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F.5 Chinese New Year Homework (2005-6) Below is your Chinese New Year Homework. There are 30 MC questions from HKCEE past paper. Work out the detail solutions for each one of them and hand them in using single-lined papers. The deadline for your work is Have a nice holiday. Mr. Li 1st February 2006 6th February 2006. page 1 of 6 F.5 Chinese New Year Homework (2005-6) Electric Circuits 1. page 2 of 6 (1) 1//1000//1x106≈ 1 (2) 1+1000+ 1x106 ≈ 1x106 (3) 1//1x106 + 1000 ≈ 1000 Hence (2) > (3) > (1); A. B. C. D. E. The answer is C. Voltmeter V1 0V 0V 1V 3V 3V Voltmeter V2 2V 3V 2V 0V 2V 3. Arrange the above networks in descending order of equivalent resistances : (1991-31) A. (1), (2), (3) B. (2), (1), (3) C. (2), (3), (1) D. (3), (1), (2) E. (3), (2), (1) 2. When S is closed, the p.d across V1 would be zero. The e.m.f. of the battery would be divided across the two resistors. V2 = [3/(10+20)] x 20 = 2 V The answer is A. In the circuit above, what happens to the readings of the three ammeters if switch S is closed ? (1992-31) A. B. C. D. E. A1 increases remains unchanged increases decreases increases A2 increases becomes zero becomes zero increases increases A3 increases increases remains unchanged becomes zero becomes zero When S is closed, the resistor connected in parallel to the switch is being short-circuited. As a result, the equivalent resistance of the circuit will decrease and no current would pass through A3. In the circuit above, what are the readings of voltmeters V1 and V2 if switch S is closed ? (1991-33) The answer is E. F.5 Chinese New Year Homework (2005-6) 4. page 3 of 6 The p.d. across PQ is = 0.6 x 5 = 3 V 6. When S is closed, the equivalent resistance of the circuit will decrease, hence I2 will increase. However, the e.m.f. of the battery remains unchanged, so the readings of both A1 and V remains unchanged. The answer is A. Note: IA1 = 2 x IA2; IA2 = 0.3 A e.m.f. of the battery = (0.3+0.6) x 10 + 3 = 12 V The answer is D. In the above circuit, the reading of ammeter A1 is 0.6 A. Find the potential difference between points P and Q. (1995-27) A. B. C. D. E. In the above circuit, the ammeters are of negligible resistance. Which of the following statements is/are true if switch S is closed ? (1995-33) 3V 6V 9V 12 V It cannot be determined since the e.m.f. of the battery is not given. 5. Since the ammeter has very small resistance, the current passing through it would be very large, producing too much heat and burn the ammeter. The answer is A. (1) (2) (3) The reading of ammeter A1 decreases. The reading of ammeter A2 increases. The reading of the voltmeter remains unchanged. A. B. C. D. E. (1) only (3) only (1) and (2) only (2) and (3) only (1), (2) and (3) 7. 0.3 x 6 = I4 x 4; I4 = 0.45 A A1 = 0.3 +0.45 = 0.75 A A student uses an ammeter and a voltmeter to find the resistance of a light bulb. He incorrectly connects the circuit as shown above. Which of the following is the most probable outcome ? (1994-26) A. B. C. D. E. The ammeter burns out. The voltmeter burns out. The light bulb burns out. The reading of the ammeter is zero. The reading of the voltmeter is zero. The answer is A. In the above circuit, the reading of ammeter A2 is 0.3 A. Find the reading of ammeter Al. (1996-27) A. B. C. D. E. 0.75 A 0.6 A 0.5 A 0.45 A It cannot be determined since the e.m.f. of the battery is not given. F.5 Chinese New Year Homework (2005-6) page 4 of 6 8. 9. Pbulb = Ptotal - Presistor = V x I – 18 = 9 x 5 – 18 = 27 W . The answer is D. (2000-34) A resistor and a bulb are connected in parallel to 9 V battery as shown above. The reading of the ammeter is 5 A. If the power dissipated by the resistor is 18 W, find the power dissipated by the bulb. A. 9 W. B. 18 W C. 22.5 W D. 27 W E. 45 W. P should be connected to wire 1 so that the ammeter is connected in series with the bulb. Wire 2 has a higher potential than wire 3, so Q wire 2 and R wire 3. The answer is A. (2000-36) F.5 Chinese New Year Homework (2005-6) 10. (1999-36) page 5 of 6 11. (2002-33) (1) (1) is wrong; the voltage applied to the bulb is 12 V. (2) is wrong; Rbulb = V2 / P = 62 / 6 = 6 After the bulb is connected in parallel to the 6 resistor the combined resistance becomes 3 . Hence the voltage applied to the bulb = [12/(3 +6)] x 3 = 4 V (3) is right; the voltage applied to the bulb = 12/2 = 6 V The answer is B. is wrong; since R = V/I, i.e. the slope of the graph;so Ry > Rx. (2) is worng; since Ry + Rx > Ry, so the combined resistance would has a steeper slope than Y. (3) is right; since any two resistors connected in parallel would always has a equivalent resistance less than either one resistor. The answer is B. F.5 Chinese New Year Homework (2005-6) page 6 of 6 13. (2003-33) 12. (2003-32) Let the resistance of each sides of the square be R. Then X= R//3R = 0.75 R, Y = 2R//2R = R, and Z = R//3R = 0.75 R. Hence X = Z < Y. The answer is C. The resistance of a wire would increase if it becomes thinner, so the slope of the its V-I graph would increase. The answer is D. F.5 Chinese New Year Homework (2005-6) 14. (2004-30) page 7 of 6 15. (2004-31) In circuit B, the motor would be shortcircuited when the switch is closed. In both circuits C and D, a.c. supplies are used. The answer is A. The resistance of the rheostat is, 40 (3/4) = 30 the ammeter reading = 6/30 = 0.2 A The answer is B. F.5 Chinese New Year Homework (2005-6) page 8 of 6 3. Electromagnetic Induction 1. PIN = 200 x 15 = POUT; no.of bulb = POUT/ Pbulb = POUT/ 40 =75 A. B. C. D. E. The answer is E. In the above circuit diagram, the transformer is 100% efficient. What is the maximum number of identical light bulbs, each of rating ‘40 W, 10 V’ that can be connected in parallel across the secondary coil without blowing the fuse ? (Assume that the fuse will blow if the current flowing through it exceeds 15 A.) (1991-38) A. B. C. D. E. 2. Which of the following can increase the efficiency of a transformer ? (1992-35) (1) Increasing the number of turns of the secondary coil. (2) Using a laminated iron core. (3) Using thicker copper wires to make the coils. (1) only (3) only (1) and (2) only (2) and (3) only (1), (2) and (3) (1) is wrong; it can only increase the Vout. (2) is right; it can reduce the heat loss (due to eddy current ) by the iron core. (3) it can reduce the loss due to the resistance of the wires. The answer is D. 4. According to Lenz’s law, the end of the coil near P is first N and then S. 2 3 20 60 75 N S Pin = 200 x 0.5 = 100 W Pout = 20 x 4 = 80 W Efficiency = Pout/ Pin = 80 % Turns ratio = 200 / 10 = 20:1 (step down) The answer is E. In the above circuit, the rating of each light bulb is ‘20 W, 10 V’. The current in the primary coil is 0.5 A. If all the bulbs work at their rated values, find the turns ratio and efficiency of the transformer. (1992-34) Turns ratio Efficiency A. 5:1 step down 40% B. 5:1 step down 80% C. 20:1 step down 20% D. 20:1 step down 40% E. 20:1 step down 80% The answer is A. A magnet is initially placed between the ends of a soft iron core as shown above. The magnet is then quickly rotated clockwise through one complete revolution. Which of the following statements correctly describes the induced current flowing through the resistor R ? (1994-36) A. B. C. D. E. The current flows through R from P to Q, and then reverses its direction. The current flows through R from Q to P, and then reverses its direction. The current flows through R from P to Q. The current flows through R from Q to P. There is no current flowing through R. F.5 Chinese New Year Homework (2005-6) page 9 of 6 5. 6. Since it is a simple d.c. generator, the correct V-t graph should be B. The answer is B. The above diagram shows an electricity generator. Which of the following graphs shows the variation of the e.m.f. produced by the generator ? Ans.: B (1993-37) The primary coil of a transformer has 3600 turns and is connected to a 200 V a.c. supply. The secondary coil has 180 turns, which can be tapped at different points as shown above. A ‘40 W, 10 V’ light bulb is connected to the transformer so that it works at its rated value. Which of the following statements is/are correct ? (1996-36) (1) (2) (3) A. B. C. D. E. The bulb should be connected to points P and Q. The current through the bulb is 4 A. If the efficiency of the transformer is 80%, the current in the primary coil is 0.25 A. (1) only (2) only (1) and (3) only (2) and (3) only (1), (2) and (3) (1) is wrong; (200/3600) x 90 =5V (2) is right; I = P/V = 40/10 =4A (3) is right; Pin = 40/80% = 50 W Iin = 50/200 = 0.25 A The answer is D. F.5 Chinese New Year Homework (2005-6) page 10 of 6 7. 8. In the above circuit, the rating of the light bulb is ‘40 W, 10 V’. The efficiency of the transformer is 80%. If the bulb works at its rated value, find the current in the primary coil. (1994-35) A. B. C. D. E. A metal ring is released and falls vertically around a magnet as shown above. Which of the following diagrams/statement correctly describes the directions of the induced current, if any, in the ring at positions X and Y ? (1997-34) 0.16 A 0.2 A 0.25 A 1.6 A 2.5 A Pin = Pout /80% = 40/80% = 50 W Iin = 50/200 = 0.25 A The answer is C. 9. According to Lenz’s law, X induced magnetic field N N Y In the circuit above, the efficiency of the transformer is 80%. If the two lamps are to work at their rated values, what is the current in the primary coil and what kind of transformer is being used ? (1990-37) The answer is B. Current in primary coil E. There are no induced currents in either case. A. B. C. D. E. 0.2 A 0.4 A 0.5 A 0.4 A 0.5 A Pin = (40+40)/80% = 100 W Iin = 100/200 = 0.5 A Turns ratio = 200 / 10 = 20:1 (step down) The answer is C. Transformer 20 : 1 step down 20 : 1 step down 20 : 1 step down 10 : 1 step down 10 : 1 step down F.5 Chinese New Year Homework (2005-6) page 11 of 6 10. 11. According to Lenz’s law, A. N B. S C. S D. N Pin = 200 x 0.5 = 100 W Pout = 20 x 4 = 80 W Efficiency = Pout/ Pin = 80 % Turns ratio = 200 / 10 = 20:1 (step down) The answer is E. (2000-38) E. N The answer is C. (2001-34) F.5 Chinese New Year Homework (2005-6) page 12 of 6 12. 13. (2002-38) Iout = Pout / Vout = 24 / 6 = 4 A Current flowing through the power lines is, Icable = Iout / 20 = 0.2 A Power loss in the cables = 0.22 x 10 = 0.4 W The answer is A. E.m.f is induced in the coil when the magnet is entering (t1) and coming out (t2) from the coil. As both t1 and t2 are short, the waveforms of the corresponding induced should be quite sharp. Secondly, according to Lenz’s law, the polarities of the voltage at t1 and t2 should be opposite to each other. Lastly, since t1 > t2. Accordingly to Faraday’s law, the magnitude of induced e.m.f. at t 2 should be larger than that produced during t 1. The answer is A. (2001-35) F.5 Chinese New Year Homework (2005-6) page 13 of 6 14. (2004-35) 15. (2004-36) Firstly, there is no induced e.m.f before the coil cuts the magnetic flux lines. Secondly, the polarities of the induced e.m.f. produced during entering and leaving the Bfield should opposite to each other. Thirdly, there is noinduced e.m.f. while the coil travelling within theB-field. The answer is C. (1) is right; since it is an a.c. generator. (2) is wrong; N x . S According to Fleming’s Right-hand Rule, the current is flowing from Q to P. (3) is wrong; when the coil is vertical, the flux cutting rate is zero momentarily, hence the induced current is zero. The answer is A.