1. P.1 Book 4 Section 3.1 Electrical power and energy Which one is brighter? Electrical power Power rating Check-point 1 Electrical energy Check-point 2 3.1Electrical power and energy

2. P.2 Book 4 Section 3.1 Electrical power and energy Which one is brighter? Connect two light bulbs rated ‘220 V, 60 W’ and ‘220 V, 100 W’ in series. Surprisingly, the 60 W bulb glows brighter!

3. P.3 Book 4 Section 3.1 Electrical power and energy 1 Electrical power Electrical appliances convert energy to other forms of energy. Putting your hands near a lighted bulb, you feel warm. Conductors heat up when current passes through.  heating effect of electric current

4. P.4 Book 4 Section 3.1 Electrical power and energy 1 Electrical power In light bulb, internal energy of filament is non-useful. However, for heaters, irons and boilers, internal energy is useful.

5. P.5 Book 4 Section 3.1 Electrical power and energy 1 Electrical power Power: the rate at which energy is converted from one form to another 1 W = 1 J s –1 unit: watt (W) The power of a battery Example 1 Power = energy time P = EtEt

6. P.6 Book 4 Section 3.1 Electrical power and energy Example 1 The power of a battery 1.5-V Battery maintains a 1-A current. Power of the battery = ? Current of 1 A  charges pass through the battery at 1 C per second. E.m.f. of 1.5 V  1.5 J of energy is imparted to each coulomb of charge. Rate at which battery supplies electrical energy to the circuit = 1  1.5 J per second i.e. power of the battery = 1.5 W

7. P.7 Book 4 Section 3.1 Electrical power and energy 1 Electrical power Power of a source = e.m.f.  current P = e.m.f.  I Similarly, power of a load resistor is given by Power of a load = p.d.  current (P = VI )

8. P.8 Book 4 Section 3.1 Electrical power and energy 1 Electrical power For a load or resistor of resistance R, P = VI = IR  I = I 2 R P = =VI =I 2 R EtEt = V 2 R  or P = VI = V  = VRVR V 2 R

9. P.9 Book 4 Section 3.1 Electrical power and energy 1 Electrical power In each formula, power is expressed in terms of two of the three quantities, V, I and R : Resistors in series Example 2 i.e. Once we know either two of V, I and R, we can find power P.

10. P.10 Book 4 Section 3.1 Electrical power and energy Example 2 Resistors in series Two resistors are connected in series to a battery. (a) Current through each resistor = ? I = VRVR 4.5 3 + 6 = = 0.5 A (b) (i) Power of the 3-  resistor = ? P = I 2 R = 0.5 2  3 = 0.75 W (ii) Power of the 6-  resistor = ? P = I 2 R = 0.5 2  6 = 1.5 W

11. P.11 Book 4 Section 3.1 Electrical power and energy 1 Electrical power Resistors in parallel Example 3

12. P.12 Book 4 Section 3.1 Electrical power and energy Example 3 Resistors in parallel Two resistors are connected in parallel to a battery. (a) Voltages across each resistor = ? (b) Power of each resistor = ? = V 2 R Voltages across each resistor = voltage of the battery = 6 V = 623623 = 12 W Power of the 3-  resistor = V 2 R = 626626 = 6 W Power of the 6-  resistor

13. P.13 Book 4 Section 3.1 Electrical power and energy 2 Power rating  rated value When hairdryer is operated at a voltage of 220 V, it converts 1200 J of electrical energy to heat and KE per second. There are two values marked on electric appliances: voltage rating & power rating

14. P.14 Book 4 Section 3.1 Electrical power and energy 2 Power rating Typical power ratings of some common domestic electrical appliances:

15. P.15 Book 4 Section 3.1 Electrical power and energy 2 Power rating Brightness of light bulbs Example 4

16. P.16 Book 4 Section 3.1 Electrical power and energy Example 4 Brightness of light bulbs Two identical light bulbs, L 1 and L 2 of rating ‘220 V, 100 W’, are connected in series. Ken thinks that the current will  when it flows through the circuit, so L 1 should be brighter than L 2.

17. P.17 Book 4 Section 3.1 Electrical power and energy Example 4 Brightness of light bulbs (a) Is Ken correct? Why? Ken is incorrect. The current is the same throughout the series circuit. ∵ the bulbs are identical ∴ they get the same amount of energy  both bulbs are of the same brightness

18. P.18 Book 4 Section 3.1 Electrical power and energy = 807  = 484  Example 4 Brightness of light bulbs By R =, V 2 P R1 =R1 = 220 2 100 R3 =R3 = 220 2 60 VR1 + R3VR1 + R3 I =I = 220 484 + 807 = = 0.170 A (b) Ken replaces L 2 with L 3 of rating ‘220 V, 60 W’. (i) Find the powers of L 1 and L 3.

19. P.19 Book 4 Section 3.1 Electrical power and energy Example 4 Brightness of light bulbs Power of L 1 = I 2 R 1 = 0.17 2  484 = 14.0 W Power of L 3 = I 2 R 3 = 0.17 2  807 = 23.3 W (b) (ii) Which bulb is brighter? L 3 ( ∵ larger power)

20. P.20 Book 4 Section 3.1 Electrical power and energy = 50  2 = 100 W Check-point 1 – Q1 P.d. across a bulb of resistance R = 50 V current flowing through = 2 A (a) What is the power of the light bulb? P = VI P.d. across other resistance = (b) If e.m.f. of the power supply = 220 V, power of the other resistance = ? 220 – 50 = 170 V Power of other resistance = 170  2 = 340 W

21. P.21 Book 4 Section 3.1 Electrical power and energy Check-point 1 – Q2 The power of a light bulb is 40 W when the p.d. across it is 220 V. Find the resistance of the light bulb. By P =, V 2 R R =R = V 2 P = 1210  = 220 2 40

22. P.22 Book 4 Section 3.1 Electrical power and energy Check-point 1 – Q3 Power dissipated by filament bulb A = 60 W Power dissipated by filament bulb B = 100 W Which bulb is brighter at the rated voltage? Bulb B

23. P.23 Book 4 Section 3.1 Electrical power and energy Check-point 1 – Q4 Resistances of bulbs P, Q and R are 1 , 3  and 5  respectively. Which is the brightest, and which the dimmest? Brightest : Dimmest : P R

24. P.24 Book 4 Section 3.1 Electrical power and energy 3 Electrical energy Electrical energy = power  time (E = Pt ) Using the formula for power, the electrical energy E consumed by or supplied to an appliance in a time interval t is given by E = Pt = VIt = I 2 Rt = V 2 t R Electrical energy consumed Example 5

25. P.25 Book 4 Section 3.1 Electrical power and energy Example 5 Electrical energy consumed A kettle is plugged into a mains supply of 220 V. If the heating element has a resistance of 20 , how much electrical energy is consumed in 5 min? Electrical energy consumed in 5 min = 726 kJ V 2 t R = 220 2  5  60 20 =

26. P.26 Book 4 Section 3.1 Electrical power and energy 3 Electrical energy a Measuring electrical energy Electricity is supplied to our home via kilowatt-hour meter. It measures electrical energy supplied to electrical appliances.

27. P.27 Book 4 Section 3.1 Electrical power and energy a Measuring electrical energy Unit of energy consumption measured by a kilowatt-hour meter: kilowatt-hour (kW h) One kilowatt-hour is the energy transferred in 1 hour in an appliance whose power is 1 kilowatt. 1 kW h = 1000  (60  60) = 3.6  10 6 J = 3.6 MJ Measuring electrical energy with a kilowatt-hour meter Expt 3a

28. P.28 Book 4 Section 3.1 Electrical power and energy Experiment 3a Measuring electrical energy with a kilowatt-hour meter 1.Connect a kettle to the mains socket via a kilowatt-hour meter. Switch on the kettle to boil some water. Record the time and the number of revolutions of the rotating disc on the meter.

29. P.29 Book 4 Section 3.1 Electrical power and energy Experiment 3a Measuring electrical energy with a kilowatt-hour meter 3.1 Expt 3a - Measuring electrical energy with a kilowatt-hour meter 2.Check the calibration on the meter and work out the energy in J supplied to the kettle. 3.Calculate the power of the kettle. Video

30. P.30 Book 4 Section 3.1 Electrical power and energy 3 Electrical energy b Electric bill Electric bills are calculated according to the kilowatt-hour meter reading. Try to calculate the amount you need to pay with your recent electricity bill. Cost of switching on 5 lamps Example 6

31. P.31 Book 4 Section 3.1 Electrical power and energy Example 6 Cost of switching on 5 lamps ‘Electricity’ costs $0.9 per kW h. (a) Find the cost for switching on 5 bulbs each rated 100 W connected in parallel to the mains circuit for 8 hrs. Energy consumed = Pt = 5  0.1  8 = 4 kW h Each unit costs $0.9  Total cost = 4  0.9 = $3.6

32. P.32 Book 4 Section 3.1 Electrical power and energy Example 6 Cost of switching on 5 lamps (b) Find the cost for switching on 5 compact fluorescent lamps (CFLs) each rated 20 W connected in parallel to the mains circuit for 8 hrs. Energy consumed = Pt = 5   0.02  8 = 0.8 kW h Each unit costs $0.9  Total cost = 0.8  0.9 = $0.72 ‘Electricity’ costs $0.9 per kW h.

33. P.33 Book 4 Section 3.1 Electrical power and energy A 0.1 J B 100 J C 48.4 kJ D 360 kJ Check-point 2 – Q1 What is the energy consumed by a bulb of 484  when it is connected to the mains (220 V) and switched on for 1 hour?

34. P.34 Book 4 Section 3.1 Electrical power and energy Check-point 2 – Q2 A 120 J B 120 kW h C 432 kJ D 432 kW h How much energy is consumed in switching on a lamp rated at 60 W for 2 hours?

35. P.35 Book 4 Section 3.1 Electrical power and energy Check-point 2 – Q3 How much does it cost to switch on a TV rated at 150 W for 2 hours? Each unit of ‘electricity’ costs $0.9. Energy consumed = 0.15  2 = 0.3 kW h Total cost = energy  unit cost = 0.3  0.9 = $0.27 = power (in kW)  time (in h)

36. P.36 Book 4 Section 3.1 Electrical power and energy The End