the internal resistance of the battery ii) the e.m.f. of the battery r = -gradient E = c = intercept of V-axis = 0.26Ω = 1.5 V7 A graph of R against 1/I shown in figure was obtained R/Ω from an experiment to determine the electromotive force, 1.3 e.m.f., E and internal resistance, r of a cell. From the graph, determine a) the internal resistance of the cell E = I(R + r) 1 (A ) Rearrange -1 :R= 0.5 I - r, Hence, r = -gradient = -(-0.2) = 0.2Ω - 0.2 b) the e.m.f. of the cell e.m.f. = gradient = 3 V - 40 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity7.5 ELECTRICAL ENERGY AND POWERElectrical Energy1. Energy Conversion(a) battery (b) battery (chemical energy) (chemical energy) current current current current Light and heat Energy Conversion: Energy Conversion: Electrical energy → Light energy Electrical energy → Kinetic + Heat energy energy2. When an electrical appliance is switched on, the flows and the .............................. energy supplied by the source is ................................... to other forms of energy.3. Therefore, we can define electrical energy as : ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ - 41 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityElectrical Energy and Electrical Power1. Potential difference, V across two points is the ............................ dissipated or transferred by a coulomb of charge, Q that moves across the two points.2. Therefore, Potential difference, V = Electrical energy dissipated, E Charge, Q3. Hence, E = VQ4. Power is defined as the rate of energy dissipated or transferred.5. Hence, Power, P = Energy dissipated, E time, t Electrical Energy, E Electrical Power, P From the definition of potential Power is the rate of transfer of electrical energy, difference, V V = VQ P = VQ t Electrical energy converted, E P = VQ E = VQ ; where Q = It t Hence, ; where V = IR E = VI t P = VI 2 Hence, E = I2R ; where I = V P= I R R t Hence, V2 t P = I2 R E= R -1 SI unit : Joule (J) SI unit : Joule per second // J s // Watt(W) - 42 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityPower Rating and Energy Consumption of Various Electrical Appliances1. The amount of electrical energy consumed in a given period of time can be calculated by Energy consumed = Power rating x Time E = Pt where energy, E is in Joules power, P is in watts time, t is in seconds2. The unit of measurement used for electrical energy consumption is the ………………………………………... 1 kWh = 1000 x 3600 J = 3.6 x 106 J = 1 unit3. One kilowatt-hour is the electrical energy dissipated or transferred by a ….. kW device in ……... hour4. Household electrical appliances that work on the heating effect of current are usually marked with, ……………… and …………………..5. The energy consumption of an electrical appliance depends on the ……………… and the………………………., , E = Pt6. Power dissipated in a resistor, three ways to calculate: R= 100Ω, I=0.5 A, P=? R= 100Ω, V=50 W, P=? V=50 V, I=0.5 A, P=? P = I2R P = (V/R)2 R P = I2(V/I) = (0.5)2 100 = V2/R = IV = 25 watts = (50)2 /100 = (0.50)50 = 2500/100 = 25 watts = 25 watts - 43 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityCost of energy Energy Appliance Quantity Power / W Power / kW Time Consumed (kWh) Bulb 5 60 8 hours Refrigerator 1 400 24 hours Kettle 1 1500 3 hours Iron 1 1000 2 hoursElectricity cost: RM 0.28 per kWhTotal energy consumed, E = (0.48 + 9.6 + 4.5 + 2.0) = 16.58 kWh Cost = 16.58 kWh x RM 0.28 = RM 4.64Comparing Various Electrical Appliances in Terms of Efficient Use of Energy1. A tungsten filament lamp changes ...............................to useful ................ energy and unwanted ................energy2. A fluorescent lamp or an ‘energy saving lamp’ produces less heat than a filament lamp for the same amount of light produced.3. a) Efficiency of a filament lamp : Efficiency = Output power x 100 Input power = 3 x 100 60 = 5% b) Efficiency of a fluorescent lamp and an ‘energy saving lamp’ Efficiency = Output power x 100 Input power = 3 x 100 12 - 44 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityExercise 7.51. How much power dissipated in the bulb? (a) R = 10Ω 5V (b) R = 10Ω R = 10Ω 5V2. V= 15V I R1=2Ω R3=4Ω R2=4ΩCalculate (a) the current, I in the circuit (b) the energy released in R 1 in 10 s. (b) the electrical energy supplied by the battery in 10 s. - 45 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity2. A lamp is marked “12 V, 24 W”. How many joules of electrical energy does it consume in an hour?3. A current of 5A flows through an electric heater when it is connected to the 24 V mains supply. How much heat is released after 2 minutes?4. An electric kettle is rated 240 V 2 kW. Calculate the resistance of its heating element and the current at normal usage.5. An electric kettle operates at 240 V and carries current of 1.5 A. (a) How much charge will flow through the heating coil in 2 minutes. (b) How much energy will be transferred to the water in the kettle in 2 minutes? (c) What is the power dissipated in the kettle? - 46 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity6. An electric kettle is labeled 3 kW, 240 V.(a) What is meant by the label 3 kW, 240 V? The electric kettle dissipates electrical power 3 kW if it operates at 240 V(b) What is the current flow through the kettle?(c) Determine the suitable fuse to be used in the kettle. 12 A(d) Determine the resistance of the heating elements in the kettle.7. Table below shows the power rating and energy consumption of some electrical appliances when connected to the 240 V mains supply. Appliance Quantity Power rating / W Time used per day Kettle jug 1 2000 1 hour Refrigerator 1 400 24 hours Television 1 200 6 hours Lamp 5 60 8 hours Electricity cost: RM 0.218 per kWh Calculate (a) Energy consumed in 1 day Energy consumed = Quantity x Power rating (kW) x Time used Kettle jug, =1x2x1 = 2 kWh Refrigerator = 1 x 0.4 x 24 = 9.6 kWh Television = 1 x 0.2 x 6 = 1.2 kWh Lamp = 5 x 0.06 x 8 = 2.4 kWh Total energy consumed = 15.2 kWh - 47 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity (b) How much would it cost to operate the appliances for 1 month? Cost = 16.58 kWh x 30 x RM 0.218 = RM 108.438. A vacuum cleaner consumes 1 kW of power but only delivers 400 J of useful work per second. What is the efficiency of the vacuum cleaner?9. An electric motor is used to lift a load of mass 2 kg to a height 5 m in 2.5 s. If the supply voltage is 12 V and the flow of current in the motor is 5.0 A, calculate (a) Energy input to the motor (b) Useful energy output of the motor (c) Efficiency of the motor - 48 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityReinforcement Exercise Chapter 7Part A: Objective Questions1. What is the unit of electric charge? 3. Which of the following graphs shows A. Ampere, A the correct relationship between the B. kelvin,K potential difference, V and current, I C. Coulomb, C for an ohmic conductor? D. Volt, V A.2. Which of the following diagrams shows the correct electric field? A. B. B. C. C. D. - 49 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity4. A small heater operates at 12 V, 2A. How much energy will it use when it is C. run for 5 minutes? A. 90 J B. 120 J C. 1800 J D. 7200 J D.5. The electric current supplied by a battery in a digital watch is 3.0 x 10-5 A. What is the quantity of charge that flows in 2 hours? A. 2.5 x 10-7 C B. 1.5 x 10-5 C 7. Why is the filament made in the C. 6.0 x 10-5 C shape of a coil? D. 3.6 x 10-3 C A. To increase the length and produce E. 2.2 x 10-1 C a higher resistance. B. To increase the current and produce6. Which of the following circuits can be used to determine the resistance of the more energy. bulb? C. To decrease the resistance and A. produce higher current D. To decrease the current and produce a higher potential difference 8. Which of the following will not affect the resistance of a conducting wire. B. A. temperature B. length C. cross-sectional area D. current flow through the wire - 50 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity 9. The potential difference between two 12. Which two resistor combinations have points in a circuit is the same resistance between X and Y? A. the rate of flow of the charge from one point to another B. the rate of energy dissipation in moving one coulomb of charge from one point to another C. the work done in moving one coulomb of charge from one point to another D. the work done per unit current flowing from one point to another A. P and Q B. P and S10. An electric kettle connected to the C. Q and R 240 V main supply draws a current D. R and S of 10 A. What is the power of the E. kettle? A. 200 W B. 2000 W C. 2400 W D. 3600 W E. 4800 W 13. In the circuit above, what is the11. An e.m.f. of a battery is defined as ammeter reading when the switch S is turned on? A. the force supplied to 1 C of charge A. 1.0 A B. the power supplied to 1 C of charge B. 1.5 A C. the energy supplied to 1 C of C. 2.0 A charge D. 9.0 A D. the pressure exerted on 1 C of E. 10.0 A charge - 51 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity 14. A 2 kW heater takes 20 minutes to 16. An electric motor lifts a load with a heat a pail of water. How much potential difference 12 V and fixed energy is supplied by the heater to current 2.5 A. If the efficiency of the the water in this period of time? motor is 80%, how long does it take 6 A. 1.2 x 10 J to lift a load of 600 N through a 6 B. 1.8 x 10 J vertical height of 4 m C. 2.4 x 106 J A. 20 s D. 3.6 x 106 J B. 40 s E. 4.8 x 106 J C. 60 s D. 80 s 15. All bulbs in the circuits below are E. 100 s identical. Which circuit has the smallest effective resistance? 17. The kilowatt-hour (kWh) is a unit of A. measurement of A. Power B. Electrical energy C. Electromotive force B. C. D. 18. The circuit above shows four identical bulbs to a cell 6 V. Which bulb labeled A, B, C and D is the brightest? - 52 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity 19. A 24 Ω resistor is connected across the terminals of a 12 V battery. 20. Which of the following quantities can Calculate the power dissipated in the be measured in units of JC-1 resistor. A. Resistance A. 0.5 W B. Potential difference B. 2.0 W C. Electric current C. 4.0 W D. 6.0 W E. 8.0 W - 53 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityPart B: Structured Questions1. The figure above shows a graph of electric current against potential difference for three different conductors X, Y and Z. (a) Among the three conductors, which conductor obeys Ohm’s law? Conductor Y (b) State Ohm’s law. The potential difference across a conductor is directly proportional to the current that flows through it, if the temperature and other physical quantities are kept constant. (c) Resistance, R is given by the formula R = V/I. What is the resistance of X when the current flowing through it is 0.4 A? Show clearly on the graph how is the answer obtained. From the graph I against V; resistance, R = reciprocal of gradient, 1/gradient 1 = 0.11 = 9.09 Ω (d) Among X, Y and Z, which is a bulb? Explain your answer. X, because as I increases, the gradient decreases. Hence, the resistance X increases as I increases which is a characteristic of a bulb. - 54 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity2. The figure below shows an electric kettle connected to a 240 V power supply by a flexible cable. The kettle is rated “240 V, 2500 W”. The table below shows the maximum electric current that is able to flow through wires of various diameters. diameter of wire / mm maximum current / A 0.80 8 1.00 10 1.20 13 1.40 15 (a) What is the current flowing through the cable when the kettle is switched on? P = IV I = P/V = 2500 / 240 = 10.4 A (b) Referring to the table above, i. What is the smallest diameter wire that can be safely used for this kettle? 1.20 mm ii. Explain why it is dangerous to use a wire thinner than the one selected in b(i) As resistance is inversely proportional to cross-sectional area, a thinner wire will have a higher resistance thus the wire will become very hot. This could probably cause a fire to break out. - 55 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity (c) State one precautionary measure that should be taken to ensure safe usage of the kettle. Do not operate kettle with wet hands. (d) Mention one fault that might happen in the cable that will cause the fuse in the plug to melt. Short circuit might occur if the insulating materials of the wires in the cable are damaged. - 56 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityPart C: Essay Questions1. Figure 1 shows the reading of the voltmeter in a simple electric circuit Figure 2 shows the reading of the same voltmeter (a) What is meant by electromotive force (e.m.f.) of a battery? (b) Referring to figure (a) and figure (b), compare the state of the switch, S, and the readings of the voltmeter. State a reason for the observation on the readings of the voltmeter. (c) Draw a suitable simple electric circuit and a suitable graph, briefly explain how the e.m.f. and the quantity in your reason in (b) can be obtained. (d) The figure above shows a dry cell operated torchlight with metal casing (i) What is the purpose of the spring in the torchlight? (ii) Why it is safe to use the torchlight although the casing is made of metal? (iii) What is the purpose of having a concave reflector in the torchlight? - 57 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityAnswer1. (a) The work done by a battery to move a unit charge around a complete circuit. (b) - Switch in figure 1 is turned off - Switch in figure 2 in turned on - Reading of voltmeter in figure 1 is higher than in figure 2 - This is due to the presence of an internal resistance in the battery (c) e.m.f = intercept on the v-axis internal resistance = -(gradient of the graph) (d) (i) To improve the contact between the dry cells and the terminals of the torchlight (ii) Current flowing through the torchlight is very small, will not cause electric shock (iii) To converge the light rays to obtain increase the intensity of the light rays projected by the torchlight. - 58 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity 2. A group of engineers were entrusted to choose a suitable cable to be used as the transmitting cable for a long distance electrical transmission through National Grid Network. Four different cables and their characteristic of the cables were given. The length and diameter of all the cables are similar. (a) Define the resistance of a conductor. (b) The table below shows the characteristic of the four cables, A, B, C and D. Maximum load Resistivity / Density / Rate of before breaking/ Ωm kgm-3 expansion N A 0.020 500 2800 Low B 0.056 300 3200 Low C 0.031 400 5600 Medium D 0.085 200 3800 High Base on the above table: (i) Explain the suitability of each characteristic of the table to be used for a long distance electricity transmission (ii) Determine the most suitable wire and state the reason (c) Suggest how three similar bulbs are arranged effectively in a domestic circuit. Draw a diagram to explain your answer. Give two reasons for the arrangement. (d) An electric kettle is rated 2.0 kW. (i) Calculate how long would it take to boil 1.5 kg of water from an initial temperature of 280 C. [specific heat capacity of water = 4200 J kg-1 0C-1] (ii) What is the assumption made in the calculations above? - 59 -
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: ElectricityAnswer2.(a) Resistance is the ratio of potential difference to current flowing in an ohmic conductor. (b) Characteristics Explanations A low resistivity Energy loss during transmission is reduced Max load before Mass or weight reduced. Can be supported by transmission braking is high tower A low density Cable will not slag when it heated during transmission Cable A is chosen because it has low resistivity, high max load before breaking, low density and low expansion rate. (c) (i) If one bulb is burnt the others is still be lighted up (ii) Each bulb can be switch on and off independently (d) (i) Pt = mcθ (2000)(t) = (1.5)(4200)(100-28) t = 226.8 s (ii) No heat is lost to the surroundings and absorbed by the kettle END OF MODULE - 60 -
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