18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 245 Quick Check 3 1. A pure capacitor is connected to an alternating voltage as shown in Figure (a). Figure (b) shows the variation of current I in the circuit with time t. Figure (a) Figure (b) ~ C t I 0 Which of the following is the graph of voltage V against time t for the capacitor? A 0 t V C 0 t V B 0 t V D 0 t V 2. A sinusoidal voltage of frequency 100 Hz and r.m.s. value 12.0 V is connected across a 2.2 μF capacitor. What is the r.m.s. current in the circuit? A 5.5 PA C 2.6 mA B 26 PA D 17 mA 3. A sinusoidal alternating voltage of period T is connected across a pure capacitor. Which of the following is the graph of power P against time t for the capacitor? A t T P 0 B t T P 0 C t T P 0 D t T P 0 4. A parallel plate capacitor is connected to an a.c. supply and an a.c. ammeter as shown in the diagram. A A piece of insulator is then inserted into the space between the plates of the capacitor. Which row correctly describes the change, if any, in the reactance in the circuit and the reading of the ammeter? Reactance Ammeter reading A Increases Increases B Increases Decreases C Decreases Increases D Decreases Decreases 5. Sketch a graph to show how the reactance of (a) a conductor (b) a capacitor depends on the frequency of the alternating voltage supply. Label your graphs clearly. 6. A sinusoidal potential difference 240 V r.m.s. and frequency 50 Hz is connected across (a) a resistor of resistance 1.2 k:, (b) a capacitor of capacitance 2.2 PF. Calculate the r.m.s. current in each circuit.
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 246 18.4 R-C and R-L Circuits in Series Students should be able to: GHÀQHLPSHGDQFH XVHWKHIRUPXODZ = R2 + (XL – XC) 2 VNHWFKWKHSKDVRUGLDJUDPVRIR-C and R-LFLUFXLWV Learning Outcomes Defi nition of Impedance 1. The impedance of an alternating current circuit is the total opposition to the fl ow of alternating current produced by the combination of components such as resistors, inductors and capacitors. 2. Since the potential differences across the resistor, inductor and capacitor are not in phase, a phasor diagram must be draw to determine the impedance. 3. A phasor is a rotating vector used to represent a quantity that varies sinusoidally. Sinusoidal alternating current and voltage can be represented by phasors. Resistor and Capacitor in Series 2007/P2/Q12 C R VR VC V V = IZ V I C = IR VC = IXC Z R XC (a) (b) (c) Figure 18.13 1. Figure 18.13 (a) shows a resistor and a capacitor connected in series with a sinusoidal alternating voltage V. Since the resistor and capacitor are in series, the same current I fl ows in the resistor and capacitor. 2. The potential difference across the resistor, VR = IR and is in phase with the current I. The potential difference across the capacitor, VC = IXC . VC lags behind the current I by 90o or S 2 radians. 3. The phasor diagram, Figure 18.13(b) shows the phase differences between I, VR, VC and the applied voltage V = IZ where Z is the impedance. 4. From the phasor diagram (Figure 18.143(b)), Figure 18.13(c) is obtained to show the relationship between R, XC and Z. Impedance, Z = R2 + X2 C = R2 + 1 1 ωC2 2 2015/P2/Q20, 2017/P2/Q17, 2018/P2/Q17 INFO +PMMLYLUJLIL[^LLU 9*HUK93*PYJ\P[
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 247 5. The phase difference φ between the applied voltage, V and the current, I is given by tan φ = VC VR = XC R The current I leads the voltage V. 6. The variation of the impedance Z with frequency f of the alternating current is as shown in Figure 18.14. For high frequency, the impedance approaches the resistance R of the resistor. Impedance, Z Current I R Impedance, Z f 0 Figure 18.14 7. When the frequency is zero, the current is zero as the capacitor blocks the direct currents. As the frequency increases, the impedance decreases and the current increases as shown in Figure 18.14. Example 9 A 12 k: resistor and a 0.50 PF capacitor are connected in series with a sinusoidal a.c. supply of angular frequency ω = 100 rad s–1. The r.m.s. current in the circuit is 5.0 mA. Calculate (a) the r.m.s. potential difference across the resistor, (b) the r.m.s. potential difference across the capacitor, (c) the r.m.s. value of the sinusoidal voltage, (d) the impedance in the circuit, (e) the phase difference between the current and the applied voltage. Solution: VC V R VR = 12 k C = 0.50 F V = IZ V I R = IR VC = IXC φ 2018/P2/Q17
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 248 (a) R.m.s. potential difference across resistor, VR = IrmsR = (5.0 u 10–3)(12 u 103 ) = 60 V (b) R.m.s. potential difference across capacitor, VC = IrmsXC = Irms 1 1 ωC2 = (5.0 u 10–3) 1 (100)(0.50 u 10–6) = 100 V (c) R.m.s. value of sinusoidal voltage, V = V2 R + V2 C = 602 + 1002 = 117 V (d) Impedance, Z = R2 + X2 C = (12 u 103 )2 + 3 1 (100)(0.50 u 10–6)4 2 = 23 k: (e) tan φ = VC VR = 100 50 Phase difference, φ = 59° Example 10 A fi lament lamp and a capacitor are connected to a sinusoidal a.c. supply. The lamp is lit to normal brightness. Describe and explain any change in the brightness of the lamp when another identical capacitor is connected in series with the lamp. Capacitor Filament lamp A.c. Solution: When another capacitor is connected in series, the effective capacitance is halved. Impedance in the circuit, Z = R2 + 1 1 ωC2 2 The impedance Z increases. The current in the circuit decreases. Hence the lamp is lit but dim.
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 249 Resistor and inductor in series 2015/P2/Q15 1. Figure 18.15(a) shows a resistor and an inductor connected in series with a sinusoidal alternating voltage V. The same current I is in the resistor and inductor. R VR VL V L Inductor Resistor V = IZ V I R = IR VL = IL φ (a) (b) Figure 18.15 2. The potential difference across the resistor, VR = IR and is in phase with the current I. The potential difference across inductor, VL = IXL. and leads the current by 90° or S 2 radians. 3. The phasor diagram in Figure 18.15(b) shows the phase differences between I, VR, VL and the applied voltage V = IZ, where Z is the impedance. 4. From the phasor diagram (Figure 18.15(b)), (IZ)2 = (IR)2 + (IXL )2 Impedance, Z = R2 + X2 L = R2 + (ωL)2 5. The phase difference φ between the applied voltage V and the current I is given by tan φ = VL VR = XL R The voltage V leads the current I. Example 11 A solenoid with a resistance of 5.0 : and an inductance of 5 mH is connected to a 12 V, 50 Hz a.c. supply. (a) What is the impedance in the circuit? (b) What is the phase difference between the voltage and the current? (c) What is the r.m.s. current? Solution: (a) Impedance, Z = R2 + XL 2 (XL = ωL = 2SfL) = R2 + (2SfL)2 = 5.02 + 2S u 50 u 5 u 10 –3)2 = 5.24 : 2017/P2/Q17
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 250 (b) V V I R VL φ Phase difference between V and I, φ = tan–11 VL VR 2 = tan–11 IXL IR 2 = tan–11 2SfL R 2 = tan–1 2S(50)(5 u 10–3) 5.0 = 17° (c) R.m.s. current, Ir.m.s. = Vr.m.s. Z = 12 7.9 A = 1.5 A Resistor, Capacitor and Inductor in Series 1. Figure 18.16 shows a resistor, a capacitor and an inductor in series with a sinusoidal alternating voltage V. R VR VL VC V L C Capacitor Resistor I V= IZ VL VC VC VR VL φ (a) (b) Figure 18.16 2. The potential difference across the resistor VR is in phase with the current I. The potential difference across the inductor VL leads the current I by 90° or S 2 radians. The potential difference across the capacitor VC lags behind the current I by 90° or S 2 radians. Figure 18.16(b) is the phasor diagram.
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 251 3. The voltage of the supply V is related to VR, VL and VC by the equation V = VR 2 + (VL – VC)2 Impedance, Z = R2 + (XL – XC)2 = R2 + 1ωL – 1 ωC2 2 (ω = 2Sf) 4. Variations of R, XL , XC and Z with frequency f of the a.c. supply are shown in Figure 18.17. f Z R 0 XL XC f 0 f I V R 0 f 0 Figure 18.17 Figure 18.18 5. The impedance Z is minimum when XL = XC 2Sf 0 L = 1 2Sf 0 C f 0 = 1 2S LC The frequency f 0 is known as the resonant frequency. 6. The variation of current with frequency is as shown in Figure 18.18. The maximum current is I max = V R . It occurs when the frequency is f 0 = 1 2S LC , the resonant frequency. Example 12 A radio uses an inductor of resistance 200 : and inductance 2.5 PH, and a variable capacitor in series as a tuning circuit. The capacitance of the capacitor is varied until the resonant frequency of the R-L-C circuit equals the frequency of the selected radio station. (a) What is the capacitance of the capacitor when the radio is tuned to an FM radio station of frequency 101 MHz? (b) What is the impedance of the circuit at this frequency? (c) Explain why broadcast from the FM station can be heard loud and clear. Solution: (a) Resonant frequency of R-L-C circuit, f 0 = 1 2S LC Capacitance, C = 1 4S2 f 0 2 L = 1 4S2 (101 u 106 )2 (2.5 u 10–6) = 9.93 u 10–13 F
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 252 (b) At the resonant frequency f 0 , inductive reactance, XL = XC, capacitive reactance Impedance in the circuit, Z = R2 + (XL – XC)2 = R = 200 : (c) The broadcast can be heard loud and clear because at the resonant frequency, the current in the tuning circuit is maximum. Quick Check 4 1. A 800 : resistor and a 4.0 PF capacitor are connected in series with a 6.0 V, 50 Hz a.c. supply. Calculate (a) the impedance in the circuit (b) the r.m.s. current. 2. A solenoid which has a resistance of 4.0 : and inductance of 7.5 mH is connected to a 12.0 V a.c. supply of frequency 50 Hz. Calculate (a) the impedance in the circuit (b) the r.m.s. current. 3. A resistor of resistance R = 25 :, a capacitor of capacitance C and an inductor of inductance L are connected to a 12.0 V a.c. voltage of frequency 50 Hz. The potential differences across the capacitor and inductor are 15.0 V and 5.5 V respectively. R C L (a) Draw a phasor diagram to show the potential difference VR, VC, VL and the a.c. voltage V. (b) Use the phasor diagram to fi nd VR the potential difference across the resistor. (c) Calculate (i) the r.m.s. current in the circuit, (ii) the impedance, (iii) the capacitance, (iv) the inductance. Important Formulae 1. A.c. in pure resistor: If I = I0 sin ωt, then V = V0 sin ωt Mean power = — 1 2 I0 V0 = —1 2 I2 0 R = V0 2 —– 2R = Ir.m.s.Vr.m.s. = I2 r.m.s.R = V 2 r.m.s. R Root-mean-square current, Ir.m.s. = I —–0 Ƽ2 and root-mean-square voltage, Vr.m.s. = V —–0 Ƽ2 2. A.c. in pure inductor: V leads I by —1 2 Srad. If I = I0 sin ωt, then V = V0 cos ωt Instantaneous power, P = IV = —1 2 I0 V0 sin 2ωt and mean power = 0 Reactance, XL = 2SfL 3. A.c in pure capacitor: I leads V by —1 2 Srad. If V = V0 sin ωt then I = I0 cos ωt Instantaneous power, P = IV = —1 2 I 0 V0 sin 2ωt and mean power = 0 Reactance, XC = ——1 2SfC
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 253 STPM PRACTICE 18 1. Which graph shows the phase relationship between the voltage V and current I when a sinusoidal current fl ows through a pure inductor? A C I, V V I 0 t I, V I V 0 t B D I, V V I 0 t I, V V I 0 t 2. A 60 PF capacitor is connected to an alternating voltage V = 4 sin 200t, where V is in volts and t is in seconds. What is the peak current in the circuit? A 240 PA C 48 mA B 12 mA D 3.33 A 3. When an alternating current of 2.0 A is in a resistor, heat is produced at a rate of P. When a steady direct current I is in the same resistor, the rate of heat produced is also P. What is the value of I? A 1.4 A C 2.8 A B 2.0 A D 4.0 A 4. When a capacitor is connected to an alternating voltage V = 250 sin 150t, the current in the circuit is given by I = 0.5 cos 150t. If V is in volts, I is in amperes and t is in seconds, what is the capacitance of the capacitor? A 13 PA C 83 PA B 17 PA D 120 PA 5. An alternating current I = 3.0 sin 50t is connected to an inductor of 40 mH. What is the peak value of the back-e.m.f. induced in the inductor? A 1.2 V C 3.0 V B 1.5 V D 6.0 V 6. When a 6.0 : resistor is connected to an alternating voltage, the average power dissipated from the resistor is 12.0 W. What is the peak voltage across the resistor? A 6.0 V C 18.0 V B 12.0 V D 24.0 V 7. A voltage V that varies with time t as shown by the graph below is applied across a capacitor. V 0 t The variation of current I in the circuit with time t is shown in? A t I 0 C t I 0 B t I 0 D t I 0 8. A sinusoidal alternating current I = 2.0 sin (50t) fl ows in a 50 : resistor. What is the mean power dissipated from the resistor? A 10 W B 20 W C 50 W D 100 W
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 254 9. An alternating voltage V is applied across a component. The variation with time t of the voltage V and current I in the circuit is represented in the graph below. 0 V, I V I t Which statement is true? A The component is a capacitor. B Power is dissipated from the component. C The reactance of the component depends on the frequency of the voltage. D The voltage V and current I are in antiphase. 10. An alternating voltage V =V0 sin (2Sft) is applied across a capacitor of capacitance C. What is the peak current in the circuit? A fCV0 C 2SfCV0 B SfCV0 D CV0 /2Sf 11. The output voltage of a generator is given by V = V0 sin 2Sft The peak voltage V0 is directly proportional to the frequency f. The output voltage is connected across a resistor and the frequency f is varied. The peak current is proportional to A f C — 1 f B f 2 D —1 f 2 12. When a 240 V r.m.s. a.c. supply is connected to the terminals PQ in the circuit shown, the fuse F breaks the circuit if the current just exceeds 13 A r.m.s. F P Q When the a.c. supply is replaced with a 120 V d.c. supply, an identical fuse breaks the circuit if the current just exceeds A —– 13 Ƽ2 A B 13 A C 13Ƽ2A D 26 A 13. The power dissipated in a variable resistor is P when the current I is constant. The resistance of the variable resistor is halved, and an alternating current flows in the resistor. If the power dissipated in the resistor is also P, what is the r.m.s. value of the alternating current? A —1 2 I C Ƽ2 I B —– 1 Ƽ2 I D 2I 14. The power dissipated in a resistor is the same when the resistor is connected to a constant potential difference V, or a sinusoidal potential difference of peak value V0 . What is the relation between V and V0 ? A V0 = V —2 B V0 = —– V Ƽ2 C V0 = V D V0 = Ƽ2V 15. An alternating voltage V which varies with time t as shown in the diagram is connected to a resistor of resistance R. V t V0 –V0 0 What is the rate of heat loss from the resistor? A V0 2 —– 2R C V 0 2 R B V0 2 —– R D V0 R2
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 255 16. The variation of the current I through, and of the potential difference across an electrical component with time is shown below. I V T 0 t – + Which graph best respresents the variation of the power P in the component with time t? A t T P 0 B t T P 0 C t T P 0 D t T P 0 17. A resistor and an inductor are connected in series with an alternating voltage V. The current in the circuit is I. Which of the following is the correct phasor diagram showing the phase difference between I and V? A I V C I V B I V D I V 18. A resistor and a capacitor are connected in series with a sinusoidal alternating voltage of variable frequency. Which graph shows the variation of the current when the frequency f of the alternating voltage is increased? A 0 f I C 0 f I B 0 f I D 0 f I 19. There is a sinusoidal alternating current in a resistor. The mean power dissipated in the resistor is A —1 2 u maximum power. B —– 1 Ƽ2 u maximum power. C 2 u minimum power. D Ƽ2 u minimum power. 20. Which of the following relates correctly the dependence of the reactance of a capacitor, and that of an inductor to the frequency f of the alternating current? Capacitor Inductor A Constant Directly proportional to f B Inversely proportional Constant to f C Inversely Directly proportional to f proportional to f D Directly proportional Inversely to f proportional to f
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 256 21. An electrical component is connected to an a.c. supply of voltage V = 12 sin 500t. The current in the circuit is I = 0.50 sin 1500t – —S 2 2 where V is in volts, I in amperes and t in seconds. (a) Sketch the phasor diagram for V and I. (b) Identify the electrical component. Explain your answer. (c) Find the reactance of the component. 22. (a) An alternating current I =I0 sin 2Sft flows in an inductor of inductance L. Derive the expression for the reactance of the inductor. (b) A sinusoidal voltage of peak value 120 V and frequency 80 MHz is applied across an inductor of inductance 0.55 mH. Determine (i) the peak current in the circuit, (ii) the peak power dissipated supplied to the inductor. 23. (a) A 13 PF capacitor is connected to a voltage V = 220 sin 50St. Calculate (i) the reactance of the capacitor, (ii) the r.m.s. current in the circuit. (b) A 250 : resistor is connected in series with capacitor and supply voltage. (i) Draw a phasor diagram to show the phase difference between the potential differences across the resistor and capacitor, and the supply voltage. Use the phasor diagram to calculate (ii) the impedance in the circuit, (iii) the phase difference between the voltage and the current in the circuit. 24. An electric kettle which can be used with different voltages is rated 700 W, 240 V a.c. What will be its power output if used on (a) a 120 V a.c. supply? (b) a 120 V d.c. supply? 25. A pure capacitor of capacitance C is connected to an alternating voltage V = V0 sin (2Sft). (a) Write expressions for (i) the reactance in the circuit, (ii) the current in the capacitor, (iii) the maximum energy stored in the capacitor. (b) On the same axes, sketch graphs to show how (i) reactance and (ii) current in the circuit vary with the frequency f. 26. An alternating current is represented by the equation I = I0 sin ωt. (a) On the same time-axis, sketch graphs of (i) I against t (ii) I2 against t (b) What is the mean value of (i) I (ii) I2 over a complete cycle? 27. (a) The graph shows the variation of a periodic current with time. t /s I / A 0 123 2 4 What is the mean value of the current? (b) Sketch a graph of I2 against t for the current. From the graph, deduce the root-mean-square value of the current. (c) The periodic current passes through a resistor and heat is dissipated at a certain rate. What steady current through the same resistor would produce heat at the same rate? 28. (a) Distinguish between direct current and alternating current. (b) A 27 μF capacitor and a 40 : resistor are connected to a voltage supply V = 28 sin 340t where t is in second. (i) What is the r.m.s. voltage of the supply? (ii) What is the frequency of the voltage?
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 257 (iii) What is the impedance in the circuit? (iv) Calculate the r.m.s current in the circuit. (v) Find the r.m.s. voltage across the resistor VR , and voltage across the capacitor . (vi) Hence determine the phase difference between VR and VC. 29. The alternating current in a pure inductor is I = I0 sin 2Sft, where I0 = peak current, f = frequency of a.c. (a) Derive an equation for the potential difference V across the inductor. (b) Hence, deduce an expression for the reactance of the inductor. (c) Sketch a graph to show the variation of the reactance of the inductor with frequency f. (d) Explain why the mean power dissipated from a pure inductor is zero. 30. (a) What is meant by (i) the peak value, (ii) the root-mean-square value of an alternating current? Write an equation to relate the two values of current for a sinusoidal alternating current. (b) In the circuit shown, A1 , A2 , and A3 are centre-zero moving-coil ammeters. V is a moving-coil voltmeter and a.c. is a low frequency alternating current source. C is a capacitor, R a resistor, and L an inductor, all of comparable reactance. L R C A1 A2 A3 ~ V A.c. Sketch on the same axes, graphs to show the variation of the readings of (i) A1 , and V (ii) A2 , and V (iii) A3 , and V with time. Explain for the shape of the graphs. 31. (a) A filament lamp is rated 120 W, 110 V. What is the resistance of the filament under normal working conditions? (b) The lamp can be operated using a 240 V, 50 Hz a.c. supply by connecting a suitable resistor in the circuit. How should the resistor be connected in the circuit, and what is its resistance? (c) Alternatively, the resistor is replaced by a suitable capacitor. (i) What is the reactance of the capacitor? (ii) Calculate the capacitance of the capacitor. What is the advantage of using the capacitor instead of the resistor? 32. The Figure (a) shows an R-L-C circuit. The r.m.s. voltage of the a.c. supply is 6.0 V and its frequency is 50 Hz. The impedance in the circuit is 3.6 k:. The phasor diagram for the circuit is shown in Figure (b). R C = 2.7 F L Figure (a) V= IZ = 6.0 V VR = IR 30° Figure (b) (a) What is meant by impedance? (b) Calculate (i) the r.m.s. current in the circuit, (ii) the resistance of the resistor, (iii) the r.m.s. potential difference across the capacitor, (iv) the inductance of the inductor.
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 258 33. A power generator produces a current of 60 A at a voltage of 2.0 kV. A transformer steps up the voltage to 360 kV for transmission. (a) Calculate the output current of the transformer. (b) The resistance of transmission cables is 40 :. Determine the percentage power loss in the transmission cables. 34. (a) What is meant by peak and r.m.s. values of a sinusoidal alternating current? Write an equation relating the peak and r.m.s. values. (b) When a sinusoidal alternating current fl ows in a resistor of resistance R, the mean power dissipated is P. Derive the expression for the peak current I0 . (c) An alternating voltage V = V0 sin Zt is applied to a pure inductor of inductance L. (i) Derive the expression to show how the voltage V across the inductor varies with time t. (ii) Derive an expression for the reactance of the inductor. Hence describe how the reactance varies when the frequency is increased. (iii) Draw a phasor diagram relating V and I in the circuit. (iv) Calculate the r.m.s. voltage across a 60 mH inductor when a sinusoidal current of frequency 50 Hz and r.m.s. value 0.64 A flows in the circuit. 1 1. B 2. C 3. C 4. B 5. D 6. C 7. C 8. A 9. B 10. C: For a resistor, V and I are in phase. 11. (a) 5.0 A (b) 3.5 A (c) 25 Hz (d) 3 — 4 S radians or 135° 12. D ۙ R = —— Vr.m.s. – Ir.m.s. (b) 2 880 W P = I r.m.s. u Vr.m.s. (c) 5 760 W P max = 2P 2 1. A 2. C 3. B 4. V, P 0 t V P 3 1. D 2. D 3. A 4. C: XC = 1 2SfC' , C' > C I = V XC' , XC' > XC 5. f XR XR = R 0 0 f XC XC = 1 2πfC 6. (a) 0.20 A I r.m.s. = —— Vr.m.s. – R. (b) 0.166 A I r.m.s. = —— Vr.m.s. – XC , XC = 1 ——– 2SfC 4 1. (a) Impedance, Z = R2 + X2 C XC = 1 2Sf 0 C ANSWERS
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 259 = 1 2S(50)(4 u 10–6) = 795.8 k: Z = 8002 + (795.8)2 = 1.1 k: (b) R.m.s. current = V Z = 6 1.1 u 103 = 5.5 mA 2. (a) Impedance, Z = R2 + X2 L , XC = 2Sf 0 L = 2S(50)(7.5 u 10–3) = 2.36 : Z = 2.362 + 42 , = 4.6 : (b) R.m.s. current = V Z = 12 4.6 = 2.6 A 3. (a) VL = 5.5 V VC VL = 9.5 V V = 12.0 V VC = 15.0 V VR (b) VR = 12.02 – 9.52 V = 7.33 V (c) (i) I = VR R = 7.33 25 = 0.293 A (ii) Z = V I = 12 0.293 = 41 : (iii) VC = 15.0 V = IXC = 1 2SfC C = 62 μF (iv) VL = 5.5 V = IXL = I(2SfL) L = 60 mH STPM PRACTICE 18 1. A: Pure inductor: V leads I by 1 4 cycle or 90° 2. C: I0 = V0 XC = V0 (ωC) = (4)(200)(60 u 10–6) = 48 mA 3. %ffl 5PVFXUUHQWKDVWKHVDPHKHDWLQJHͿHFWDV the d.c. 4. A: XC = 1 ωC = V0 I0 = 250 0.5 = 500 : C = 1 (150)(500) = 13 μF 5. D: Eback = –LdI dt = –(0.040)(3.0 u 50) cos 50t = –6.0 cos 50t Peak value of Eback = 6.0 V 6. B: P = V0 2 2R, V0 = 2(12.0)(6.0) = 12.0 V 7. D: For capacitor, current I leads V by 1 4 cycle. 8. A: 1 2 I 2 0 R = 1 2 (22 )(5) = 10 W 9. C: XL = 2f L 10. C: XC = —– V0 I0 = ——– 1 2SfC I0 = 2SfCV0 11. A 12. B 13. C 14. D 15. B 16. D 17. C 18. D 19. A 20. C 21. (a) I V (b) Inductor, Reason: V leads I by S 2 rad. (c) X : V0 I0 = 12 0.50 = 24 : 22. (a) I = I0 sin 2Sft V = L—– dI dt = 2SfL I0 cos 2Sft (b) (i) XL = —– V0 I0 = 2SfL I 0 = ——– V0 2SfL = ——–———————— 120 2S(80 u 106 )(0.55 u 10–3) A = 0.43 mA (ii) Instantaneous power, P = IV = (I0 sin 2Sft)(V0 cos 2Sft) = —1 2 I 0 V0 sin 4ۛft Peak power = —1 2 I 0 V0 when sin 4Sft = 1 = —1 2 (0.43 u 10–3)(120) W = 0.026 W 23. (a) (i) XC = 1 ωC = 1 (50S)(13 u 10–6) : = 490 : (ii) Ir.m.s. = —— Vr.m.s. – XC = V0 2XC Ƽ = 220 Ƽ 2(490) = 0.317 A
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 260 (b) (i) V = IZ V I R = IR VC = IXC φ (ii) Impedance, Z = R2 + X2 0 = 2502 + 4902 = 550 : LLL 3KDVHGLͿHUHQFHEHWZHHQV and I, I = tan–11 XC R 2 = tan–1 1 490 250 2 = 63.0° 24. (a) 175 W P = V2 ——r.m.s. – R (b) 175 W 25. (a) (i) XC = 1 2SC (ii) I = 2SfCV (iii) Maximum energy stored = 1 2 CV0 2 (b) XC I 0 f (i) (ii) 26. (a) I0 2 I 2 I I 0 –I0 t T T 0 2 (b) (i) Zero (ii) 1 —2 I0 2 27. (a) Imean = 1.0 A (b) t /s I 2 / A2 I 2 0 1 ( I 2 ) mean 8 12 16 A1 A2 Use A2 = A1 Ir.m.s. = Ƽ(I 2 ) mean Ir.m.s. = 2.0 A (c) 2.0 A 28. (a) Refer page 232 (b) (i) 20 V (ii) 54 Hz (iii) Z = R2 + 1—–1 ωC2 2 = 116 ۙ (iv) Irms = ——– Vrms z = 0.172 A (v) VR = Irms R = 6.88 V VC = Irms(—–1 ωC) = 18.7 V (vi) VR leads Vc by φ = tan–1 —– VC VR = 70o 29. (a) Derive: V = 2SfIo L cos 2Sft (b) Deduce: XL = 2SfL (c) f XL 0 (d) Refer to text. 30. (a) V0 = Ƽ 2 Vr.m.s. (b) (i) t V A1 0 V leads I by S —2 rad. (ii) t V A2 0 V and I in phase. (iii) t V A3 0 I leads V by S —2 rad.
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 261 31. (a) P = V2 Z , R = 101 : (b) Resistor in series with lamp. R = (240 – 110) I V, I = 120 110 A R = 119 : (c) (i) VR = 110 V V= 240 V VC From the phasor diagram VC = V2 – VR 2 = 2402 – 1102 = 213 V XC = VC I = 213 (120/100) = 195 : (ii) XC = 1 2SfC , C = 195 : C = 16.3PF Advantage: No power dissipated from the capacitor. Power is dissipated as heat from the resistor. 32. (a) ,PSHGDQFHfflWRWDOUHVLVWDQFHWRÁRZRIDFLQ circuit produced by R, C and L. (b) (i) I = V Z = 6.0 3.6 u 103 = 1.7 mA (ii) VR = V cos 30° = 6.0 cos 30°= 5.2 V = IR R = 3.1 k: (iii) VC = IXC = I 2SfC = 1.7 u 10–3 2S(50)(2.7 u 10–6) = 2.0 V (iv) VL – VC = 6.0 sin 30° = 3.0 V VL = (3.0 + VC) = IXL= I(2SfL) (3.0 + 2.0) = I(2SfL) L = 5 2S(50)(1.7 u 10–3) = 9.4 H 33. (a) Io Vo = Ii Vi Io = (60)(2000) ——–—— 360 u 103 A = 0.333 A (b) Percentage of power loss = I2 o R ——– I i Vi = (0.333)2 (40) ——–——— (60)(2000) (100) % = 0.0037% 34. (a) Refer to text (b) Mean power, P = I2 rms R = 1 I ——0 2 2 2 R I0 = 2P —–– R (c) (i) I = I0 sin Zt V = L—– dI dt = ZL I0 cos Zt (ii) XL = —– V0 I 0 = ZLI ——–0 I 0 = ZL ω = 2Sf, when f increases, reactance XL increases. (iii) Phasor diagram V I (iv) Vrms = I rmsXL = (0.64)(2S)(50)(0.060) V = 12 V
262 STPM Model Paper (960/2) Paper 2 Kertas 2 (1 hour 30 minutes) (1 jam 30 minit) Section A [15 marks] Bahagian A [15 markah] Instructions There are fi fteen questions in this section. Each question is followed by four choices of answer. Select the best answer. Answer all questions. Marks will not be deducted for wrong answers. Arahan Ada lima belas soalan dalam bahagian ini. Setiap soalan diikuti dengan empat pilihan jawapan. Pilih jawapan yang terbaik. Jawab semua soalan. Markah tidak akan ditolak bagi jawapan yang salah. 1. The electric potential at a point distance r from a point charge is V. The electric fi eld strength E at that point is given by Keupayaan elektrik di suatu titik pada jarak r dari suatu cas titik ialah V. Kekuatan medan elektrik E di titik itu diberi oleh A E = rV B E = r2 V C E = V r D E = V r2 2. Two point charges of +2Q and -Q are separated by a distance of r. Jarak di antara dua cas titik yang bercas +2Q dan -Q ialah r. +2Q –Q The electric fi eld potential at the point P between the charges is zero. What is the distance of the point P from the charge of +2Q? Keupayaan elektrik di suatu titik P yang terletak di antara dua cas itu ialah sifar. Berapakah jarak titik P dari cas titik +2Q? A r 4 B r 3 C r 2 D 3r 3 3. A parallel-plate capacitor is charged to a potential difference V and the charge in the capacitor is Q. The capacitor is disconnected from the charging voltage and the dielectric is inserted between the plates of the capacitor. The new potential difference is V1 and the charge is Q1. Which of the following is correct? Suatu kapasitor plat selari dicas sehingga beza keupayaan merentasinya ialah V dan cas pada kapasitor ialah Q. Kapasitor itu ditanggalkan daripada voltan mengecas dan sekeping dielektrik diselitkan di antara plat-plat kapasitor. Keupayaan elektrik merentasi kapasitor menjadi V1 dan cas padanya Q1. Antara berikut, yang manakah betul? A V1 > V, Q1 > Q C V1 = V, Q1 > Q B V1 > V, Q1 = Q D V1 < V, Q1 = Q
Physics Term 2 STPM Model Paper Term 2 263 4. Two capacitors of capacitance C1 = 18 μF and C2 = 6 μF are connected to 6.0 V battery as shown in the diagram. Dua kapasitor yang berkapasitans C1 = 18 μF dan C2 = 6 μF disambungkan kepada sebuah bateri 6.0 V seperti ditunjukkan dalam gambar rajah d bawah. C1 C2 6.0 V The potential differences across the capacitors are V1 and V2 respectively. What are the values of V1 and V2? Beza keupayaan elektrik merentasi kapasitor itu adalah masing-masing V1 dan V2. Berapakah nilai V1 dan V2? A V1 = 1.5 V, V2 = 4.5 V B V1 = 3.0 V, V2 = 3.0 V C V1 = 4.5 V, V2 = 1.5 V D V1 = 6.0 V, V2 = 6.0 V 5. A 32 mF capacitor which is charged to a potential difference of 6.0 V is connected in series with a 500 : resistor and a switch S as shown in the diagram. Suatu 32 mF kapasitor yang dicas sehingga beza keupayaan merentasinya 6.0 V disambungkan secara sesiri dengan suatu perintang 500 : dan satu suis S seperti ditunjukkan dalam gambar rajah. C = 32 mF S 500 1 What is the value of the potential difference across the capacitor 8.0 seconds after the switch S is closed? Berapakah nilai beza keupayaan merentasi kapasitor 8.0 s selepas suis S ditutup? A 3.00 V B 3.64 V C 4.02 V D 4.80 V 6. The resistivity of copper increases with increase in temperature. Which explanation is correct? Kerintangan kuprum bertambah apabila suhu bertambah. Penjelasan yang manakah betul? A The drift velocity of free electrons increases. Halaju hanyut elektron bebas bertambah. B The number density of free electrons increases. Ketumpatan nombor elektron bebas bertambah. C The mean free path of free electrons increases. Jarak bebas min elektron bebas bertambah. D The mean time between collisions between free electrons and copper ions increases. Masa min antara perlanggaran elektron bebas dengan ion kuprum bertambah.
Physics Term 2 STPM Model Paper Term 2 264 7. When a tungsten wire is stretched, its thins uniformly and its length becomes xl(x t 1). The volume of the wire remains unchanged. Which graph represents the variation of the resistance R of the wire with the value of x? Apabila seutas dawai tungsten diregangkan, dawai itu menipis secara seragam dan panjangnya bertambah kepada xl(x t 1). Isi padu dawai tidak berubah. Graf yang manakah mewakili ubahan rintangan R dawai itu dengan nilai x? A x R 0 1 C x R 0 1 B x R 0 1 D x R 0 1 8. Three resistors are connected to a battery of e.m.f. 6.0 V and internal resistance 1 : as shown in the diagram. Tiga perintang disambungkan kepada sebuah bateri 6.0 V dan rintangan dalamnya 1 Ω seperti ditunjukkan dalam gambar rajah. 6 1 3 1 2 1 I 1 I 2 I İ = 6.0 V, r = 1 1 What are the values of the current I1, I2 and I? Berapakah nilai arus I1, I2 dan I? I1 I2 I A 0.20 A 0.40 A 0.60 A B 0.40 A 0.20 A 0.60 A C 0.40 A 0.80 A 1.20 A D 0.80 A 0.40 A 1.20 A
Physics Term 2 STPM Model Paper Term 2 265 9. Two cells of e.m.f., E1 and E2 are connected to three resistors of resistances R1, R2 and R3. The currents in the circuit I, I1 and I2 are as shown in the diagram. Dua sel dengan d.g.e. E1 dan E2 disambungkan kepada tiga perintang berintangan R1, R2 dan R3. Arus di dalam litar I, I1 dan I2 adalah seperti ditunjukkan dalam gambar rajah. R1 I 1 E1 R2 R3 I 2 I E2 Which equation is correctly obtained using Kirchhoff's laws? Ungkapan yang manakah diperoleh dengan betul dengan menggunakan hukum-hukum Kirchhoff? A E1 = I1(R1 + R3) + I2R3 B E1 = I1R1 + I2(R1+ R2) C E2 = I1(R1 + R3) + I2R3 D E2 = I1R1 + I2(R1 + R2) 10. A plane coil of wire which has N turns and diameter d carries a current I. If μ0 is the permeability of free space, which expression is for the magnetic flux density B at a distance centre of the coil? Suatu gegelung dawai bulat satah mempunyai N lilitan dan diameternya d membawa arus I. Jika μ0 ialah ketelapan ruang bebas, yang manakah ialah ungkapan yang betul bagi ketumpatan fluks magnet di pusat gegelung? A B = μ0 N1 d B B = μ0 N1 2d C B = μ0 N1 2πd D B = μ0 I 2πNd 11. Two positive ions, X and Y enter a region of uniform magnetic field B. The radii of curvature of the paths of X and Y are r and 2r respectively as shown in the diagram. Dua ion yang bercas positif X dan Y memasuki suatu medan magnet seragam B. Jejari kelengkungan lintasan X dan Y dalam medan itu adalah masing-masing r dan 2r seperti ditunjukkan dalam gambar rajah. B Y X If other variables are the same for X and Y, which statement is not correct? Jika pemboleh ubah yang lain adalah sama bagi X dan Y, pernyataan yang manakah tidak betul? A Charge of Y is twice charge of X. Cas Y dua kali cas pada X. B Mass of Y is twice mass of X. Jisim Y dua kali jisim X.
Physics Term 2 STPM Model Paper Term 2 266 C Speed of Y is twice speed of X. Laju Y dua kali laju X. D Specific charge of X is twice specific charge of Y. Cas tentu X dua kali cas tentu Y. 12. A metal rod of length L is placed between two parallel conducting rails on a horizontal surface in a uniform magnetic field of flux density B which acts vertically as shown in the diagram. The rod is moved from rest with constant acceleration a. Sebatang rod logam berpanjang L diletakkan di atas landasan pengkonduki yang selari dan mengufuk di dalam suatu medan magnet mencancang berketumpatan fluks B seperti ditunjukkan dalam gambar rajah. Rod itu digerakkan dari pegun dan bergerak dengan pecutan seragam a. L a B The e.m.f. E induced in the rod after a time t is D.g.e. aruhan E dalam rod itu selepas masa t ialah A E = BLa B E = BLat C E = BLat2 D E = 1 2BLat2 13. A solenoid is moved towards a short bar magnet as shown in the diagram. Sebuah solenoid digerakkan menuju sebatang magnet bar seperti ditunjukkan dalam gambar rajah. P Q N S What are the directions of the induced current in the wire PQ when the solenoid approaches the S-pole of the magnet, and as the solenoid leaves the N-pole of the magnet? Apakah arah arus aruhan dalam dawai PQ ketika solenoid menuju ke kutub-S magnet dan ketika solenoid menjauhi kutub-U magnet? Solenoid approaches S-pole Solenoid leaves N-pole Solenoid menuju kutub-S Solenoid menjauhi kutub-U A PQ PQ B PQ QP C QP PQ D QQ QP
Physics Term 2 STPM Model Paper Term 2 267 14. A sinusoidal voltage V = 3 sin 50πt is applied across a pure capacitor. Which graph is the correct for the I in the circuit? Suatu voltan sinusoidal V = 3 sin 50πt disambungkan merentasi sebuah kapasitor tulen. Graf yang manakah adalah arus dalam litar I lawan masa t yang betul? A t I 0 C t I 0 B t I 0 D t I 0 15. An inductor which has a resistance is connected in series with a sinusoidal alternating voltage V. Which phasor diagram is correct for the current I and the voltage V? Sebuah induktor yang mempunyai rintangan disambungkan secara sesiri dengan suatu voltan ulang-alik sinusoidal V. Gambar rajah fasor manakah yang betul bagi arus I dan voltan V? A I V B I V C I V D I V
Physics Term 2 STPM Model Paper Term 2 268 Section B (15 marks) Bahagian B (15 markah) Answer all the questions in this section. Jawab semua soalan dalam bahagian ini. 16. Three point charges each -5.0 μC are fixed at the vertices of an equilateral triangle as shown in the diagram. The distance of the charges from the centre C of the triangle is 2.0 cm. Tiga cas titik, setiap satu bercas -5.0 μC ditetapkan di bucu suatu segitiga sisi sama seperti ditunjukkan dalam gambar rajah di bawah. Jarak cas-cas titik itu dari pusat segitiga C ialah 2.0 cm. –5.0 +C –5.0 +C C –5.0 +C (a) Determine the magnitude and direction of the electric field strength at the centre C of the triangle. Hitungkan magnitud dan arah kekuatan medan elektrik di pusat C segitiga itu. [3 marks / 3 markah] (b) Find the electric potential at C. Cari keupayaan elektrik di titik C. [2 marks / 2 markah] (c) What is the electric potential energy of the system of three point charges? Berapakah tenaga keupayaan elektrik sistem tiga cas itu? [3 marks / 3 markah] 17. An alternating voltage, V = 5.0 sin 100πt is applied across a 25 μF capacitor. Suatu voltan ulang-alik V = 5.0 sin 100πt disambungkan merentasi sebuah kapasitor 25 μF. (a) What is the reactance of the capacitor? Berapakah reaktans kapasitor? [2 marks / 2 markah] (b) What is the value of the peak current in the circuit? Berapakah nilai puncak arus di dalam litar? [2 marks / 2 markah] (c) Derive the expression for the instantaneous power P. State the value of the mean power. Terbitkan ungkapan bagi kuasa seketika P. Nyatakan nilai kuasa min. [3 marks / 3 markah]
Physics Term 2 STPM Model Paper Term 2 269 Section C (30 marks) Bahagian C (30 markah) Answer two questions in this section. Jawab mana-mana dua soalan dalam bahagian ini. 18. (a) (i) Define capacitance of a capacitor. Takrifkan kapasitans sebuah kapasitor. (ii) A parallel-plate air capacitor consists of two plates separated by a distance d and each of the area is A. If ε0 is the permittivity of free space, derive an expression for the capacitance of the capacitor. Sebuah kapasitor plat selari terdiri daripada dua plat selari yang dipisahkan sejauh d dan luas setiap plat itu ialah A. Jika ε0 ialah ketelusan ruang bebas, terbitkan ungkapan bagi kapasitans itu. [4 marks / 4 markah] (b) A 24 μF capacitor is connected to a two way switch S. To charge the capacitor, the switch S is closed at A. After a time t 1 , the capacitor is fully charged. Sebuah kapasitor 24 μF disambungkan kepada suatu suis dua hala S. Suis S ditutup di A untuk mengecaskan kapasitor. Selepas suatu masa t1 , kapasitor telah dicas penuh. C = 24 +F 6.0 V 25 k1 25 k1 A B S What is the value of the potential difference across the capacitor (i) immediately, and (ii) 1.50 s after the switch S is closed at A? Berapakah nilai beza keupayaan merentasi kapasitor (i) sejurus, dan [1 mark / 1 markah] (ii) 1.50 s selepas suis S ditutup di A? [3 marks / 3 markah] (c) The switch S is then closed at B to discharge the capacitor. What is the value of the current (i) immediate the switch S is closed at B and (ii) 0.30 s later? Suis S kemudian ditutup di B untuk menyahcas kapasitor. Berapakah nilai arus (i) sejurus suis S ditutup di B dan [2 marks / 2 markah] (ii) 0.30 s kemudian? [2 marks / 2 markah] (d) Draw a graph to show the variation of potential difference V across the capacitor during charging and discharging. Mark the time t 1 on the time-axis. Lukiskan graf untuk menunjukkan ubahan beza keupayaan merentasi kapasitor dengan masa semasa mengecas dan semasa menyahcas. Tandakan masa t1 pada paksi masa. [3 marks / 3 markah]
Physics Term 2 STPM Model Paper Term 2 270 19. (a) (i) Describe the mechanism of electrical conduction in metals. Perihalkan mekanisma konduksi elektrik dalam logam. [3 marks / 3 markah] (ii) Sketch a graph to show the variation of conductivity σ of a metal with temperature θo C. Explain qualitatively the change in the conductivity of a metal when temperature increases. Lakarkan satu graf untuk menunjukkan ubahan kekonduksian elektrik σ suatu logam dengan suhu dalam θo C. Jelaskan secara kualitatif ubahan kekonduksian logam dengan kenaikan suhu. [5 marks / 5 markah] (b) The potential difference across a piece of copper wire of length 20 m and cross sectional area of 1.8 u 10-6 m2 is 0.50 V. The current in the wire is 2.6 A. The number density of free electrons in copper is 8.5 u 1028 m-3. Beza keupayaan merentasi seutas dawai kuprum yang berpanjang 20 m dan luas keratan rentasnya 1.8 u 10-6 m2 ialah 0.50 V. Arus dalam dawai ialah 2.6 A. Ketumpatan nombor elektron bebas dalam kuprum ialah 8.5 u 1028 m-3. (i) What is the current density in the wire? Brapakah ketumpatan arus dalam dawai? [2 marks / 2 markah] (ii) Find the drift velocity of the free electrons in copper wire. Tentukan halaju hanyut elektron bebas dalam dawai kuprum. [2 marks / 2 markah] (iii) Determine the electrical conductivity of copper. Hitungkan kekondusian elektrik kuprum. [3 marks / 3 markah] 20. (a) A proton enters with a velocity v at right angles to (i) a uniform gravitational field and (ii) a uniform electric field. Draw diagrams to show the paths of the proton in each of these fields. Show the directions of v and the field. Suatu proton memasuki dengan halaju v yang serenjang dengan (i)medan graviti yang seragam, dan (ii) medan elektrik yang seragam. Lukiskan gambar rajah untuk menunjukkan lintasan proton dalam setiap medan itu. Tunjukkan arah v dan medan itu. [4 marks / 4 markah] (b) A particle P of mass 6.64 u 10-27 kg and charge +3.20 u 10-19 C enters a uniform magnetic field B with a velocity of 5.40 u 105 m s-1. The direction of the field is normal to the direction of motion of the particle. The path of the particle in the field is a semicircle of diameter 0.058 m as shown in the diagram. Suatu zarah P yang berjisim 6.64 u 10-27 dan bercas +3.20 u 10-19 C memasuki suatu medan magnet seragam B dengan halaju 5.40 u 105 m s-1. Arah medan adalah serenjang dengan arah gerakan zarah. Lintasan zarah dalam medan magnet adalah satu bulatan separa yang diameternya 0.058 m panjang seperti ditunjukkan dalam gambar rajah.
Physics Term 2 STPM Model Paper Term 2 271 B 0.058 m (i) Determine for the magnitude and direction of the magnetic fi eld. Tentukan magnitud dan arah medan magnet itu. [4 marks / 4 markah] (ii) An electric fi eld E which combined with the magnetic fi eld B causes the particle P to move through the combined fi eld without defl ected. Determine the magnitude and direction of the electric fi led E. Draw a diagram to show the direction of v, B and E. Suatu medan elektrik E apabila digabungkan dengan medan magnet B menyebabkan zarah P melalui medan gabungan tanpa dipesong, Tentukan magnitud dan arah medan elektrik E itu. Lukiskan gambar rajah untuk menunjukkan arah-arah v, B dan E. [4 marks / 4 markah] (c) A second particle Q of half the mass and half the charge of particle P enters the same fi eld with the same velocity. Calculate the diameter of its path in the fi eld. Suatu zarah yang kedua Q yang jisim dan casnya separuh nilai jisim dan cas zarah P memasuki medan yang sama dengan halaju yang sama. Hitungkan diameter lintasan zarah Q dalam medan itu. [3 marks / 3 markah] ANSWERS PAPER 1 Section A 1. C: V = Q ——— 4Sε0 r' , E = Q ——— 4Sε0 r2 2. C: +2Q ——— 4Sε0 x + –Q ——— 4Sε0 (r-x) = 0 3. D: Charge is conserved, C increases V1 = Q C1 < V 4. C: V1 = ( C2 C1 + C2 )6.0 V, V2 = ( C2 C1 + C2 )6.0 V 5. B: V =6.0 e-t/CR = 3.64 V 6. D 7. D: R = ρl A , Volume = lA = (xl)A' R' = ρ(xl) _ A x + v x2 8. C: 1 R1 = 1 6 + 1 3 gives R1 = 2 : I = 6.0 2+2+1 A = 1.20 A I1 = ( 3 3+6 )(1.20)A = 0.40 A 9. A: I = I1+ I2 E1 = I1R1 + (I1+ I2)R3 10. A: B = μ0NI 2R = μ0NI d 11. A: F = qvB = mv2 r r = mv qB 12. B: E = BLv = BL(at) 13. B: Apply Lenz's law. 14. C: I leads V by 1 4 cycle. 15. A: 16. (a) –5.0 +C –5.0 +C E1 E3 E2 –5.0 +C
Physics Term 2 STPM Model Paper Term 2 272 E2= E2 = E3 = Q 4SHx2 in the directions shown Resultant E = ( Q 4SHx2 ) – 2( Q 4SHx2 )cos 60o 0 (b) At C, V = 3( -5.0 u 10-6 4SH (2.0 u 10-2) ) V = -6.74 u106 V (c) Side of triangle, a = 2 (2.0 sin 60o ) Electric potential energy at C = 3( Q2 4πεa ) = 19.5 J V XR XL I 17. (a) XC = Q2 2πfc = 127 : (b) I0 = 39.4 mA (c) P = IV = (0.0394 cos100πt)(5.0 sin 100πt) = 0.197 sin 200 πt Mean power = 0 18. (a) (i) and (ii) Refer to page 45. (b) (i) V = 0 V (ii) V = E(1 - e-t/CR) = 5.51 V (c) (i) I0 = 6.0 R = 0.24 mA (ii) I = I0e-t/CR = 0.146 mA (d) V t 1 t 0 19. (a) (i) Refer to page 82 (ii) 0 m e°C Explanation: Refer to page 92. (b) (i) J = 1 A = 1.44 u 106 A m-2 (ii) I = nAve, gives v =1.06 u 10-4 m s-1 (iii) J = ֆ( and E = V l ֆ = J V/l = 5.76 u 107 S m-1 20. (a) Path: Both parabola. (i) g + v (ii) E + v (b) (i) F = qvB = mv2 r gives B = 0.077 T perpendicular out of the page. (ii) v = E B gives E = 4.2 u 105 V m-1 E B v (c) qvB = mv2 r gives r = mv qb v m q (m/2) ( q 2) = m q hence same radius r. Diameter = 0.058 m
Summary of Key Quantities and Units QUANTITY KUANTITI BASE QUANTITY KUANTITI ASAS SYMBOL SIMBOL UNIT UNIT Mass Jisim m kg Length Panjang l m Time Masa t s Electric current Arus elektrik l A Thermodynamic temperature Suhu termodinamik T K Amount of substance Amaun bahan n mol Other Quantities Kuantiti Lain Distance Jarak d m Displacement Sesaran s, x m Area Luas A m2 Volume Isi padu V m3 Density Ketumpatan ρ kg m–3 Speed Laju u, v m s–1 Velocity Halaju u, v m s–1 Acceleration Pecutan a m s–2 Force Daya F N Weight Berat W N Work Kerja W J Energy Tenaga E, U J Potential energy Tenaga keupayaan U J Heat Haba Q J Power Kuasa P W Pressure Tekanan p Pa Period Kala T s Frequency Frekuensi f, v Hz Wavelength Panjang gelombang λ m Electric charge Cas elektrik Q, q C Current density Ketumpatan arus J A m–2 Electric potential Keupayaan elektrik V V Electric potential difference Beza keupayaan elektrik V V Electromotive force Daya gerak elektrik ε, E V Resistance Rintangan R : Resistivity Kerintangan ρ : m Conductance Konduktans G S = :–1 Conductivity Kekonduksian σ S m–1 = :–1 m–1 Electric fi eld strength Kekuatan medan elektrik E N C–1 Capacitance Kapasitans C F Magnetic fl ux Fluks magnet φ Wb Magnetic fl ux density Ketumpatan fl uks magnet B T Self inductance Swainduktans L H Mutual inductance Induktans saling M H Reactance Reaktans X : Impedance Impedans Z : Force constant Pemalar daya k N m–1 Temperature Suhu T, θ °C 273
274 Physics Term 2 STPM