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YCK2 Form 7 Physics First Assessment 2004-05 Section A (30) 1. 2. 3. 4. 5. C B B D C…… 11. 12. 13. 14. 15. 6. 7. 8. B B D 16. A 17. B 18. A 26. D 27. D 28. A 9. D 10. D 19. B 20. A 29. D 30. D Section B 1. C B D C A.......... 21. 22. 23. 24. 25. A A C C C.............. (32) (a) (1) (1) At coil position shown, the magnetic flux linking each turn of coil is = B A cos t ( 12 ) By Faraday law of Induction, the induced e.m.f., d V = N = B A N sin t dt or V = Vo sin t ( 12 ) ( 12 + 12 ) where Vo = B A N (the maximum induced e.m.f.) From diagrams in (b), V is the maximum when d/dt is maximum and the coil is horizontal; V is zero when d/dt = 0 and the coil is vertical. (1) 1 (b) Diagram Values of L, C and R roughly correct (c) (1) Transformer The primary and the secondary windings are wound on the same core, thus the rate of change of magnetic flux, d/dt at any time is the same. ( 12 ) Vp = Np (d/dt) (equal to back e.m.f.) ( 12 ) Vs = Ns (d/dt) (when no current is drawn) ( 12 ) Hence (2) (2) (1) Vs N s Vp N p ( 12 ) Full-wave rectification Rectifiers offer low resistance against current flow in one direction but high resistance in the opposite direction. ( 12 ) ( 12 ) In the positive half cycle current passes through the rectifier in one direction; while in the negative half cycle, current passes through the rectifier in the opposite direction. ( 12 ) The circuit is connected such that current always passes through the load in one direction. Thus the output voltage becomes d.c. (unidirectional) ( 12 ) (3) Filter ( 12 ) 2 The smoothing of the output voltage is achieved by separating the d.c. component from the a.c. component of the signal. ( 12 ) A capacitor connected across the load is in fact a storage capacitor which stores charge and energy when it is charged up and then releases charge and energy through the load. ( 12 ) ( 12 ) An inductor is connected in series with a capacitor act as a potential divider. ( 12 ) The inductor offers a great impedance to the a.c. component whereas the capacitor offers a great resistance to the d.c. component. ( 12 ) Thus the a.c. component (unwanted ripples) mostly drops across the inductor; whereas the d.c. component appears across the capacitor. ( 12 ) For good smoothing action, the time-constant RC must be larger than 1/2f for the first capacitor. For 50 Hz, R is about 1 kHz, C is about 10 F. If L and C series circuit is also used, L is about 10 H, C is about 47F ( 12 ) 2. (a) (i) (ii) (b) (i) Electric field, E, in the circuit is the force per coulomb acting on the free electrons which move in a net direction along the connecting wires and across the resistor. (1) The direction of electric field is along the direction of current (from higher potential to lower potential). (0.5) (suitable figure: 0.5) Potential difference across the resistor is the energy converted to some other forms from electric potential energy per coulomb of charge passing through the resistor. (1) According to (a)(ii), the energy converted in R when charge Q passes through it is given by E=QV where V is the potential difference across R (0.5) The rate of heat conversion is 3 dE d dQ (QV) V VI I 2 R dt dt dt (0.5+0.5) (ii) IX V IR In an a.c. circuit with reactance X, V I where I and V are r.m.s. values; 2 2 R X (I is the current through R and V is the voltage across R) (c) (i) (0.5) The rate of heat conversion is still I2 R. There is a time delay in the lighting up of the light bulb intensity. It is because a large induced e.m.f. is created in the inductor. The induced e.m.f. opposes the current through the inductor. Thus the current rises slowly to its maximum value. Since induced e.m.f. dI dt (0.5) to full (0.5) (0.5) (0.5) (0.5) Thus, induced e.m.f. decreases with time and becomes zero when current reaches its maximum. (0.5) (ii) When the d.c. supply is switched off, there is a large induced e.m.f. in the inductor. (0.5) This opposes the collapse of current. (0.5) This is large enough to light up the neon lamp. (0.5) (d) a.c. V (small R) Suppose L I I o sin t (0.5) dI LI o cos t (0.5) dt If R is small, V = (1) Thus, V leads I by /2 or T/4 (0.5) 4 Figures: (1+0.5) X L L R R For small R, 90 Also, Y1 for measuring V Y2 for measuring IR, or I. tan (0.5) (0.5) (0.5) (0.5) //(6) Section C (40) 1. By Fleming left hand rule, the charge carriers drift to the top surface. Thus the charge carrier has positive charge. (0.5+0.5) V 6 10 3 6 V m 1 3 d 10 The drift velocity, v , is given by, (at equilibrium) qvB=qE The electric field is, v or E E 6 6 m s 1 B 1 (0.5) (0.5) (0.5) The current is given by I nAvq Thus, 2. n (0.5) I 10 1.04 10 25 m 3 19 6 qvA 1.6 10 6 10 (a) The motor does not rotate. Thus there is no back e.m.f. Current V 3 0 .6 A R 5 (1) //(4) (1) (1) (b) Mechanical power output = m g v = 0.05 10 0.3 = 0.15 W Let I be the current through the coil. Power lost in resistive heating = I2 R = 5 I2 Power input = I V = 3 I At constant speed, power is conserved, (0.5) 3 I = 5 I2 + 0.15 or 5 I2 3 I + 0.15 = 0 Solve for I, we get I = 0.545 A or I = 0.055 A (c) Consider the voltage of the circuit, (0.5) (1) (0.5) (0.5) V=IR+ (where is the back e.m.f. ) (1) For I = 0.545 A, = 0.275 V (0.5) For I = 0.055 A, = 2.755 V (0.5) Since is proportional to the speed of motor, this shows that the motor rotates faster when the current is smaller. (1) //(8) 5 3. (a) The capacitor blocks all d.c. and the equivalent circuit is as shown in the right. (0.5) Let R' be the equivalent resistance of the two shunt resistors, then 3 R' (0.5) 6 10 6 R' = 0.5 M Thus, (b) 10 6 R 10 R 6 R (0.5) (0.5) 05 10 6 (0.5) Hence, R = 1 M When the input is a.c., the equivalent circuit is as shown in (0.5) the right. Let Z be the equivalent impedance of the black box parallel with the 1 M resistor. Then 3 10 9 3 (0.5) Z (1) 10 6 Z Thus, Z = 0.33 k (0.5) This impedance is the result of C in parallel with 0.5 M resistor. Yet Z << 0.5 M. This means that the 0.5 M resistor can be ignored and Z is essentially the reactance of C. (1) 1 Z Thus, C 1 1 or (1) //(7) C 96 pF 2 f Z 2 (5 10 6 ) (0.33 10 3 ) 4. (a) Voltage across RL = 6 1.5 = 4.5 V Thus, RL 4.5 1.5 k 0.003 (1) (b) Voltage across RB = 6 0.7 = 5.3 V I 0.003 IB C 1.5 10 5 A and 200 Thus, RB 5.3 1.5 10 5 (0.5) 353 k 6 (0.5) (0.5) (1) (c) 5. IB 5.3 0.106 mA 50 10 3 I C 200 I B 21.2 mA (a) Thus, If (1) (0.5) //(5) VCC Vout I C R L (0.5) Vin VBE I B R B RL Vout VCC (Vin VBE ) RB 130 R L Vout 6 (Vin 0.7) RB (0.5) Vout = 2.5 V , Then Vin = 0.88 V (1) (1) (b) The smallest value of Vout is zero. Thus the maximum variation is 2.5 V and the maximum peak-to-peak voltage of Vout is 5 V. (1) Vout RL (c) The voltage gain, (0.5) Vin RB When Vout = 5 V, Vin = 0.26 (d) At saturation, VCC 0.2 I CS R L (0.5) (0.5) 6 0.2 (12 10 3 ) R L RL = 483 Vout RL Vin RB Since 10 (1) (0.5) 130 483 RB RB = 6279 6. (a) G (b) Rf 100 10 R in 10 (0.5+1) Saturated at positive side means Vout = 13 V Thus, (c) (1) I in Vout 13 1.3 V G 10 1.3 0.13 mA 10 10 3 Vin Vin R in 7 (0.5) (0.5+1) (0.5+1) //(8) (d) 1.5 1.5 (3) //(8) 8