The heating effect of current increases with the applied voltage across the wire. Microscopically, when a voltage is applied across the ends of a metal wire, an electric field is set up between the ends. The free electrons are accelerated by the field and collide frequently with the metal ions in the wire. Thus, some energy is transformed into the internal energy of the wire. When the applied voltage increases, the electric field is stronger. The electrons collide harder as well as more frequently with the ions.
Thus, more energy is transformed into the internal energy. 2. The heating element is usually wound into a coil to facilitate heat transfer from the heating element to the targeted area. The coiling of the heating element results in a long enough wire that gives a reasonable resistance. This is to protect the source against short circuit. 3. No. The resistance off copper wire is much smaller than that off Monochrome wire of the same length. If we replace the wire with a copper wire, the battery will be short-circuited and become very hot in a short time.
It may even explode. Checkpoint (p. 200) 1. A) The equation defines the average electrical power P as the amount of electrical energy E transformed in a time interval t. It is applicable (but not restricted) to all kinds of electrical energy transformation. (b) The equation defines the electrical power P supplied to or dissipated by an electrical component. When the voltage across the component is V and the current flowing through the component is l, the electrical power P = VI. The equation is applicable to all kinds of electrical components. C) The equation defines the electrical power P supplied by a source. When a source of e. M. F. E delivers a current I too component, the power supplied P = El. The equation is applicable to sources only. (d) The power P dissipated by a resistor of resistance R is given by P = ERR where I is the current flowing through the resistor. The equation is applicable to all resistive components. (e) The power P dissipated by a resistor of resistance R is given by where V is the voltage across the resistor. The equation is applicable to all resistive components. 2. A) Applying , the total current I passing through the circuit is Applying , the power of the 2 Q resistor is Similarly, the power of the 4 Q resistor is ND the power of the 6 Q resistor is Thus, the total power = 2 +4 12 W. (b) Applying , we have the power of the 2 Q resistor, the power of the 6 Q resistor. Thus, the total power = 72 + 36 + 24 Exercise (p. 200) Applying, we have for choice A, the wire resistance; for choice B, the wire resistance; for choice C, the wire resistance; = 132 w. For choice D, since the wire has a radius of (d/2) and a length larger than 21, the resistance.
By , when the same voltage V is applied across the wires, the wire with the smallest resistance has the greatest heating effect. Thus, the answer is choice A. Applying , we have The power supplied by the cell is 1. 2 mm. Let I be the total current delivered by the battery, r be the resistance of each resistor and UP, PC and PR be the power dissipated in resistors P, Q and R respectively. Applying P = ERR, we have UP = err. By symmetry, the currents through resistors Q and R are equal to. Thus, we have the total power. It is given that the total power delivered by the battery is 12 W, we have Therefore, the powder dissipated in R is.
Let R be the total resistance of the circuit. Under a constant supplied voltage E, the power P consumed in the circuit is given by. From the equation, a smaller R will result in a larger P. The smallest total resistance is obtained when the two resistors are connected in parallel to the battery as shown. The smallest total resistance. Thus, the largest power consumed. Since Al and LA are identical and are connected in parallel, they have the same brightness. As the equivalent resistance of Al and LA connected in parallel is smaller than the resistance of LA, the p. D. Across LA is larger than that across Al and LA.
Thus, LA is the brightest. 6. A If the filament of LA is burnt out, the p. D. Across Al and that across LA are the same. Thus, they have the same brightness. Since the equivalent resistance across the two ends of Al increases, the p. D. Across Al increases. As the e. M. F. Of the battery is constant, Al becomes brighter than before while LA becomes dimmer. 7. Let Earthier and Patchier be the resistance and power of the thicker wire while Retainer BY, . (a) If they are connected in series to the supply, the same current I passes through the two wires. By , we have and Therefore, Patchier : Pithiness = 1 : 4. B) If they are connected in parallel to a supply of negligible resistance, the voltages across the two wires are the same. By, we have and Therefore, Patchier : Pithiness = 4 : 1. 8. (a) Applying , the voltage V supplied to the heater is (b) Applying , the operating resistance R of the heater is 9. First, use the electronic balance to find the mass m of the water. Second, use the thermometer to measure the initial temperature of the water. Third, connect the immersion heater to the power supply and immerse the heater into the water. Forth, heat the water for a period of time t with the help of the stop-watch.
Fifth, switch off the heater and use the thermometer to measure the final temperature of the water. Assume that there is no loss of energy to the surroundings. The energy E transferred to the water is given by where c is the specific heat capacity of water and AT is the difference of the final and initial temperatures. Thus, the power output of the heater is given by 10. Let Pl and UP be the power dissipated in RI and RE respectively. Consider RI and RE connected in parallel to the source. Since , by, we have Consider RI and RE connected in series to the source. The same current passes through the resistors.
By, we have Thus, the power dissipated in RI to that in RE=2 : 1. (a) The bulb can provide a desired output power by closing different switches. 11. When only switch A is closed, only 1 resistor is connected to the supply to provide a power of 50 W. When only switch B is closed, another resistor of smaller resistance is connected to the supply to provide a power of 100 W. When both switches are closed, the two resistors are connected in parallel to the supply to provide a desired power of 150 W. (b) When switch A is closed, the resistance of the 50 W filament is When switch B is closed, the resistance of the 100 W filament is (c) Yes.
As the filaments are connected in parallel to the supply, the meltdown of one filament does not affect another. The bulb can still glow. 12. A) Since the operating voltage of the fan is 24 V, the fan should be connected in parallel to the 24 V d. C. Power supply. The switch should be installed in the main branch to resistance R of the heating element is given by (it) Applying P = VI, the current through the heating element is given by . Similarly, the current through the fan is given by Thus, the total current drawn from the supply = 6. 25 + 0. 8333 7. 08 A (c) The two possible circuits are Checkpoint (p. 07) 1. (a) correct (b) incorrect (c) incorrect 2. No. The bulbs may have different power ratings, so they may have different brightness. 3. Electrical appliance rower rating / W operating voltage / V operating current / A operating resistance / Q electric kettle 2000 110 18. 2 6. 05 toaster 1100 220 5 ray box lamp 24 12 2 6 waffle iron 1180 5. 37 41 torch 1. 2 0. 4 7. 5 Checkpoint (p . 210) Statement (3) is incorrect because the cost also depends on the operating time of the appliance. 2. The time taken for the fan to use up 1 k h of electrical energy is 3. The power of the air conditioner is.
The cost of running it per month Exercise (p. 211) The resistance of the heater. When the heater is connected to a 110 V main supply, the power consumed by the heater -? Statement (1) is correct. Applying, the operating resistance of the cooker is Statement (2) is incorrect. Applying P = VI, the operating current of the cooker is Statement (3) is correct. The electrical energy consumed – Statement (1) is correct. From the figures, the power of the hair dryer is twice as large as that of the electric kettle. By P = VI, the current drawn by the hair dryer is also twice as large as that drawn by the electric kettle.
Statement (2) is incorrect. The appliances will operate at a lower power output. Statement (3) is incorrect. By , the resistance of the hair dryer is only half of that of the electric kettle. . B According to, the power output is the smallest if the equivalent resistance of the heating elements is the largest. Thus, the heating elements should be connected in series to the a. C. Source by connecting terminals 2 and 3. 5. B From the given energy efficiency label, 840 k h of electrical energy is used in 1200 hours. Thus, the power P of the air conditioner is . Thus, the energy used in 8 hours = 0. X 5. 6 k h. 6. (a) Applying P = VI, the mains voltage in Australia (b) Applying P = ERR, the operating resistance R is 7. (a) Applying , the operating resistance R of the electric boiler is . B) Applying P = VI, the current drawn is (c) The energy consumed = 1000 x 5000 W h = 5 k h (d) The cost of running it for 5 hours = 5 x 0. 90 = $4. 50 8. Power / W operating time per day / h energy consumed / k h cost / $ lamps 5. 4 microwave oven 0. 25 air conditioner 1 oho 180 162 LCD TV 150 22. 5 20. 25 refrigerator 108 97. 2 iron 750 5. 625 5. 0625 kettle 0. 5 27 Checkpoint (p . 19) (b) A fault in one lamp will not affect the operation of the others. 2. (a) Kilowatt-hour (b) The main fuse should be connected to the live wire. 3. (a) A: neutral wire; B: earth wire; C: fuse; D: live wire (b) Exercise (p. 220) (b) Originally, the lamps work at their normal rating of ‘220 V, 60 W’. We have where VI and Pl -60 W. If the circuit is connected to a 110 V mains supply (I. E. VI 12), the power consumed by each lamp is The power consumed and hence the brightness of the lamps will decrease by 4 times. 6. (a) P (b) For socket Q, there is no earth wire connected.
For socket R, the live and neutral pin-holes are wrongly connected. (c) Yes, sockets Q and R can still be used to power an appliance. For socket Q, the live and neutral pin-holes are connected correctly. If there is no current leakage problem, the appliance will still work properly. For socket R, although the live and neutral pin-holes are connected wrongly, the p. D. Across the pin-holes is the same as that across a normal socket. A normal a. C. Current will flow through an appliance connected across the pin-holes. However, there is risk that the appliance is at high potential even when it is switched off. . (b) A fault in one lamp will not affect the operation of the others. (c) The chance of overloading the circuit is reduced as the current flows from the consumer unit to the sockets via two paths. Thus, thinner and cheaper wires can be used. Checkpoint (p. 228) 1. (a) Incorrect. Fuses are used to protect circuits against overloading or short circuit. (b) Incorrect. The rating of a fuse should be slightly higher than the operating current of the appliance. (c) correct (d) correct (e) Incorrect. Electrical appliances with double insulation do not need an earth wire. 2. A) The fuse and switch should be connected to the live wire instead of the neutral wire. (b)(I) incorrect correct incorrect (iv)lunchroom. Since the neutral and earth wires are both at zero potential, there is no current flowing between the two wires. (v) incorrect Thus, a fuse of 5 A rating should be used. For the microwave oven, the operating current . Thus, a fuse of 5 A rating should be used. For the electric kettle, the operating current. Thus, a fuse of 10 A rating should be used. (b) No. The total operating current = 4. 545 + 3. 636 + 8. 18182 16. 4 A > 13 A. It is unsafe to run all the appliances the same time.
Exercise (p. 230) 5. (a) When the switches are open, the bulbs are not at high potentials. (b) 0. 795 A Applying P = VI, the total current drawn from the mains is (c) No. The fuse rating of the fuse is much higher than the total current drawn from he mains. It is unsafe to run the lighting circuit. Even the current drawn from the mains is much higher than the sum of the operating currents of the bulbs, the circuit is still not broken. 6. (a) The operating current . (b) The fuse of a rating of 5 A is the most suitable. (c) Yes. Even there is no earth wire, an alternating current still flows to and fro in the live and neutral wires. D) The appliances have double-insulation. 7. (a) Water from the wet hands may form unintended paths that short the live parts of the appliances. Thus, we may get electric shocks if we touch electrical appliances hen our hands are wet. (b) The wires or conducting parts of the appliances may get loose and touch the casing of the appliances accidentally. Electric shock will be received when a person touch the appliances if its earth wire and insulation fails to work. (c)The current drawn from the mains may be too high that overloading may be resulted.
This may cause a fire if the fuses or circuit breakers installed fail to work. Chapter Exercise l. C There is no difference in the heating effect whether the conductor is wound into a coil or not. The difference is the transfer of heat from the conductor to the targeted area. 3. C The operating current I = Thus, fuse of 6 A rating should be used. The operating current I Applying P = VI, the estimated power of the AMP player The circuit can be redrawn as When more bulbs are connected in parallel to the battery, the size of the total current drawn from the battery increases.
Thus, the potential drop across the internal resistance of the battery increases while the p. D. Across each bulb decreases. Therefore, statement (1) is incorrect while statement (3) is correct. According to P = VI, the total output power of the battery increases too. Therefore, tenement (2) is correct. 8. C The lamps are connected in series to the 220 V mains. Statement (1) is correct because each lamp shares a p. D. Of Statement (2) is incorrect as the circuit will be broken. By, the resistance of the 60 W lamp is larger than that of the 100 W lamp.