Question .1. What do you understand by 1 kilowatt hour?
1 kilowatt hour is the energy consumed by an electric device of power 1 kilowatt when used for 1 hour.
Question .2. An electric bulb has a filament and gases filled at low pressure.
(i) What is the material commonly used for the filament?
(ii) Name the gases used for filling the bulb.
(i) The material of the filament is tungsten.
(ii) The gas used is inert gas argon and traces of nitrogen.
Question .3. Name the colour of the three core cable along with their corresponding electrical connections to a socket.
As per internationally accepted convention brown is for live, light blue for neutral and green or yellow for earth.
Question .4. Which part of an electric appliance is earthed?
The metallic body of an electrical appliance is earthed.
Question .5. Define 1 watt and 1 kilowatt.
(i) 1 watt : 1 watt is the power of an electrical appliance if a current of 1 ampere flows through it under a potential difference of 1 volt.
(ii) 1 kilowatt : 1 kilowatt = 1000 watt.
Question .6. There are three pins in an electric plug top. Answer the following :
(i) How would you identify the earth pin?
(ii) In which of the three connecting wires should the electric switch be connected?
(iii) Explain why a switch should not be touched with wet hands.
(i) The earth pin is thicker and longer than the other two pins.
(ii) The electric switch should be connected in the ‘Live’ wire.
(iii) Wet hand is a good conductor of electricity and there is a chance of getting electric shock. Hence the switch should not be touched with wet hands.
Question .7. Give one example each of material suitable for making (i) fuse wire and (ii) heater element.
(i) Alloy of copper and lead or copper and tin or lead and tin.
Question .8. Score off the incorrect and write correctly the following : In wiring a three pin plug, the brown sleeved wire is connected to the material/live pin and the green/ blue sleeved wire is connected to the metal body of the appliance.
In wiring the three pin plug, the brown sleeved wire is connected to the ‘live’ pin and the green sleeved wire to the metal body of the appliance.
Question .9. You have just paid the electricity bill for your house
(i) what was it that your family consumed, for which you had to pay?
(ii) In what unit was it measured?
(i) My family consumed electrical energy.
(ii) It was measured in kilowatt hour (kWh)
Question .10. What are high tension wires? Give two characteristics of these wires.
High tension wires are wires transmitting electricity at high voltage. The characteristics of these wires are:
(i) They have low resistance and
(ii) are non-corrosive.
Question .11. Why is the filament of an electric iron or kettle put between mica sheets?
Mica is a poor conductor of electricity and provides good insulation. It is good conductor of heat and does not prevent heat to conduct to the base plate.
Question .12. Explain briefly how a fuse protects an electric circuit?
Fuse has low melting point. Whenever the current exceeds the pre determined limit , the fuse melts and the circuits are broken.
Question .13. What is earthing? What is its use?
What is meant by earthing of an electrical appliance? Why is it essential?
Earthing is a process of connecting the metallic body of an appliance to the earth through a conductor. It protects the user of an appliance from electric shock.
Question .14. What is meant by earthing of an electrical appliance? How does earthing offer protection?
Earthing is a process of connecting the metallic body of an appliance to the earth through a conductor.
Protection : When an appliance is properly earthed then any accidental contact of live wire with its metallic body results in a flow of heavy current through the appliance. Earthing results in safe and easy flow of electric current to the earth. Thus user is protected from any fatal electric shock.
Question .15. Name the material used for making fuse wire. State two properties of the material of fuse wire which make it suitable for use.
Material used for making fuse wire : 37% lead and 63% tin.
Properties of fuse wire :
(i) low melting point and
(ii) high resistivity
Question .16. Under what circumstances does one get an electric shock from an electric appliance?
When a live wire come in contact of the metallic body of the electric appliance and then to the body of the user, it results in an electric shock.
Question .17. How does the power rating of a device help us to decide about the type of leads (connecting wires) to be used for that?
As P = V I or I = P / V . Thus we see that current flowing through a device is directly proportional to its power. By knowing the power-voltage rating of a device, we know the current flowing through it and it help to decide about the type of leads to be used. If current flowing is more than 5A we use thick wire otherwise thin wires(leads).
Question .18. Explain briefly the function of the following in the household wiring (i) a three pin plug and (ii) main switch.
(i) In three pin plug, one big plug is connected to the metallic body of the appliance. This plug is connected to the earth and the user of the electrical appliance is protected against accidental electric shocks.
(ii) The main switch is connected at the starting point of the wiring of the house. This enable us to switch off (or on) the supply of electricity according to our need.
Question . 19. Make a table with names of 3 electrical appliances used in your home in one column, their power, voltage rating and approximate time for which each is used in one day in the other column.
|Name of appliance|
Question .20. Which of the two wires of similar dimension, copper or nichrome, would you use for the electric heater. Give reasons to justify your answer.
We should use nichrome wire for making electric heater element. Copper will not be used as its resistivity is very small and will draw very large current resulting in large amount of heat to be produced. This heat may cause fuse to be blown and heater element itself would melt.
Question .21. Two fuse wires of the same length are rated 5 A and 20 A. Which of the two fuse wires is thicker and why?
The fuse wire rated 20 A is thicker of the two. The thicker wire would have low resistance and produce less heat for a given heat. It can, therefore, carry a much larger current through it before melting down.
Question .22. Electric power P is given by the expression P = (Q × V)
(i) What do the symbol Q and V represent?
(ii) Express power in terms of current and resistance explaining the symbols used their in.
Q is electric charge and V is potential difference.
P = I2 R, where I stands for electric current and R for resistance.
Question .23. State the purpose of a fuse in an electric circuit. Name the material used for making a fuse wire.
The fuse helps to control the flow of electric current to the electrical appliances. When the current in that circuit exceeds a maximum value, it melts and saves the electrical / electronic device from the circuit.
Fuse wires are made from an alloy of tin (63%)and lead(37%).
Question .24. In a three-pin plug, why is the earth pin made longer and thicker than the other two pins?
It is made so to ensure that the body of the appliances has a good contact with the earth and it is long to ensure its connection is made first.
Question .25. Draw a labelled diagram of a three-pin socket.
Consult a Text Book.
Question .26. Of the three connecting wires in a household circuit :
(i) Which two of the three wires are at the same potential?
(ii) In which of the three wires should the switch be connected?
(i) Of the three wires, the Earth wire and the Neutral wire are at the same potential of zero.
(ii) The switch be always connected in the Live wire.
Problems for Practice
Numerical – 1. An electric heater is rated 500 kVA, 220 V. If the heater is operated for 1 hour, calculate the energy consumed
(i) in kWh and
(ii) in joule.
Here, P = 500kVA = 500 kW,
(i) Energy consumed in 1 hour = E = 500kW × 1 hour = 500kWh.
(ii) Energy consumed in joule = 500 × 3.6 × 106 J. ( 1 kWh = 3.6 × 106 J)
= 1.8 × 109 J.
Numerical – 2. Convert 1 kWh into joule.
1 kWh = 1 kW × 1 hour = 1000 W × 3600 s = 3.6 × 106 J.
Numerical – 3. An electric bulb is marked 220V – 100 W. The bulb is connected to a 220 V supply.
(i) Calculate (a) the resistance of the filament. (b) Current flowing through the filament.
(ii) is the filament resistance of the glowing bulb greater, smaller or same as compared to its resistance when it is not glowing?
Here, P = 100 W, V = 220 V
(i) Using P = V2/R, we get
R = V2/P = (220)2/100 = 484 Ω
(ii) Current flowing through the filament :
P = V I => I = P / V = 100/220 A = 0.454 A.
(iii) The filament resistance of the ‘on’ bulb is greater because it is at a higher temperature when it is ‘on’ as resistance of metals increases with an increase in temperature.
Numerical – 4. A refrigerator is marked 80 W, 220 V. (i) How much energy does it consume in one day if on an average it is used for 20 hours a day? (ii) What is likely to happen if the voltage drops to 50 V.
Here, P = 80 W, V = 220 V, t = 20 hours, E = ?
(i) E = P × t = (80 × 20) / 1000 kWh = 1.6 kWh.
(ii) If voltage drops a large current will flow through it and damage the refrigerator.
Numerical – 5. What quantity of heat will be produced in a coil of resistance 80 ohm if current of 3 A is passed through it for 4 second.
Here, R = 80 ohm, I = 3 A, t = 4 s.
Using, H = I2 R t = (3)2 × 80 × 4 = 2880 J.
Numerical – 6. Calculate the electrical energy in SI units consumed by a 100 W bulb and a 60 W fan connected in parallel in 5 minutes.
Here, P1 = 100 W, P2 = 60 W, t = 5 × 60 = 300 s,
Total power in parallel = P1 + P2 = 100 + 60 = 160 W = P (say)
E = P × t = 160 × 300 = 48000 J.
Numerical – 7. A family uses a light bulb of 100 W, a fan of 100 W and an electric heater of 1000 W each for 8 hours daily. If the cost of electricity is Rs2.00 per unit, what is the expenditure for the family per day on electricity?
Here, P1 = 100 W, P2 = 100 W, P3 = 1000 W, t = 8 hours,
Cost per kWh = Rs2.00.
Total power = P = P1 + P2 + P3 =100 + 100 + 1000 = 1200 W,
Total electrical energy consumed = P × t = 1200 × 8 = 9600Wh = 9.6 kWh
Total cost = Rs2.00 × 9.6 = Rs19.60.
Numerical – 8. A bulb is marked 100 W, 220 V and an electric heater is marked 2000 W, 220 V
(i) What is the ratio of their resistances?
(ii) In which of the above a thicker connecting wire or lead is required?
(i) Using P = V2 / R we get R = V2 / P
Resistance of the bulb marked 100 W, 220 V we have
RB = (220)2/100 = 484 Ω
Resistance of the heater marked 2000 W, 220 V we have
RH = (220)2/2000 = 24.2 Ω
RB/RH = 484/24.2 = 20/1
(ii) Current flowing through the bulb = 100/220 = 0.454 A and
current flowing through the heater = 2000/220 = 9.08 A = 20 times current through bulb.
Hence, thicker wire or lead is required for the electric heater.
Numerical – 9. A geyser has a label 2 kW, 240 V. What is the cost of running it for 30 minute, if the cost of electricity is Rs 3.00 per unit?
Here, P = 2 kW, V = 240 V, t = 30 minutes = 0.5 h, cost = Rs 3.00 per unit,
E = P × t = 2 kW × 0.5 h = 1 kWh
Cost = 1 × Rs 3.00 = Rs 3.00.
Numerical – 10. An electric kettle is rated 2.5 kW, 250 V. Find the cost of running the kettle for two hours at 60 paisa per unit.
Here, P = 2.5 kW, V = 250 V, t = 2 h, cost per unit = 60 paisa,
E = P × t = 2.5 × 2 = 5 kWh.
Cost = 5 × Rs 0.60 = Rs 3.00.