Question Bank. Electric Energy, Power and Household Circuits

Electric Energy, Power and Household Circuits 1. (a) What do you understand by the term electric work? (b) State the SI unit of electric work and define it. (c) Name two bigger units of electric work. How are these units related to the SI unit of electric work? Ans. (a) Electric work is said to be done, when a charge flows through a conductor at some potential difference. (b) The SI unit of electric work is joule. When a current of 1A flows through a conductor at a p.d. of 1 volt for one second, the electric work is said to be 1 joule. (c) (i) Kilojoule = 103 joules, (ii) megajoule = 106 joules.. State expressions for electric work connecting : (i) Current, resistance and time. (ii) Current, potential difference and time. (iii) Potential difference, resistance and time. Ans. (i) W = I.Rt. (ii) W = IVt. (iii) W = V t. R 1

3. (a) State three factors which determine the quantity of heat produced in a conductor. (b) Can heat produced in a conductor be called a measure of electric work? If so, give a reason in support of your answer. Ans.(a) (i) Heat produced in a conductor is directly proportional to the square of magnitude of current. (ii) Heat produced in a conductor is directly proportional to the resistance of the conductor. (iii) Heat produced in a conductor is directly proportional to the time. (b) Yes, heat produced in a conductor is the measure of electric work. It is because, by the law of conservation of energy, energy can neither be created, nor be destroyed, even if it changes its form, from one kind to another kind. 4. (a) What do you understand by the term electric power? (b) State the SI unit of electric power and define it. (c) Name two bigger units of electric power. How are these units related to the S.I units of electric power? Ans. (a) The rate of doing electric work is called electric power. (b) The SI unit of electric power is watt. When a current of 1A flows through a conductor, whose ends are maintained at a p.d of one volt, the power is said be one watt. (c) (i) 1 kilowatt = 103 W (ii) 1 megawatt = 106 W.

5. State expression for electric power connecting (i) Current and resistance. (ii) Current and potential difference. (iii) Resistance and potential difference. Ans.(i) P = IR (ii) P = IV (iii) P = R 6. (a) Name and define the smallest commercial unit of electric energy. (b) Name and define the standard commercial unit of electric energy. V (c) How many joules of energy are there in one kilowatt hour? Ans. (a) The smallest commercial unit of electric energy is watt hour. When a power of one watt is maintained in a conductor for one hour, the energy used is said to be one watt hour. (b) The standard unit of electric energy is kilowatt hour. When a power of one kilowatt is maintained in a conductor for one hour, the energy used is said to be one kilowatt hour. (c) 1 kwh= 3600000 J 7. (a) Distinguish between kilowatt and kilowatt hour. (b) What do the following units measure? (i) coulomb, (ii) kilowatt hour. Ans. (a) Kilowatt is the unit of electric power, whereas kilowatt hour is the unit of electric energy. (b) (i) Coulomb measures the quantity of electric charge. (ii) Kilowatt hour measures the electric energy. 3

8. Explain the meaning of statement the power of an appliance is 100 W. Ans. For the statement 100 W, it implies that a given appliance is using power at the rate of 100 Joules/second. 9. State the SI unit of : (i) electric power, and (ii) electric energy. Ans. (i) SI unit of electric power is watt. (ii) SI unit of electric energy is Joule. 10. (i) Define household unit of electricity. (ii) What is the voltage of electricity that is generally supplied to a house? Ans. (i) The unit of house hold electric energy is kilowatt hour. It is defined as the energy consumed when a power of 1000 W (1 kw) is drawn for one hour or the product of power drawn in watts and time in hours is 1000 Wh. (ii) The voltage of electricity that is generally supplied to a house is 0 V. 11. (a) What is the function of fuse? (b) What requirement is kept in mind while choosing material of fuse? (c) Out of the three fuses available, with current rating 5A, 10A and 13 A, which one should be used in a circuit containing geyser rated 3 kw 50 V? Give one reason. Ans. (a) Fuse is a device which melts and breaks electric circuit when circuit is overloaded or gets short-circuited. (b) (i) The material of fuse should have low melting point, around 00 C. (ii) The material of fuse should have high resistance. 4

(c) Current in geyser, I = P = 3000 W = 1 A. V 50 V Thus, required fuse should be of 13 A. It is because 5A and 10A fuses will melt, as the current in circuit is higher than the fuses can tolerate. 1. (a) Rewrite the following sentence, using correct alternative. A fuse is connected in parallel / series to earth /neutral /live wire. (b) Which of the two cables, one 5 A and other 15 A, will be thicker? Give one reason for your answer. Ans. (a) A fuse is connected in series to live wire. (b) 15 A cable is thicker than 5A cable. It is because thicker the cable, less is its resistance, and hence more the current it can tolerate, without getting hot. 13. (a) Where is a fuse wire placed in an electric circuit? (b) Name the material suitable for making fuse wire. Ans. (a) Fuse is placed in series in live wire, before the switch of an electric appliance. (b) An alloy of 50% lead and 50% tin (fusible alloy) is used for making fuse wire. 14. (a) What do you understand by the term earthing? (b) Which part of an electric appliance is earthed? (c) How does earthing protect user from receiving electric shock? (d) How is a household circuit earthed? (e) Explain how the fuse melts when a short-circuited appliance gets earthed. 5

Ans. (a) Connecting the metallic body of an electric appliance with a thick wire of copper, whose other end is burried deep in earth and is surrounded by the common salt and charcoal is called earthing. (b) The metallic body of electric appliance is earthed. (c) The metallic bodies of electrical appliances are earthed. It is because in case of short circuit, the metal body gets connected with live wire. This in turn gives electric shock, if metal body of appliance is touched. During short circuit the rise in the magnitude of current overloads circuit and hence fuse in that circuit melts. On melting of fuse, the short-circuited appliance will not receive any current and hence user is protected from electrical shock. (d) Refer to answer 14 (a). (e) Refer to answer 14 (c). 15. (a) What is the function of a switch in an electric appliance? (b) Why is switch placed in the live wire? Ans. (a) An electric switch is used to open or close an electric circuit. (b) When switch is in the live wire, no current will enter or leave the appliance, when switch is in off position. Thus, the user will not get electric shock. 16. (a) Why is household wiring done in parallel? Give at least two reasons. (b) What are the disadvantages of wiring a household circuit in series? Ans.(a) (i) In parallel circuit, e.m.f. is a constant quantity. Thus, every appliance works at the same p.d. (ii) In parallel circuit, each appliance can be switched on or off independently. 6

(b) (i) In series circuit, p.d. is not a constant quantity. Thus, different appliances will be at different potential differences. (ii) In series circuit, each appliance cannot be switched on / off independently. 17. Draw the circuit diagram for the distribution of power from poles to main switch and label each part clearly. State the function of each part. Ans. (i) Pole fuse protects the kwh meter during overloading. (ii) kwh meter records the energy consumed. (iii) Mains fuse protects the household circuit during overloading. (iv) Double pole main switch is used to cut off complete electric supply to household circuit during emergency. 18. Name two systems of distribution of power in a house. Give the advantages and disadvantages of these systems. Ans. (a) Tree system, (b) Ring system. Advantages of tree system : The power coming from main circuit is distributed into smaller circuits, with independent fuses. In case of short circuit in some particular sub-circuit, only fuse of that circuit melts. Disadvantages of tree system : 7

1. As all sub-circuits originate from main circuit at the main switch, therefore, longer length of wiring is required.. When a fuse in a particular sub-circuit melts, all appliances in that sub-circuit are disconnected. Advantages of ring system : 1. Every appliance has its own fuse. Therefore, during short circuit, only defective appliance is switched off.. As a single ring feeds the whole house, therefore, length of wiring used is considerably small. 19. Draw a circuit diagram, showing two electric bulbs, a three pin socket and a fan connected in a household circuit. Clearly show the position of fuses; switches and earth. Ans. 8

0. (a) Draw a circuit diagram for the distribution of current in ring system. Show a heater connected in this system. (b) State two advantages of ring system over tree system. Ans. (a) (b) Refer to answer of question 18. 1. (a) State the colour of : (i) live wire, (ii) neutral wire (iii) earth wire according to international connections. (b) State the positions of : (i) earth pin, (ii) neutral wire, (iii) neutral in an electric plug. (c) What is a three core flexible wire? Why short circuit takes place, if more current is made to flow in this wire? (d) How is the difficulty mentioned in (c) overcome? Ans. (a) Old system New system Live wire Red Brown Neutral wire Black Light Blue Earth wire Green Yellow or Green (b) (i) Earth pin is connected to green wire. It is on the top of plug. (ii) Live pin is on right side of plug. It is connected to brown (or red) wire. (iii) Neutral pin is on left side of plug. It is connected to black (or blue) wire. 9

(c) An electric cable having three insulated wires, running parallel to each other is called three core insulated wire. When more current flows through three core wire, the insulation over them melts. Thus, as the live wire comes in contact with neutral or earth wire, a short circuit takes place. (d) A fuse is placed in the live wire. The fuse melts, as soon as circuit gets overloaded. Thus, three core wire does not burn.. Why is the earth terminal in a plug made : (i) thicker? (ii) longer as compared to live or neutral terminals? Ans.(i) The earth pin being thicker can never enter in live or neutral sockets, and hence, electrical appliance can get connected only in proper electrical position. (ii) The earth pin is made longer, so that it gets connected to earth first, before the live or neutral wire. This ensures that user will not get electric shock as the fuse will melt, if the given electric appliance is short circuited. 3. (i) What do you understand by the term earthing with reference to electrical appliances? (ii) Describe briefly, why earthing is essential. (iii) The flexible leads, used for connecting electrical appliances to mains are made up of three separate wires coloured, red; black and green. Which wire is meant for earthing? Ans.(i) Refer to answer 14 (a). (ii) Refer to answer 14 (c). (iii) Green wire. 10

4. Name the marking of three pin plug. Give the colour code of connecting leads. Ans. Refer to answers 1 (a) and (b). 5. (a) Name the colours of three core cable, along with their corresponding electrical connection to socket. (b) Which wire is connected to earth pin? Ans. Refer to answers 1 (a) and (b). 6. There are three pins in an electric plug. Answer the following questions : (i) How can you identify the earth pin? (ii) In which of the three connecting wires should an electric switch be placed? (iii) Explain why an electric switch should not be touched with wet hands. Ans. (i) Earth pin is longer and thicker as compared to other pins. (ii) Live wire. (iii) It is because, natural water generally contains minerals, which makes it conducting. Thus, if moisture on hands comes in contact with terminals of switch, it forms conducting path and hence, one receives a severe shock. 7. (a) Name four devices which make use of only heating effect of current. (b) Name two devices which in addition to heat, produce light energy. Ans.(a) (i) Electric heater (ii) Electric oven (iii) Electric iron (iv) Electric geyser (b) (i) Electric bulb (ii) Arc lamp. 11

8. Two bulbs of 100 W and 5 W are connected in series to 00 V - A.C. mains. It is found that the 5 W bulb glows brightly. Explain the observation. Ans. Resistance of 100W bulb, R100 V 00 00 = = = 400 Ω. P 100 Resistance of 5 W bulb, R5 V 00 00 = = = 1600 Ω. P 5 As both the bulbs are connected in series, therefore the magnitude of current flowing in each bulb is the same. The heat produced and hence light emitted by a bulb is given by the expression I R.t. As resistance of 5W bulb is more than 100W bulb, therefore, it will produce more light. 9. Write whether the following statements are true or false. (i) A wire with blue sleeve connects to live pin of a plug. (ii) A wire with brown sleeve connects to neutral pin of a plug. (iii) A wire with yellow sleeve connects to earth pin of a plug. Ans.(i) False, (ii) False, (iii) True. 30. Fill in the blanks to complete the sentences. (i) A fuse is a piece of wire of low and high. (ii) The colour of insulation sleeve of earth wire is, neutral wire is and live wire is. 1

(iii) Kilowatt-hour is the unit of electrical, whereas kilowatt is the unit of electric. (iv) Energy spent in 1 kilo-volt ampere hour = 5 Amp Volts hour 1000 (v) An electric fuse wire has a low, and is made of an alloy of and. If the current in circuit is too high, the melts, thus stopping and safeguarding the wire and electric appliances from getting. Ans.(i) Small; melting point; resistance. (ii) Green; blue ; brown (iii) Energy; power. (iv) 5 amp 00 volts 1 hour 1000 (v) Melting point; lead and tin; fuse; current; burnt. 31. Explain the following : (i) Wires used as leads of an electric oven are thicker than those used in the leads of a table lamp. (ii) The wires leading current to an electric bulb are thicker, while its filament is made of thin wire. (iii) The heating element of a room-heater becomes red hot, while lead wires remain cold. Ans. (i) The magnitude of current in case of oven is very high, as compared to a table lamp. As heat produced in cables is given by the expression I Rt, therefore in 13

order to lower the value of heat produced in cables the R should be as small as possible. Thus, the heater cables are made thicker, as thicker the cable lesser the resistance. (ii) Heat produced is directly proportional to resistance. As the heat is to be produced in the filament and not in the lead wires, therefore resistance of lead wires is kept less by making them thick, whereas resistance of filament is increased by making it thin. (iii) The magnitude of current is the same in the lead wires as well as heating element. As the resistance of heating element is very high as compared to lead wires, therefore heat produced in heating element is very large, as compared to lead wires. Thus, heating element gets red hot, but not the lead wires. 3. (a) Name the material used for making fuse wire. State two properties of material of fuse wire which make it suitable for use. (b) What is meant by earthing of an electrical appliance? How does earth offer protection? Ans. (a) Fusible alloy containing 50% lead and 50% tin is used as material for fuse wire. Properties of fuse. 1. It has a low melting point, around 00 C.. It offers more electrical resistance than connecting wires. (b) Refer to answers 14(a) and 14(c). 14

33. Why is the fuse wire fitted in a porcelain casing? Ans. Porcelain is (i) bad conductor of electricity, (ii) fire resistant. Thus, when fuse wire melts due to overloading or short circuiting, the circuit does not catch fire. 34. For earthing an electrical appliance, one has to remove paint from the metal body of the appliance, where electrical contact is made. Explain the reason. Ans. Paint is a bad conductor of electricity. Thus, in order to have a proper electrical contact between the metal body and the earth wire, we have to remove the nonconducting paint from that point. 35. Explain the meaning of the statement, the current rating of a fuse is 5A. Ans. From the above statement, it implies that fuse wire can withstand a current strength of 5A. If the current rating exceeds 5A, then circuit will get over loaded and the fuse wire will melt. 15

Numerical Problems. 1. Calculate the energy released by a heater, which draws a current of 5 A at 0 V for 1 minute. Ans. I = 5A; p.d. = 0 V; t = 1 min = 60 s. Energy released = I Vt = 5 0 60 = 66000 J.. An electric appliance consumes 4500 J of energy in 30 minutes, while operating at 4V. Calculate the current drawn from cell. Ans. Energy = 4500 J; t = 30 min = 1800 s; p.d. = 4 V; I =? E 4500 I = = = 0.104 A. Vt 4 1800 3. An electric iron is rated 750 W-50 V. Calculate the energy consumed by iron in 16 hours. Ans. P = 750 W; t = 16 hrs; p.d = 50 V; Energy consumed = P t = 750 W 16 h = 1000 Wh = 1 kwh. 4. An electric appliance having a resistance of 00 Ω. is operated at 00 V. Calculate the energy consumed by appliance in 5 minutes. Ans. R = 00Ω; p.d. = 00 V; t = 5 min = 300 s; E =? V. t 00 00 300 E= = = 60,000J. R 00 16

5. A bulb is joined to a battery of e.m.f. 6V. A steady current of 0.5 A flows through circuit. Calculate the total energy provided by battery in 5 minutes. Ans. e.m.f = 6 V; I = 0.5 A; t = 5 min = 300 s; E =? E = IVt = 0.5 6 300 = 900 J. 6. An electric bulb is rated 100 W -0V. It is connected to 0 V supply. Calculate : (i) Resistance of filament (ii) current flowing through the filament. Ans. P = 100 W; p.d. = 0 V; R =? I =? (i) (ii) V 0 0 R= = = 484W. P 100 P 100 I= = = 0.45 A. V 0 7. An electric kettle draws a current of 4A for.5 minutes. If the resistance of heating element is 100 Ω, calculate the electric energy drawn by the kettle in kilojoules. Ans. I = 4 A; t =.5 min = 150 s; R = 100 Ω; E =? E = IR.t = (4) 100 150 = 1600 150 = 40,000 J = 40 kj. 17

8. A soldering iron draws an energy of 4500 J in 4 minutes when current flowing through its element is 6 A. Calculate the resistance of its heating element. Ans. E = 4500 J; t = 4 min = 40 s; I = 6 A; R =? E=I R. t R= I 4500 R= 6 6 40 E t = 0.5 Ω. 9. Calculate the heat energy given out by the filament of an electric bulb in 0 s, when resistance of it is 4 Ω and p.d. across its terminals is 1V. Ans. E =? t = 0 s; R = 4 Ω; p.d. = 1 V. V t 1 1 0 E= = = 70J R 4 10. An electric appliance gives out 66000 J of heat energy in 1 minute, when current flows through it, at a p.d. of 4 V. Find the resistance of the device. Ans. E = 66,000 J; t = 1 min = 60 s; p.d. = 4 V; R =? R= E= V R t V t 4 4 60 = = 0.5 Ω. E 66000 18

11. An electric heater draws a current of 3.5 A at a p.d. of 50 V. Calculate the power of 4 such heaters. Ans. I = 3.5 A; p.d. = 50 V; No. of heaters = 4 Power of one heater, P = I V = 3.5 50 = 875 W Power of four heaters = 4 875 = 3500 W. 1. An electric bulb is rated 500 W-00V. Calculate the magnitude of the current. Ans. P = 500 W; p.d. = 00 V; I =? P = I V 500 = I 00 I =.5 A. 13. An electric heater of power 1000 W, draws a current of 4.5 A. Calculate the line voltage. Ans. P = 1000 W; I = 4.5 A; p.d. =? V = P I = 1000 4.5 =. volts. 14. An electric heater has a resistance of 40 Ω and draws a current of 4 A. Calculate : (i) its power (ii) p.d. at its ends. Ans. R = 40 Ω; I = 4 A; P =?; V =? (i) P = I.R = (4) 40 = 16 40 = 640 W. (ii) V = I.R. = 4 40 = 160 V. 19

15. An electric heater of power 1600 W, has a resistance of 36 Ω. Calculate the magnitude of current and p.d. at its ends. Ans. P = 1600 W; R = 36 Ω; I =?; p.d. =? P = I.R. I = V = I.R. = 40 6 P 1600 40 = = = 6.66 A. R 36 6 36 = 40 V. 16. An electric motor of power 750 W, operates at 50 V. Calculate the resistance of motor and current flowing through it. Ans. P = 750 W; p.d = 50 V; R =?; I =? R= V 50 50 = = 83.33 Ω. P 750 I = V 750 R 50 = 3 A. 17. An electric device operates at 4 V and has a resistance of 8 Ω.. Calculate the power of device and current flowing through it. Ans. p.d = 4 V; R = 8 Ω; P =?; I =? V 4 4 P= = = 7W. R 8 V 4 I= = = 3A. R 8 0

18. An electric bulb is rated 00W-00V. It is immersed in 00 g of oil (S.H.C. 0.8 Jg 1 C 1) at 10 C. The bulb is switched on for minutes. If all the electric energy is absorbed in the form of heat energy by the oil, calculate : (i) Resistance of filament of bulb, (ii) Current flowing through bulb, (iii) Final temperature. Ans. P = 00 W; p.d. = 00 V; m = 00 g; S.H.C. = 0.8 Jg 1 C 1; Initial temp. = 10 C; t = min = 10 s; R =?; I =? Final temp. =? (i) R = V 00 00 = = 00 Ω. P 00 (ii) I = V = 00 = 1 A. R 00 (iii) Energy flowing through the bulb = P t = 00 10 Heat energy absorbed by oil = mcθr Comparing (i) and (ii) 160 θr = 4000 J = 4000 J (i) = 00 0.8 θr = 160 θr (ii) 1

θr = 150 C Final temp. = Initial temp. + θr = (10Cº + 150º) = 160 C. 19. An electric kettle is rated 1000 W-50V. It is used to bring water at 0 C to its boiling point. If the kettle is switched on for 1 minute, calculate : (i) Resistance of the element of kettle, (ii) Current flowing through element, (iii) Mass of water inkettle. Ans. P = 1000 W; p.d. = 50 V; θr = (100 0) = 80 C; t = 1 min = 60s; c = 4. Jg 1 C 1; R =?; I =? m =? V 50 50 (i) R = = = 6.5 Ω. P 1000 V 50 (ii) I= = = 4A. R 6.5 (iii) Electric energy consumed by kettle = P t = 1000 60 = 60,000 J (i) Heat energy absorbed by water = mcθr = m 4. 80 = 336 m (ii) Comparing (i) and (ii) 336 m = 60,000 m = 60000 = 178.57 g. 336

0. Calculate the resistance of a nichrome wire, which will bring 00 g of water at 0 C to its boiling point in 8 minutes, when current flowing through wire is 4 A. Ans. R =?; m = 00 g; θinitial = 0 C; θfinal = 100 C; θr = (100 0) = 80 C I = 4A; t = 8 min = 480 s Heat absorbed by water = mcθr = 00 4. 80 = 6700 J (i) Electric energy consumed by nichrome wire = I.R.t = 4 4 R 480 = 7680 R (ii) Comparing (i) and (ii) 7680 R = 6700 R = 8.75 Ω. 1. Calculate the current flowing through an electric drill, connected to 00 V supply, if it drills a hole in metal plate of mass 500 g, such that its temperature rises from 10 C to 60 C in 5 minutes, assuming all electric work done is converted into heat energy. [specific heat capacity of metal is 0.6 Jg 1 C 1]. Ans. I =?; p.d. = 00 V; m = 500 g; θr = (60 10) = 50 C; t = 5 min = 300 s S.H.C. = 0.6 Jg 1 C 1. 3

Electric energy consumed by drill = IVt = I 00 300 = 60000 I (i) Heat energy produced = mcθr Comparing, (i) and (ii) = 500 0.6 50 = 15000 J (ii) 60000 I = 15000 I = 0.5 A.. A bulb is joined to a battery of e.m.f. 4 V. and internal resistance of.5 Ω. A steady current of 0.5 A flows through circuit. Calculate : (i) Total energy provided by battery in 10 minutes. (ii) Heat dissipated by bulb in 10 minutes. Ans. e.m.f. = 4 V; r =.5 Ω; I = 0.5 A; t = 10 min = 600 s. (i) Energy provided by battery = IVt = 0.5 4 600 = 100 J. (ii) Let R be the resistance of bulb. I = R E + r 0.5 = 4 R +.5 0.5 R + 1.5 = 4; 0.5 R =.75; R = 5.5 Ω Heat energy dissipated in bulb = I.R.t = 0.5 0.5 5.5 600 = 85 J. 4

3. A battery of 1 V and negligible internal resistance is connected to an external circuit, consisting of three resistors of 6 Ω, 3 Ω and Ω in parallel, which is further connected to a resistance of 3 Ω. The resistance of 3 Ω is immersed in 50 g of oil of S.H.C 0.8 Jg 1 C 1, when the temperature rises by 60 C. (i) Draw the labelled circuit diagram, (ii) Calculate the value of current in main circuit, (iii) Calculate the current flowing in Ω resistor in parallel, (iv) Calculate the time for which current is switched on. Ans. (i) The figure is shown below. (ii) Equivalent resistance of 6 Ω, 3 Ω and Ω in parallel, 1 1 1 1 6 = + + = = 1 R 6 3 6 1 1 R1 = 1Ω Total resistance in external circuit, R = 1 + 3 = 4 Ω. Current in main circuit, I = V = R (iii) p.d. across parallel circuit, 1 4 = 3 A. 5

V = I.R. = 3 1 = 3 V. Current in Ω resistor, I = V = 3 = 1.5 A. R (iv) Let t be the time for which electric circuit is switched on. Electrical energy consumed = I.Rt = (3) 3 t = 7 t (i) Heat energy absorbed by oil = mcθr = 50 0.8 60 Comparing (i) and (ii) = 400 J (ii) 7 t = 400 t = 400 7 = 88.89 s. 6

4. Two bulbs are rated 60 W 0 V and 60 W 110 V. Calculate the ratio of their resistances. Ans. Let R1 be the resistance of bulb working at 0 V and R working at 110 V. V 0 0 R 1 = = ; P 60 R 1 V 110 110 = = P 60 0 0 110 110 R :R = : =4:1 60 60 5. An electric kettle is rated 0 V and can bring certain amount of water to its boiling point in 5 minutes. If it is connected to the voltage supply of 00V, calculate the time in which the same amount of water will reach its boiling point. Ans. At 0 V, V. t 0 0 5 min. E= = (i) R R At 00 V, V. t 0 0 t E= = R R (ii) Comparing (i) and (ii) 00 00 t 0 0 5 min. = R R 0 0 5 t = 00 00 = 6.05 minutes. 7

6. An immersion heating rod is rated 0 V and can bring certain amount of water to its boiling point in 15 minutes. When this immersion rod is actually connected to an electric circuit, it brings the water to boil in 18.15 minutes. Calculate the line voltage. Ans. At 0 V, V. t 0 0 15 E = = (i) R R V. t V 18.15 At voltage V, E = = (ii) R R Comparing (i) and (ii) V 18.15 0 0 15 = R R 0 0 15 V = = 40000 18.15 V = 00 volts. 7. A geyser is rated 1500 W - 50 V. If this geyser is connected to 50 V mains, calculate : (i) Current drawn, (ii) Energy consumed in 50 hours, (iii) Cost of energy consumed in 50 hours at 30 paise per kilowatt hour. Ans. P = 1500 W; p.d = 50 V; I =?; t = 50 h. (i) I = P V = 1500 50 = 6 A. (ii) Energy consumed by geyser in 15 hrs = P t = 1500 W 50 h 8

= 75000 Wh = 75 kwh. 30 paise (iii) Cost of energy = 75 kwh kwh = 50 paise = Rs.50. 8. An electric oven is marked 1000 W 00 V. Calculate: (i) Resistance of its element (ii) Energy consumed by oven in 1 h in joules. (iii) Time in which it will consume 15 kwh of energy. Ans. P = 1000 W; p.d. = 00 V; R =? (i) R = V 00 00 = = 40 Ω. P 1000 (ii) Energy consumed by geyser in 1/ h = P t = 1000 1/ 3600 = 1800000 J. (iii) Energy consumed = P t. 15 kwh = 1000 W t t = 15 hours. 9. A geyser is rated 000 W and operates hours a day on 00 V mains. Calculate the monthly bill for running geyser, when energy costs Rs. 1.90 per kwh. Ans. Power of geyser = 000 W. 9

Energy consumed by geyser in a day for h = P t = 000 = 4000 Wh Energy consumed in a month = 30 4000 Wh = 10,000 Wh = 10 kwh. Monthly bill = 10 kwh Rs1.90 kwh = Rs. 8. 30. An electric oven of resistance 0 Ω, draws a current of 10 A. It works 3 hours daily. Calculate the weekly bill, when energy costs Rs. 1.60 per kvah. Ans.Power of electric oven = I.R = (10) 0 = 000 W. Energy consumed by heater @ hr a day = 000 W h = 4000 Wh. Energy consumed by heater in a week = 4000 7 = 8000 Wh = 8 kwh = 8 kvah Monthly bill = 8 kvah Rs1.6 kvah = Rs. 44.80. ( 1 kwh = 1 BOT) 30

31. An electric bulb draws a current of 0.8 A and works on 50 V, on the average 8 hours a day. If the energy costs Rs. 1.50 per board of trade unit, calculate the monthly bill. Ans. Power of bulb = IV = 0.8 50 = 00 W. Energy consumed by bulb @ 8 hr a day = 00 W 8h = 1600 Wh. Energy consumed by bulb in a month = 1600 30 = 48000 Wh = 48 kwh. Monthly bill = 48 kwh Rs1.5 kwh = Rs. 7.00. 3. 4 tubelights of 40 W each and fans of 100 W each are connected to 00V mains and operate on the average 8 hours a day. If the energy costs Rs. 1.50 per kwh, calculate: (i) Monthly bill, (ii) Minimum fuse rating. Ans.(i) Power of four tubelights = 4 40 = 160 W. Power of two fans = 100 = 00 W. Total power of all appliances = 160 + 00 = 360 W. Energy consumed by all appliances @ 8 hr a day = 360 8 = 880 Wh. Energy consumed by all appliances in a month 31

= 880 30 = 86400 Wh = 86.40 kwh. Monthly bill = 86.40 kwh Rs1.50 kwh = Rs. 19.60. (ii) Minimum fuse rating = Total power of appliances p.d. = 360 00 = 1.8 A. 33. A boys hostel has following appliances, when energy is supplied at 00 V and costs Rs. 1.75 per kilowatt hour. (a) 40 bulbs of 100W each, working 8 hours a day, (b) 0 fans, each drawing a current of 0.8 A and working 15 hours a day. Calculate: (i) Monthly bill (ii) Amongst the fuses 35A and 37A, which one will you use and why? Ans.(i) Power of 40 bulbs = 40 100 = 4000 W. Energy consumed by 40 bulbs @ 8 h a day = 4000 8 = 3000 Wh. Power of 0 fans = 0 I V = 0 0.8 00 = 300 W. Energy consumed by 0 fans @ 15 h a day = 15 300 = 48,000 Wh Total energy consumed by bulbs and fans in a day = (3,000 + 48,000) 3

= 80,000 Wh. Total energy consumed by bulbs and fans in a month = 30 80,000 = 400000 Wh = 400 kwh. Monthly bill = 400 kwh 1.75 Rs kwh = Rs. 400. (ii) Total power of 40 bulbs and 0 fans = (4000 + 300) = 700 W. Minimum fuse rating = Total power p.d. = 700 00 = 36 A. Thus, 37 A fuse is needed as fuse as 35 A rating is likely to melt, when all appliances are switched on. 34. Power is supplied at 00 V and at the rate of Rs. 1.50 kwh to the following electric appliances. (i) Two T.V. sets, each of resistance 00 Ω and working 4 hours a day. (ii) Two electric motors of 1.5 H.P. each working 4 hours a day. Calculate : (i) Monthly bill, (ii) Minimum fuse rating. Ans.(i) Power of T.V. sets = = V R 00 00 00 = 400 W. 33

Energy consumed by T.V. sets @ 4 hr a day = 400 W 4 h = 1600 Wh. Power of electric motors = 1.5 = 3 HP = 3 750 W = 50 W. Energy consumed by motors @ 4 h a day = 50 W 4 h = 9,000 Wh Total energy consumed by T.V. sets and electric motors in a day = (1600 + 9000) = 10600 Wh Total energy consumed by T.V. sets and motors in a month = 10600 30 = 318000 Wh = 318 kwh. Monthly bill = 318 kwh Rs1.50 kwh (ii) Minimum fuse rating = Rs. 477. = Total power of all appliances p.d. = (400 + 50) W 00 V = 650 00 = 13.5 A. 34

35. Calculate the electric energy in the SI units, consumed by 100 W bulb and 60W fan, connected in parallel for 5 minutes. Ans. Total power of bulb and fan = 100 W + 60 W = 160 W = 160 Js 1 Time = 5 min = 300 s Energy consumed = P t = 160 Js 1 300 s = 48,000 J. 35