JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity CHAPTER 7: ELECTRICITY 7. 1 CHARGE AND ELECTRIC CURRENT Van de Graaf 1. What is a Van de Graaff generator? Fill in each of the boxes the name of the part shown. A device that produces and store electric charges at high voltage on its dome + + + + roller + + + + + + Metal dome dome rubber belt roller motor -1- JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 2.

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 rubber belt. ? This causes the rubber belt to against the roller and hence becomes charged. ? The charge is then carried by the moving belt up to the metal dome where it is collected. ? A large amount of charge is built up on the dome ? Positive 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 deflection of the pointer of the meter. ? This indicates an electric current flow. + + + + + -2- JPN Pahang Teacher’s Guide Physics Module Form 5 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, + + + + + sparkling occurs. ? An electric current flow.

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 attracts the hair and the hair stand upright. ? This is because of each strand of hair receives positive charges and repels each other. 6.

The flow of electrical charges produces electric current. JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Electric Current 1. Electric current consists of a flow of electrons 2. The more charges that flow through a cross section within a given time, the larger is the current. 3. Electric current is defined as the rate of flow of electric charge 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: where I = electric current I=Q t Q = charge t = time 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) (iv) By rearranging the above formula, Q = ( It / t I t / I ) 4.

If one coulomb of charge flows past in one second, then the current is one ampere. 5. 15 amperes means in each second, 15 coulomb of charge through a cross section of a conductor. 6. In a metal wire, the charges are carried by electrons. 7.

Each electron carries a charge of 1. 6 x 10-19 C. 8. 1 C of charge is 6. 25 x 1018 electrons. JPN Pahang Teacher’s GuidePhysics Module Form 5 Chapter 7: Electricity Electric Field a) An electric field is a region in which an electric charge experiences a force. b) An electric field can be represented by a number of lines indicate both the magnitude and direction of the field c) The principles involved in drawing electric field lines are : (i) electric field lines always extend from a positively-charged object to a negatively-charged object to infinity, or from infinity to a negatively-charged object, (ii) electric field lines never cross each other, (iii) electric field lines are closer in a stronger electric field.Demo 1 : To study the electric field and the effects of an electric field.

Apparatus & materials Extra high tension (E. H. T) power supply (0 – 5 kV), petri dish, electrodes with different shapes (pointed electrode and plane electrode), two metal plates, talcum powder, cooking oil, polystyrene ball coated with conducting paint, thread and candle. Method DEMO A) 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- JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 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- JPN Pahang Teacher’s Guide Physics Module Form 5 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- JPN Pahang Teacher’s GuidePhysics Module Form 5 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 two metal plates. 2. Switch on the E. H. T and displace the polystyrene ball slightly so that it touches one of the metal plates and experiences a repulsive force. ? The ball will then move to the positively charged plate.

When the ball touches the plate, the ball loses some of its negative charges to the plate and becomes positively charged. ? It then experiences a repulsive force. This process continues. -8- JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity EFFECT OF AN ELECTRIC FIELD ON A CANDLE FLAME C) 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 polystyrene ball with a lighted candle. ) Sketch the flame observed when the E. H. T.

is switched on. ? The heat of the flame ionizes the air molecules to become positive and negative charges. ? 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 region where an electric charge experiences a force.

2. Like charges repel each other but opposite charges attract each other. . Electric field lines are lines of force in an electric field. The direction of the field lines is from positive to negative. -9- JPN Pahang Teacher’s Guide Physics Module Form 5 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? = It = Q/t = 5 / 10 = 0. 5 A A charge of 300 C flow through a bulb in every 2 minutes. What is the electric current in the bulb? Q I = = = = It Q/t 300 / 120 2. 5 A Q I 2. 3.

The current in a lamp is 0. 2 A. Calculate the amount of electric charge that passes through the lamp in 1 hour. Q = = = It 0. 2 (60 x 60) 720 C . 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) Q = = = It 0. 8 (60) 2880 C Convert: 1 minute = 60s 1. 6 x 10 -19 C of charge ? 1 electron. Hence, 2880 C of charges is brought by 48 C 1. 6 x 10 -19 C = 3 x 10 20 electrons An electric current of 200 mA flows through a resistor for 3 seconds, what is the (a) (b) a) Q electric charge the number of electrons which flow through the resistor? = = = It 200 x 10-3 (3) 0. 6 C b) 1. 6 x 10 -19 C of charge ? 1 electron.

Hence, 2880 C of charges is 0. 6 C = 3. 5 x 10 18 electrons -19 1.

6 x 10 C – 10 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Ideas of Potential Difference (a) (b) X P Q Y ? ? Pressure at point P is greater than the pressure at point Q ? Water will flow from P to Q when the valve is opened. ? This due to the difference in the pressure of water ? Gravitational potential energy at X is greater than the gravitational potential energy at Y. ? The apple will fall from X to Y when the apple is released. ? This due to the difference in the gravitational potential energy. (c) Similarly, ? Point A is connected to positive terminal ?Point B is connected to negative terminal ? Electric potential at A is greater than the electric potential at B. ? Electric current flows from A to B, passing the bulb in the circuit and lights up the bulb.

? This is due to the electric potential difference between the two terminals. ? As the charges flow from A to B, work is done when electrical energy is transformed to light and heat energy. ? The potential difference, V between two points in a circuit is defined as the amount of work done, W when one coulomb of charge passes from one point to the other point in an electric field.

The potential difference,V between the two points will be given by: A Bulb B W Work V = Quantityof ch arg e = Q where W is work or energy in Joule (J) Q is charge in Coulomb (C) – 11 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Device and symbol ammeter voltmeter connecting wire Cells A V Switch Constantan wire // eureka wire bulb resistance rheostat Measuring 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 JPN Pahang Teacher’s GuidePhysics Module Form 5 Chapter 7: Electricity 1. Name the device used to measure electrical current. An ammeter 1. Name the device used to measure potential difference. A voltmeter 2.

(a) What is the SI unit for current? Amperes 2. (a) What is the SI unit for potential difference? Volts (b) What is the symbol for the unit of current? A (b) What is the symbol for the unit of potential difference? V 3. How is an ammeter connected in an electrical circuit? In series 3. How is an voltmeter connected in an electrical circuit? In parallel 4. The positive terminal of an ammeter is connected to which terminal of the dry cell?Positive 4. The positive terminal of a voltmeter is connected to which terminal of the dry cell? Positive 5.

What will happen if the positive terminal of the ammeter is connected to the negative terminal of the dry cell? The ammeter needle will deflect and show reading below zero. – 13 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Exp 1: To investigate the relationship between current and potential difference for an ohmic conductor. (a) (b) Figure (a) and figure (b) show two electrical circuits. Why do the ammeters show different readings? Why do the bulbs light up with different intensity?Referring to the figure (a) and (b) complete the following table: (a) Inference (b) Hypothesis (c) Aim The current flowing through the bulb is influenced by the potential difference across it. The higher the current flows through a wire, the higher the potential difference across it. To determine the relationship between current and potential difference for a constantan wire. (i) manipulated variable : current, I : potential difference, V : length of the wire // cross sectional area // temperature (d) Variables (ii) responding variable (iii) fixed variable Apparatus / materials JPN Pahang Teacher’s GuidePhysics Module Form 5 Chapter 7: Electricity Method : 1. 2.

Set up the apparatus as shown in the figure. Turn on the switch and adjust the rheostat so that the ammeter reads the current, I= 0. 2 A. 3. 4.

Read and record the potential difference, V across the wire. 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 0. 2 0. 3 0.

4 0. 5 0. 6 0. 7 Volt, V/V 1. 0 1.

5 2. 0 2. 5 3. 0 3. 5 Analysis of data : Draw a graph of V against I Potential difference, V /V 4. 0 – 3. 0 – 2.

0 – 1. 0 – Current, I /A 0. 2 0. 4 0.

6 0. 8 – 15 – JPN Pahang Teacher’s GuidePhysics Module Form 5 Chapter 7: Electricity 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? The gradient ? the ratio of V I is a constant as current increases.

2. The resistance, R, of the constantan wire used in the experiment is equal to the gradient of the V-I graph. Determine the value of R. 3. 5 o. 7 =5? 3.What is the function of the rheostat in the circuit? It is to control the current flow in the circuit Conclusion : 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. – 16 – JPN Pahang Teacher’s Guide Physics Module Form 5 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 or V = constant I (c) The constant is known as resistance, R of the conductor.

(d) The resistance, R is a term that describes the opposition experienced by the electrons as they flow in a conductor. It is also defined as the ratio of the potential difference across the conductor to the current, I flowing through the conductor. That is V R= I and V= IR (e) The unit of resistance is volt per ampere (V A-1) or ohm (? ) (f) An ohmic conductor is one which obeys Ohm’s law, while a conductor which does not obey Ohm’s law is known as a non-ohmic conductor JPN Pahang Teacher’s Guide Physics Module Form 5 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) the length of the conductor b) the cross-sectional area of the conductor c) type of material of the conductor d) the temperature of the conductor 5.Write down the relevant hypothesis for the factors affecting the resistance in the table below. Factors Diagram Hypothesis The longer the conductor, the higher its resistance Resistance is directly proportional to the length of a conductor Graph Length of the conductor, l The cross-sectional area of the conductor, A The bigger the cross-sectional area, the lower the its resistance Resistance is inversely proportional to the crosssectional area of a conductor The type of the material of the conductor Different conductors with the same physical conditions have different resistance The temperature of the conductorThe higher temperature of conductor, the higher the resistance – 18 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 6.

From, the following can be stated: ? Resistance of a conductor, ? Resistance of a conductor, R R ? ? length 1 cross-sectional area ? Hence, resistance of a conductor, R ? length cross-sectional area Or R? l A or R= ? l A where ? = resistivity of the substance – 19 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Exercise 7. 2 1. Tick (v) the correct answers True (a) Unit of potential difference is J C-1 (b) J C-1 ? olt, 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. (d) 2 volt is two joules of work done to move 2 coulomb of charge from one to another in an electric field.

v v v v v False (e) Potential difference ? Voltage 2. i) Electric charge, ii) Work done, Q = ( It / I t / ) t I W = (QV / V Q / ) 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 ItV 3. 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 2.

5 V = = = QV 5. 0 (V) 0. 5 V 4. 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 20 V = = = QV 6 (V) 3. 33 V – 20 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 5. 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 J Bulb . When the potential difference across a bulb is 20 V, the current flow is 3 A. How much work done to transform electrical energy to light and heat energy in 50 s? W = = = VIt 20 (3) (50) 3000 J 3A 20 V 7. What is the potential difference across a light bulb of resistance 5 ? when the current that passes through it is 0. 5 A? V = = = IR 0. 5 (5) 2. 5 V 8.

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 3. 0 R = = = IR 2. 0 (R) 1. 5 ? 9.

What is the value of the resistor in the figure, if the dry cells supply 2. V and the ammeter reading is 0. 5 A? V 2. 0 R = = = IR 0. 5 (R) 4? JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 10.

If the bulb in the figure has a resistance of 6 ? , what is the reading shown on the ammeter, if the dry cells supply 3 V? V 3. 0 R = = = IR 6 (R) 0. 5 ? 11. If a current of 0. 5 A flows through the resistor of 3 ? in the figure, calculate the voltage supplied by the dry cells? V = = R = IR 0. 5 (3) 1. 5 ? 12.

The graph shows the result of an experiment to determine the resistance of a wire. The resistance of the wire is From V-I graph, resistance = gradient = 5 = 2. 4? . 2 V/V 1. 2 0 I/A 5 13. 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 = 8 x10? 3 4 = 2.

0 x 10 -3? JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 14. Referring to the diagram on the right, calculate (a) The current flowing through the resistor. V 12 I = = = IR I (5) 2. 4 A I 5? 12 V (b) The amount of electric charge that passes through the resistor in 30 s Q = = = It 2. (30) 72 C (c) The amount of work done to transform the electric energy to the heat energy in 30 s. W = = = QV 72 (12) 864 C or W = VIt = 12(2.

4)(30) = 864 C 15. Figure shows a torchlight that uses two 1. 5 V dry cells. The two dry cells are able to provide a current of 0. 3 A when the bulb is at its normal brightness. What is the resistance of the filament? V 3.

0 I = = = IR 0. 3(R) 10? + 1. 5 V – + 1. 5 V – 16. 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 JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity V/V X 8 Y 2 0 0 2 17. The graph shows the relationship between the potential difference, V and current, I flowing through two conductors, X and Y. a) Calculate the resistance of conductor X.

From V-I graph, resistance = gradient =2 = 4? b) Calculate the resistance of conductor Y. From V-I graph, resistance = gradient =2 = 1? 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. 2 8 I/A R ? = A = ?l RA l 4( 5 . 0 x10 ? 6 ) = 1 .

2 = 1. 67 x 10-5? m 18.The graph shows a graph of I against V for three conductors, P, Q and R. i) Compare the resistance of conductor P, Q and R. Q I/A P Q Rp > RQ >Rr R ii) Explain your answer in (a) From V-I graph, resistance = gradient The greater the gradient, the greater the resistance Gradient of P > Gradient of Q > Gradient of R V/V JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 19. Figure shows a wire P of length, l with a crosssectional 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 Conductor Q R R’ = A ? l ? l’ = A’ (notes: P and R have the same resistivity, ? ) = ? ( 3l ) 2A 3 = 2R 20.

PQ, is a piece of uniform wire of length 1 m with a resistance of 10?. Q is connected to an ammeter, a 2 ? 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 ? 20 Hence, 20 cm = 100 (10) R = 2? Total resistance 2? + 2? = 4? Current, I = R = 4 3 V = 0.

75 A JPN Pahang Teacher’s Guide Physics Module Form 5 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 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 7.

3 SERIES AND PARALLEL CIRCUITS Current Flow and Potential Difference in Series and Parallel Circuit SERIES CIRCUIT PARALLEL CIRCUIT 1 the current flows through each bulb/resistor is the same I = I1 = I2 = I3 1 the potential difference is the same across each bulb/resistor V = V1 = V2 = V3 2 the potential difference across each bulb / resistor depends directly on its resistance.The potential difference supplied by the dry cells is shared by all the bulbs / resistors. 2 the current passing through each bulb / resistor is inversely proportional to the resistance of the resistor. The current in the circuit equals to the 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 battery I = I1 + I2 + I3 where I is the total current from the battery 3 If Ohm’s law is applied separately to each bulb / resistor, we get : V = V1 + V2 + V3 IR = IR1 + IR2 + IR3 If each term in the equation is divided by I, we get the effective resistance If Ohm’s law is applied separately to each bulb / resistor, we get : I = I1 + I2 + I3 V V V V R = R1 + R2 + R3 If each term in the equation is divided by V, we get the effective resistance 1 1 1 R = R1 + R2 + 1 R3 R = R1 + R2 + + R3 JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Identify series circuit or parallel circuit (a) Series (b) Parallel (c) A, B – series (d) Q, S – parallel Ammeter reading ? Current 0.

5 1 3 3 2 5 2 1. 5 Voltmeter reading ? Potential difference ? Voltage 5 1 6 2 2 3 JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: ElectricityEffective resistance, R (a) R = 20 + 10 + 5= 35 ? (b) 1/R = ? +1/5 + 1/10 = 4/5 Effective R = 1. 25 ? (c) 1/R = 1/8 + 1/8= 1/8 R=4? Effective R = 20 + 10 + 4 = 34 ? (d) =1/16 + 1/8 + 1/8 =5/16 Effective R = 3. 2 ? 1/R (e) 1/R = 1/4 + 1/2=3/4 R = 1. 33 ? Effective R = 1.

33 + 1 = 2. 33 ? (f) 1/R = 1/4 + 1/12=1/3 R=3? Effective R = 3 + 2 = 5 ? (h) (g) Effective R = 2+5+3+10 = 20 ? 1/R = 1/20 + 1/20=1/10 R = 10 ? Effective R = 10 + 10 + 5 =2 5 ? (i) 1/R = 1/5 + 1/10=3/10 R = 3. 33 ? (j) 1/R = 1/10 + 1/10=2/10 R=5? – 29 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: ElectricitySolve problems using V = IR V = IR 9 =I(18) = 0. 5? V = IR 240 = 6(R) I =40 A 1/R = 1/5 + 1/20=1/4 R=4? Effective R = 1 + 4 = 5 ? V = IR = 2(5) = 10 V 1/R = 1/10 + 1/10 =2/10 R=5? Effective R = 1 + 4 = 5 ? V = IR 12 =I(5) = 2.

4 A Exercise 7. 3 1. The two bulbs in the figure have a resistance of 2? and 3? 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 ? (b) the main current, I in the circuit V = IR 2. 5 =I(5) = 0.

5 A (c) the potential difference across each bulb. 2? : V = IR = (0. 5)(2) = 1V 3? : V = IR = (0. )(3) = 1.

5 V There are two resistors in the circuit shown. Resistor R1 has a resistance of 1?. 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) R 2 = 5? 2. JPN Pahang Teacher’s Guide 3. Physics Module Form 5 Chapter 7: Electricity The electrical current flowing through each branch, I1 and I2, is 5 A. Both bulbs have the same resistance, which is 2?. Calculate the voltage supplied. Parallelcircuit;V =V1=V2 = IR1 or = IR2 = 5(2) = 10 V 4.

The voltage supplied to the parallel is 3 V. R1 and R2 have a resistance of 5? and 20?. 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=4? (c) the main current, I in the circuit V = IR 3 =I(4) = 0. 75 A 5. In the circuit shown, what is the reading on the ammeter when switch, S (a) is open? Effective R = 6 ? V = IR 12 =I(6) I=2A 6.

Determine the voltmeter reading. (a) R = 12 ? I = 24/12 = 2A V= IR = (2)(8) = 16 V (b) R = 12 ? I = 6/12 = 0. 5A V at 9? : V= IR = (0. 5)(9) = 4. 5 V V reading : 6 – 4. 5 = 1. 5 V R =9 ? I = 4.

5/9 = 0. 5A A reading : 0. 5/2= 0. 25 A Notes: Divide 2 because the resistors have similar resistance. b) is closed? Effective R = 4 ? V = IR 12 =I(4) I=3A Determine the ammeter reading.

(a) (d) the current passing through each resistor 5? : V = IR 3 =I(5) I = 0. 6 A 20 ? : V = IR 3 =I(20) I = 0. 15 A JPN Pahang Teacher’s Guide 7. Physics Module Form 5 Chapter 7: Electricity Calculate (a) The effective resistance, R R = 12 ? (b) The main current, I I=2A (c) The current passing through 8? and 2. 5? resistors.

I=2A (d) (i) The potential difference across 8? resistor. V = IR = 2(8) = 16 V (ii) The potential difference across 2. 5? resistor. V = IR = 2(2. 5) = 5 V (e) The current passing through 6 ? resistor. V = V8 + V2.

+Vparallel 24 = 16 + 5 + Vparallel Vparallel = 3V V = IR 3 = I(6) I = 0. 5 A – 32 – JPN Pahang Teacher’s Guide 8. Physics Module Form 5 Chapter 7: Electricity 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. Suggested answer ? ? ? ? ? ? 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 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 7. 4 ELECTROMOTIVE FORCE AND INTERNAL RESISTANCE Electromotive force Figure (a) Figure (b) Voltmeter reading, e. m. f. Voltmeter reading, potential difference, V < e. m. f.

, E,r R No current flow Current flowing 1.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. JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 2.

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 = IR + Ir where VR = IR and Vr = Ir = I (R + r) – 35 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 3. a) 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 b) Determine the value of the internal resistance.

Since E = V + Ir 1. 5 = r = 1. 1 + 0. 8 ? 0. 5 r Therefore, the value of the internal resistance is 0. 8 ? ) Determine the value of the external resistor.

Since V = IR 1. 1 = R = 0. 5 R 2.

2 ? ? Therefore, the value of the external resistance is 2. 2 – 36 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Activity : To determine the values of the electromotive force (e. m. f. ) and the internal resistance, r of the cell Voltmeter V Internal resistance + Ammeter Dry cell Switch Rheostat To determine the values of the electromotive force (e.

m. f. ) and the internal resistance, r of the cell Dry cells holder, ammeter (0 – 1 A), voltmeter(0 – 5 V), rheostat (0 – 15 ? ), connecting wires, switch, and 2 pieces of 1. V dry cell. : a) Set up the circuit as shown in the figure. Aim Apparatus / materials Method 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 0. 2 0. 3 0.

4 0. 5 0. 6 0. 7 Volt, V/V 2. 6 2. 5 2. 4 2. 2 2.

0 1. 9 – 37 – JPN Pahang Teacher’s Guide Analysis of data : Potential difference, V /V Physics Module Form 5 Chapter 7: Electricity Draw a graph of V against I . 0 – 2. 0 – 1. 0 – Current, I /A 0. 2 Discussion : 0. 4 0. 6 0.

8 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 ? 38 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Exercise 7.

4 1 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 ? 2. 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 ? and the potential difference across it is 2. V, find the value of the current, I in the circuit and the internal resistance, r.

E = 3. 0 V, R = 10 ? , V = 2. 5 V Calculate current : V = IR Calculate internal resistance : E = I(R + r) r = 2. 0 ? 3 A simple circuit consisting of a 2 V dry cell with an internal resistance of 0. 5?.

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 E = I(R + r) 2 = 0. 4(R + 0.

5) R = 4. 5 ? (c) the voltmeter reading in closed circuit V = IR = 0. 4 (4. 5) = 1. V Find the voltmeter reading and the resistance, R of the resistor. E = V + Ir 12 = V + 0. 5 (1. 2) V = 11.

4 V e. m. f. V = IR 11. 4 = 0. 5 (R) R = 22. 8 ? 4 – 39 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 5 A cell of e. m.

f. , E and internal resistor, r is connected to a rheostat. The ammeter reading, I and the voltmeter reading, V are recorded for different resistance, R of the rheostat.

The graph of V against I is as shown. From the graph, determine 2 2 b) the internal resistor, r of the cell r = – gradient = – (6 – 2) 2 =2? /A 6 /V a) the electromotive force, e. . f. , E E = V + Ir Rearrange Equivalent :V = E – Ir : y = mx + c Hence, from V – I graph : E = c = intercept of V-axis =6V 6 The graph V against I shown was obtained from an experiment.

V/V 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 r = -gradient = 0.

26? ii) the e. m. f. of the battery E = c = intercept of V-axis = 1. 5 V 7 A graph of R against 1/I shown in figure was obtained R/? 1.

3 from an experiment to determine the electromotive force, e. m. f.

, E and internal resistance, r of a cell.From the graph, determine a) -1 the internal resistance of the cell E = I(R + r) 0. 5 – 0. 2 1 (A ) Rearrange I :R= E – r, I Hence, r = -gradient = -(-0. 2) = 0. 2? b) the e. m. f.

of the cell e. m. f. = gradient = 3 V JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 7. 5 ELECTRICAL ENERGY AND POWER Electrical Energy 1. Energy Conversion battery (chemical energy) battery (chemical energy) (a) (b) current current current current Light and heat Energy Conversion: Electrical energy ? Light energy + Heat energy Energy Conversion: Electrical energy ? Kinetic energy 2.

When an electrical appliance is switched on, the current flows and the electrical energy supplied by the source is transformed to other forms of energy. 3. Therefore, we can define electrical energy as : The energy carried by electrical charges which can be transformed to other forms of energy by the operation of an electrical appliance. – 41 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Electrical Energy and Electrical Power 1. Potential difference, V across two points is the energy,E dissipated or transferred by a coulomb of charge, Q that moves across the two points.Potential difference, V = Electrical energy dissipated, E Charge, Q 3.

Hence, E = VQ 4. Power is defined as the rate of energy dissipated or transferred. 5. Hence, Power, P = Energy dissipated, E 2. Therefore, time, t Electrical Energy, E From the definition of potential difference, V Electrical Power, P Power is the rate of transfer of electrical energy, V= E Q P= E t Electrical energy converted, E E = VQ ; where Q = It P = VQ t Hence, E = VI t ; where V = IR P = VI Hence, E = I2R 2 E= V t R ; where I = V R 2 P= I R t Hence, P = I2 R SI unit : Joule (J) SI unit : Joule per second // J s // Watt(W) 1 – 42 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Power Rating and Energy Consumption of Various Electrical Appliances 1.

The amount of electrical energy consumed in a given period of time can be calculated by Energy consumed E = = Power rating x Pt where Time energy, E is in Joules power, P is in watts time, t is in seconds 2. The unit of measurement used for electrical energy consumption is the kilowatt-hour, kWh. 1 kWh = = = 1000 x 3600 J 3.

6 x 106 J 1 unit 3. One kilowatt-hour is the electrical energy dissipated or transferred by a 1 kW device in one hour 4.Household electrical appliances that work on the heating effect of current are usually marked with voltage, V and power rating, P. 5. The energy consumption of an electrical appliance depends on the power rating and the usage time, E = Pt 6. Power dissipated in a resistor, three ways to calculate: R= 100? , I=0. 5 A, P=? P = I2R = (0. 5)2 100 = 25 watts R= 100? , V=50 W, P=? P = (V/R)2 R = V2/R = (50)2 /100 = 2500/100 = 25 watts V=50 V, I=0. 5 A, P=? P = I2(V/I) = IV = (0. 50)50 = 25 watts – 43 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Cost of energyAppliance Quantity Power / W Power / kW Time Energy Consumed (kWh) Bulb Refrigerator Kettle Iron 5 1 1 1 60 400 1500 1000 0. 06 0. 4 1. 5 1. 0 8 hours 24 hours 3 hours 2 hours 0. 48 9. 6 4. 5 2 Total energy consumed, E = (0. 48 + 9. 6 + 4. 5 + 2. 0) = 16. 58 kWh Cost = 16. 58 kWh x RM 0. 28 = RM 4. 64 Comparing Various Electrical Appliances in Terms of Efficient Use of Energy 1. A tungsten filament lamp changes electrical energy to useful light energy and unwanted heat energy 2. A fluorescent lamp or an ‘energy saving lamp’ produces less heat than a filament lamp for the same amount of light produced. . 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 25 % – 44 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Exercise 7. 5 1. How much power dissipated in the bulb? (a) R = 10? P = = = 5V V2 R 5 2 / 10 2. 5 W (b) R = 10? R = 10? 5V P = = = V2 R 5 2 / 10 2. 5 W 2. V= 15V I R1=2? R2 =4? R3=4? Calculate (a) the current, I in the circuit Total resistance, R = (2 + 4 + 4) ? = 10 ? IR V/R 15 / 10 1. 5 A I2Rt (1. )2 (10)(10) 225 J (b) the energy released in R 1 in 10 s. E = = = I2Rt (1. 5)2 (2)(10) 45 J V I = = = = = = = (b) the electrical energy supplied by the battery in 10 s. E – 45 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 2. A lamp is marked “12 V, 24 W”. How many joules of electrical energy does it consume in an hour? E = = = Pt 24 (1 x 60 x 60) 86 400 J 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? E = = = VI t 24 (5) (2 x 60) 144 000 J 4. An electric kettle is rated 240 V 2 kW.Calculate the resistance of its heating element and the current at normal usage. P I = = = = IV P/V 2000 / 240 8. 3 A 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. Q = = = It (1. 5) (2 x 60) 180 C (b) How much energy will be transferred to the water in the kettle in 2 minutes? E = = = QV 180 (240) 4. 32 kJ (c) What is the power dissipated in the kettle? P = = = IV 1. 5 (240) 360 W – 46 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 6. 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? P = 3000 = I = IV I (240) 12. 5 A (c) Determine the suitable fuse to be used in the kettle. 12 A (d) Determine the resistance of the heating elements in the kettle. P = 3000 = R = I2 R (12. 5)2 R 19. 2 ? 7. Table below shows the power rating and energy consumption of some electrical appliances when connected to the 240 V mains supply. Appliance Kettle jug Refrigerator Television Lamp Quantity 1 1 1 5 Power rating / W 2000 400 200 60 Time used per day 1 hour 24 hours 6 hours 8 hours Electricity cost: RM 0. 18 per kWh Calculate (a) Energy consumed in 1 day Energy consumed Kettle jug, Refrigerator Television Lamp = Quantity x Power rating (kW) x Time used =1 x2 x1 = 1 x 0. 2 x 6 = 1. 2 kWh = 5 x 0. 06 x 8 = 2. 4 kWh Total energy consumed = 15. 2 kWh = 2 kWh = 1 x 0. 4 x 24 = 9. 6 kWh – 47 – JPN Pahang Teacher’s Guide Physics Module Form 5 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. 43 8. 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?Efficiency = = = Output power x 100 % Input power 400 x 100 % 1000 40 % 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 E = = = VIt 12 (5. 0) (2. 5) 150 J (b) Useful energy output of the motor U = = = mgh 2 (9. 8) (5) 98 J (c) Efficiency of the motor Efficiency = = = Output power x 100 % Input power 98 x 100 % 150 65. 3 % – 48 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity Reinforcement Exercise Chapter 7 Part A: Objective Questions . What is the unit of electric charge? A. Ampere, A B. kelvin,K C. Coulomb, C D. Volt, V 3. Which of the following graphs shows the correct relationship between the potential difference, V and current, I for an ohmic conductor? A. 2. Which of the following diagrams shows the correct electric field? A. B. B. C. C. D. JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 4. A small heater operates at 12 V, 2A. How much energy will it use when it is run for 5 minutes? A. 90 J B. 120 J C. 1800 J D. 7200 J C. D. 5. The electric current supplied by a battery in a digital watch is 3. 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 C. 6. 0 x 10-5 C D. 3. 6 x 10-3 C E. 2. 2 x 10 -1 C 7. Why is the filament made in the shape of a coil? A. To increase the length and produce a higher resistance. 6. Which of the following circuits can be used to determine the resistance of the bulb? A. B. To increase the current and produce more energy. C. To decrease the resistance and 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 B. ire. A. temperature B. length C. cross-sectional area D. current flow through the wire JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 9. The potential difference between two points in a circuit is 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 12. Which two resistor combinations have the same resistance between X and Y? A. P and Q 10.An electric kettle connected to the 240 V main supply draws a current of 10 A. What is the power of the kettle? A. 200 W B. 2000 W C. 2400 W D. 3600 W E. 4800 W 13. In the circuit above, what is the 11. 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 B. the power supplied to 1 C of charge C. the energy supplied to 1 C of charge D. the pressure exerted on 1 C of charge A. 1. 0 A B. 1. 5 A C. 2. 0 A D. 9. 0 A E. 10. 0 A B. P and S C. Q and R D. R and S E. JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 14.A 2 kW heater takes 20 minutes to heat a pail of water. How much energy is supplied by the heater to the water in this period of time? A. 1. 2 x 10 J B. 1. 8 x 10 J C. 2. 4 x 106 J D. 3. 6 x 106 J E. 4. 8 x 106 J 6 6 16. An electric motor lifts a load with a potential difference 12 V and fixed current 2. 5 A. If the efficiency of the motor is 80%, how long does it take to lift a load of 600 N through a vertical height of 4 m A. 20 s B. 40 s C. 60 s D. 80 s 15. All bulbs in the circuits below are identical. Which circuit has the smallest effective resistance? A. E. 100 s 17. The kilowatt-hour (kWh) is a unit of 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? JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity E. 8. 0 W 19. A 24 ? resistor is connected across the terminals of a 12 V battery. Calculate the power dissipated in the resistor. A. 0. 5 W B. 2. 0 W C. 4. 0 W D. 6. 0 W 20. Which of the following quantities can be measured in units of JC-1 A. Resistance B. Potential difference C. Electric current Part B: Structured Questions – 53 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity . Gradient 0 . 6 ? 0 . 2 3 . 6 ? 0 -1 = 0. 11 A V = 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 ? (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. 2. 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”. – 54 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity The table below shows the maximum electric current that is able to flow through wires of various diameters. iameter of wire / mm 0. 80 1. 00 1. 20 1. 40 maximum current / A 8 10 13 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. c) State one precautionary measure that should be taken to ensure safe usage of the kettle. JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity 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. Part C: Essay Questions 1. Figure 1 shows the reading of the voltmeter in a simple electric circuit – 56 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: ElectricityFigure 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) (ii) (iii) What is the purpose of the spring in the torchlight?Why it is safe to use the torchlight although the casing is made of metal? What is the purpose of having a concave reflector in the torchlight? Suggested Answers 1. (a) The work done by a battery to move a unit charge around a complete circuit. – 57 – JPN Pahang Teacher’s Guide Physics Module Form 5 Chapter 7: Electricity (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) Voltmeter