NQA4B tủ tài liệu training

NUMERICALS-ANSWERS QUESTIONS An a.c source of 100 V, 50 Hz is connected across a 20 Ω resistor and a mH inductor in series Calculate (i) the impedance of the circuit, (ii) the r.m.s current in the circuit (AISSCE Delhi 1993) What is the value of the current in the a.c circuit containing R = 10Ω, C = 50µF in series across a 200 V, 50 Hz a.c source ? (AISSCE 1993) A 25µ F capacitor, a 0.10 H inductor and a 25.0 Ω resistor are connected in series with an a.c source whose e.m.f is given by E = 310 sin 314 t, where E is in volts and t in seconds (a) What is the frequency of the e.m.f ? (b) Calculate (i) the reactance of the circuit, (ii) the impedance of the circuit, (iii) the current in the circuit (AISSCE 1995) A series LCR circuit with L = 5.0 H, C = 80µF, and R = 40 Ω is connected to a variable frequency source of 230 V (i) Determine the resonance frequency of the circuit (ii) Obtain the impedance of the circuit and the amplitude of the current at resonance (AISSCE 1990C) An alternating voltage E = 200 sin 300t is applied across a series combinations of R = 10 Ω and an inductor of 800 mH Calculate (i) the impedance of the circuit, (ii) the peak value of the current in the circuit, (iii) the power factor of the circuit (AISSCE 1994) S Chand & Company Limited A capacitor, a 15 Ω resistor and 101.5 mH inductor are placed in series with a 50 Hz a.c source Calculate the capacity of the capacitor if the current is observed in phase with the voltage (AISSCE 1992) A current of 1.1 A flows through a coil, when connected to a 110 V d.c When 110 V a.c of 50 Hz is applied to the same coil, only 0.5 A current flows Calculate (i) the resistance, (ii) the impedance and (iii) the inductance of the coil (AISSCE Delhi 1992) A 100 V, 50 Hz a.c source is connected to a series combination of an inductance of 100 mH and a resistance of 25 Ω Calculate the magnitude and phase of the current (AISSCE 1991) The electric mains in a house are marked 220 V–50 Hz Write down the expression for the instantaneous voltage 10 A series LC circuit has L = 0.405 H and C = 25µF The resistance R is zero Find the frequency of resonance (AISSCE 1990) 11 A 1µF capacitor is connected to a 220 V–50 Hz a.c source Find the effective value of the current through the circuit What is the peak voltage across the capacitor ? 12 A rectangular coil of dimensions 40 cm × 20 cm, having 200 turns, rotates in a uniform magnetic field of 0.08 Wb m–2 about an axis perpendicular to the field If the coil makes 3000 revolution per minute, find the e.m.f induced when the plane of coil makes an angle with the magnetic field lines equal to (a) 0°, (b) 45°, (c) 90° 13 A coil with a reactance of Ω and a resistance of Ω is connected in series with a second coil of reactance Ω and resistance Ω Calculate the impedance of the circuit 14 The current through a coil is 10 A when it is connected to a 10 V d.c The current reduces to 1A on connecting it to V a.c Calculate the resistance, reactance and impadence of the coil 15 A circuit consists of a coil of resistance 20 Ω and inductance 0.1 H in series with an a.c ammeter The supply is 100 V, 50 Hz Calculate the current shown by the ammeter S Chand & Company Limited 16 The a.c mains in a house is marked 250 V, 50 Hz Write down the expression for the instantaneous voltage 17 Calculate the reactance of a pF capacitor at 1000 Hz 18 Calculate the frequency at which a 0.15 H coil will have an inductive reactance of 150 ohm 19 The tuning circuit of a radio receiver consists of a 0.25 × 10–3 H coil and a variable capacitor Compute the value of the capacitance required to tune the set to a broadcast station with the frequency of 10 kHz 20 An electric lamp marked 100 V (d.c.) consumes a current of A It is to be connected to a 220 V, 50 Hz a.c supply mains Calculate the inductance of the required choke 21 Calculate the value of the current flowing in a pure inductor of 63.66 mH inductance, when 10 V is applied at a frequency of 50 Hz 22 A coil has an inductive reactance of 176 Ω at a frequency of 50 Hz What is the value of its inductance ? 23 Obtain the resonant frequency of a series LCR circuit with L = 3.0 H, C = 27µF, R = 7.4 Ω 24 When an alternating voltage of 220 V is applied across a device X, a current of 0.5 A flows through the circuit and is in phase with the applied voltage When the same voltage is applied across another device Y, the same current again flows through the circuit but it leads the applied voltage by π/2 radians (a) Name the devices X and Y (b) Calculate the current flowing in the circuit when the same voltage is applied across the series combination of X and Y (AISSCE 1997) 25 When an alternating voltage of 220 V is applied across a device P, a current of 0.25 A flows through the circuit and it leads the applied voltage by π/2 radians When the same voltage is applied across another device Q, the same current again flows through the circuit and is in phase with the applied voltage S Chand & Company Limited 26 27 28 29 (a) Name the devices P and Q (b) Calculate the current flowing through the circuit when the same voltage is applied across the series combination of P and Q (AISSCE 1997) When a series combination of a coil of inductance L and a resistor of resistance R is connected across a 12 V, 50 Hz supply, a current of 0.5 A flows through the circuit The current differs in phase from the applied voltage by π/3 radians Calculate the values of L and R (AISSCE Delhi 1997) A 60 V – 10 W electric lamp is to be run on 100 V – 60 Hz mains (a) Calculate the inductance of the choke required (b) If a resistor is to be used in place of the choke coil to achieve the same results calculate its value (AISSCE Delhi 1997) A 25.0µF capacitor, 0.10 H inductor and a 25.0 Ω resistor are connected in series with an a.c source whose e.m.f is given by E = 310 sin 314t (i) What is the frequency of the e.m.f ? (ii) Calculate (a) the reactance of the circuit, (b) the impedance of the circuit and (c) the current in the circuit An LC circuit contains a 20 mH inductor and a 50µF capacitor with an initial charge of 10 mC The resistance of the circuit is neglibible Let the instant the circuit is closed be t = (a) What is the total energy stored initially Is it conserved during the LC oscillations ? (b) What is the natural frequency of the circuit ? (c) At what times is the energy stored (i) completely electrical (i.e., stored in the capacitor) ? (ii) completely magnetic (i.e., stored in the inductor) ? (d) At what times is the total energy shared equally between the inductor and the capacitor ? (e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat ? S Chand & Company Limited 30 A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = Ω, L = 25.48 mH and C = 7.96 × 10– F Find (a) the rms voltage, (b) XL and XC , (c) the impedance Z, (d) the peak current and the phase angle, (e) the rms values of the currents and voltages across the circuit elements, (f) the power dissipated in the circuit, (g) the power factor, (h) the power input, (i) the frequency of supply at which resonance occurs, (j) the impedance at resonant condition, (k) the current at resonant condition, and (l) the power dissipated at resonant condition 31 Draw the graph showing the variation of reactance of (a) a capacitor (b) an inductor with the frequency of the a.c circuit 11 kW of electric power can be transmitted to a distant station at (i) 220 V or (ii) 22,000 V Which of the two modes of transmission should be preferred and why ? Support your answer with possible calculations (AISSCE 1998) 32 A capacitor, a resistor of Ω, and an inductor of 50 mH are in series with an a.c source marked 100 V, 50 Hz It is found that the voltage is in phase with the current Calculate the capacitance of the capacitor and the impedance of the circuit (AISSCE Delhi 1999) 33 Distinguish between reactance and impedance When a series combination of inductance and resistance are connected with a 10 V, 50 Hz a.c source, a current of A flows in the circuit The voltage leads the current by π/3 radian Calculate the values of resistance and inductance (AISSCE 1999) S Chand & Company Limited 34 For a given a.c circuit, distinguish between resistance, reactance and impedance An a.c source of frequency 50 hertz is connected to a 50 mH inductor and a bulb The bulb glows with some brightness Calculate the capacitance of the capacitor to be connected in series with the circuit, so that the bulb glows with maximum brightness (AISSCE Delhi 2000) 35 An electric heater and an electric bulb are rated 500 W, 220 V and 100W, 220 V respectively Both are connected in series to a 220 V a.c mains Calculate the power consumed by (i) the heater and (ii) the electric bulb (AISSCE Delhi 1997) ANSWERS (i) Erms = 100 V, Now ν = 50Hz, R = 20 Ω, L = × 10– H Z = R +ω L 2 −3 = (20) + (2 × 3.14 × 50) × (2 × 10 ) = 20 Ω (ii) I rms = R = 10Ω, E rms 100 = =5A Z 20 C = 50 × 10– F, Erms = 200 V, Impedance Z = R + 1/ ω2 C = (10) + ν = 50 Hz (2 × 3.14 × 50) × (50 × 10 −6 )2 = 100 + 4056.8 = 4156.8 = 64.5 Ω E 200 Current I = rms = = 3.1 A Z 64.5 S Chand & Company Limited C = 25 × 10– F, L = 0.10 H, E = 310 sin 314 t R = 25.0Ω (a) Now E = E0 sin2 πνt ∴ 2π ν = 314 314 = 50 Hz × 3.14 Reactance X = XL ~ XC Frequency ν = (b) (i) = ωL ~ ωC = × 3.14 × 50 × 0.10 ~ = 31.4 ~ × 3.14 × 50 × 25 × 10 −6 106 = 95.98 Ω 314 × 25 = 96 Ω (ii) Impedance Z = R + (ωL − ) = (25)2 + (96)2 ωC = 625 + 9216 = 9841 = 99.2 Ω (iii) Current in the circuit I rms = Erms Z S Chand & Company Limited = 310 × 99.2 C = 80 × 10– F, L = 5.0 H, E = 230 V (a) At resonance ω L = ωC 1 ⇒ ω= = 2.21 A LC = R = 40 Ω (5.0 × 80 × 10 ) –6 = 20 × 10 −3 = 50 rad s– (b) Impedence at resonance, Z = R = 40 Ω Amplitude of the current I = V0 = 230 / 40 = 8.13 A R E = 200 sin 300 t, R = 10Ω , L = 800 × 10– H = 0.8 H, ω = 300 rad s−1 (i ) Z = R + ω2 L2 = (10)2 + (300 × 0.8)2 = 100 + 57600 = 57700 = 240.2 Ω (ii ) I0 = E0 200 = = 0.83 A Z 240.2 (iii ) Power factor = R 10 = = 0.042 Z 240.2 S Chand & Company Limited R = 15Ω, L = 101.5 mH, ν = 50 Hz Since the current is in phase with the voltage, the circuit is in resonance Therefore or ωL = ωC 1 , C= = or ω L (2 × π × 50) × 101.5 × 10−3 100 ì 106 = 100 àF D.C current = 1.1 A, D.C voltage = 110 V A.C current = 0.5 A, A.C voltage = 110 V, (i ) R= d.c voltage 1.10 = = 100 Ω d.c current 1.1 (ii ) Z= a.c voltage 110 = = 220 Ω a.c current 0.5 (iii) Z = R + ω2 L2 ⇒ Inductance L = Z − R2 ω = ν = 50 Hz (220) − (100) (314)2 196 = 0.624 H 314 L = 100 × 10– H = V = 100 V, ν = 50 Hz R = 25 Ω S Chand & Company Limited Z = R + ω2 L2 = (25)2 + (314 × 0.1)2 = 625 + 986 = 1611 = 40.1 Ω V 100 = 2.5 A Now I rms = = Z 40.1 ωL Also tan φ = R −1 314 × 0.1 ⇒ φ = tan = tan −1 1.256 25 = 51° 31 Erms = 220 V ν = 50 Hz Instantaneous voltage is given by E = E0 sin ωt = E0 sin 2π ν t = 220 sin (2π × 50t) = 311 sin 314 t V C = 25 × 10– F 1 = Resonance frequency ω = LC 0.405 × 25 × 10 −6 10 L = 0.405 H, = 314 rad s −1 S Chand & Company Limited 13 XL = 4Ω, XL = 5Ω, R1 = 2Ω, R2 = 3Ω, XL = XL + XL = + = 9Ω, R = R1 + R2 = + = 5Ω, Z = R + X L2 = (5)2 + (9)2 = 25 + 81 = 106 = 10.3 Ω 14 D.C current = 10 A, A.C current = A D.C voltage = 10 V, A.C voltage = V (i ) (ii ) d.c voltage 10 = =1Ω d.c current 10 a.c voltage = =5Ω Impedence Z = a.c current Resistance R = (iii ) We have Z = R + X L2 or X L = Z − R = 52 − 12 = 24 or X L = 24 = 4.9 Ω S Chand & Company Limited 15 R = 20 Ω, L = 0.1 H, Erms = 100 V ν = 50 Hz Z = R +ω L 2 = (20) + (314 × 0.1) = 400 + 986 = 1386 = 37.2 Ω Erms 100 = = 2.7 A Z 37.2 16 Erms = 250 V, ν = 50 Hz I rms = Peak voltage E0 = Erms × = 250 × = 353.5 V Instantaneous Voltage E = E0 sin2π ν t = 353.5 sin 314 t 17 C = × 10 – 12 F, ν = 1,000 Hz Capacitive Reactance 1 = XC = = = ωC 2π ν C × 3.14 × 1000 × × 10−12 109 = 3.98 × 107 Ω × 3.14 18 L = 0.15 H XL = 150 Ω = ∴ XL = 2π ν L X 150 ν= L = 2π L × 3.14 × 0.15 = 159.2 Hz = × 107 Ω S Chand & Company Limited 19 L = 0.25 × 10 – H For tuning, ν = C= or ν = 10000 Hz 2π LC π Lν 2 = × 9.8 × 0.25 × 10−3 × (10000)2 = × 9.8 × 0.25 × 10−5 = 1µF 20 D.C voltage = 100 V, I =2A A.C voltage = 200 V ν = 50 Hz d.c voltage 100 = = 50 Ω I a.c voltage 200 = = 100 Ω Z= I R= Now Z = R + ω2 L2 or 100 = (50)2 + (314 L)2 or (314 L) = (100) − (50) or (100) − (50) 86.6 = 314 314 = 0.28 H L= S Chand & Company Limited 21 L = 63.66 × 10 – H, Erms = 10 volt, ν = 50 Hz XL = ωL = 2π νL = × 3.14 × 50 × 63.66 × 10 –3 = 314 × 63.66 × 10 –3 Ω Erms 10 = XL 314 × 63.66 × 10 −3 = 0.50 A I= 22 XL = 176Ω, ν = 50 Hz, L=? XL = ωL = 2π ν L L= XL 176 = 2π ν × 3.14 × 50 = 0.50 A 23 L = 3.0 H, C = 27 × 10 –6 F, Resonant frequency R = 7.4 Ω 1 2π LC 1 = , = 17.7 Hz 2π × 27 × 10−6 ν= 24 (a) Device X is a resistor and device Y is a Capacitor (b) When X and Y are connected in series their impedance is S Chand & Company Limited Z= R + X C2 Here R = X C = ∴ 220 = 440 Ω 0.5 Z = 440 Eeff 220 = = = 0.35 A Z 440 2 25 (a) The device P is a capacitor and device Q is a resistor I eff = (b) Here R = X C = 200 = 880 Ω 0.25 Impedance of the circuit Z = R + X C2 = (880) + (880) = 880 Ω Eeff 220 = = Z 880 = 0.18 A I eff = 26 Eeff = 12V, ν = 50 Hz, Ieff = 0.5 A S Chand & Company Limited π radians E 12 = 24 Ω Z = eff = I eff 0.5 φ= Power factor cos φ = So or Also π R = 24 R = or 24 R Z cos R = 12 Ω Z = R + ω2 L2 or (24)2 = (12)2 + (314)2 L2 or (24)2 − (12)2 = (314)2 L2 36 × 12 L2 = 314 × 314 or or L= 12 = 0.066 H 314 27 (a) Here Eeff = 60 V P = 10 W S Chand & Company Limited Eeff 60 × 60 = = 360 Ω 10 P 10 P Also I eff = = = A Eeff 60 ′ = 100 V and ν = 60 Hz Applied effective voltage Eeff Resistance of lamp R = ∴ Z= ′ Eeff 100 = = 600 Ω I eff 1/ Z = R + X L2 Now or X L2 = Z − R2 or 4π2 ν L2 = Z − R or L2 = Z − R2 4π ν = or Z = R + X L2 6002 − 3602 × (3.14)2 × (60)2 230400 × 9.85 × 3600 = 1.624 or L = 1.27 H (b) Value of resistance required in place of choke = 600 – 360 = 240 Ω = 28 Here E = 310 sin 314t, C = 25 × 10 – F, L = 0.10 H and R = 25.0 Ω (i) Comparing with the standard equation E = E0 sin 2π ν t, we have 2π ν t = 314t S Chand & Company Limited 314 = 50 Hz 2π 1 ( ii ) ( a ) X C = = π ν C × 3.14 × 50 × 25 × 10 − ν= or = 127.3 Ω XL = 2π νL = × 3.14 × 50 × 0.10 = 31.4 Ω Net reactance = XC – XL = 127.3 – 31.4 = 95.9 Ω (b) Impedance Z = R + ( X C − X L )2 = (25) + (95.9) = 625 + 9126.81 = 9821.81 = 99.1 Ω (c) Current I eff = 29 Here Eeff E = = Z 2Z 310 × 99.1 = 2.22 A L = 20 mH = 20 × 10 – H C = 50 µF = 50 × 10 – F Q0 = 10 mC = 10 × 10 – C S Chand & Company Limited (a) Total energy stored initially = = (10 × 10 −3 )2 × 50 × 10−6 Q0 2C = 1J Yes, the energy is conserved (b) Natural frequency of the circuit ν = = 2π LC × 3.14 20 × 10−3 × 50 × 10−6 = 159.2 Hz × 3.14 × 10 −3 (c) During LC oscillations the charge on the plates of the capacitor at any instant is given by = 2π t T (i) The energy stored will be completely electrical when Q = ± Q0 Q = Q cos ωt = Q cos i.e., or or t = 0, cos t= 2π t = ±1 T nT or 2π t = nπ, T n = 0, 1, 2, 3T 5T 1 T ,T , , T, , ; T = = = 6.3 × 10–3 s 2 ν 159 S Chand & Company Limited (ii) The energy stored is completely magnetic when Q = 2π t π = = cos (2n + 1) T 2π t π or = (2n + 1) T or t = (2n + 1) T / 4, n = 0,1, 2,3, or t = T / 4, 3T / 4, 5T / 4, (d) The times at which the total energy is shared equally between the inductor and the capacitor are i.e., cos (0 + T / 4) (T / + T / 2) , , 2 = T / 8, 3T / 8, 5T / 8, 7T / 8, (T / + 3T / 4) , (e) The energy dissipated as heat is equal to the initial energy, i.e., 1.0 J 30 Here V0 = 283 V, R = 3Ω, ν = 50 Hz L = 25.48 × 10– H C = 7.96 × 10– F S Chand & Company Limited (a ) Vrms = V0 = 283 2 ≈ 200 V (b) X L = ωL = π νL = × 3.14 × 50 × 25.48 × 10−3 = 8Ω XC = 1 = = 4Ω ωC 314 × 7.96 × 10−4 (c ) Z = R + ( X L − X C ) = (3) + (8 − 4) = + 16 = Ω V0 283 = = 56.6 A Z X − XC = tan φ = L R −1 or φ = tan (4 / 3) = 53.13° I 56.6 (e ) I rms = = = 40 A 1.414 Now VR = I rms R = 40 × = 120 V (d ) I = VL = I rms X L = 40 × = 320 V VC = I rms X C = 40 × = 160 V R = (40)2 × = 4800 W ( f ) P = I rms (g) Power factor = cos φ = cos 53.13o = 0.6 (approx.) S Chand & Company Limited = Vrms Irms cos φ = 200 × 40 × cos 53.13° = 200 × 40 × 06 = 4800 W (h) Power input (i ) At resonance ν = 2π LC = × 3.14 25.48 × 10−3 × 7.96 × 10−4 = 35.4 Hz ( j ) The impedance Z at resonance is Z = R = 3Ω (k ) The rms current at resonance Vrms 200 = = 66.67 A R (l ) Power dissipated at resonance = P = I rms R = (66.67)2 × = 13.33 × 103 = 13.33 KW 31 For graphs see Short-Answer Q 48 Current in case (i) is P 11000 = = 50 A V 220 Current is case (ii ) is I1 = I2 = 11000 = 0.5 A 22, 000 S Chand & Company Limited Since I2 < < I1, the I 2R loss will be much less in case (ii) Therefore the second mode of transmission should be preferred 32 Since the voltage is in phase with the current, the circuit is in resonance Therefore, X L = XC or 2πνL = or C= = 2πνC 4π2 ν L × (3.14) × (50) × 50 × 10−3 = × 10-4 F Impedance Z = R = Ω 33 See Short-Answer Q 16 = = = T ∫0 T E0 I T E0 I T Z= E sin ωt I sin (ωt − φ) dt 0 T ∫ sin ωt sin (ωt − φ) dt T ∫ sin ωt (sin ωt cos φ − cos ωt sin φ) dt .(1) Veff 10 = = 10 Ω I eff S Chand & Company Limited or or R π = cos φ = cos   = Z 3 R = 10 × = Ω ωL tan φ = R R tan φ L= ω × tan (π / 3) = = 0.027 H × 3.14 × 50 34 See Short-Answer Q 16 and 18 For maximum brightness, the current should be maximum This happens when the circuit is in resonance, i.e., when ωL = ωC C= or ω L = = × 10−4 F 4π × (50) × 50 × 10−3 S Chand & Company Limited 35 Resistance of heater, R1 = Vrms (220)2 = = 96.8 Ω 500 P1 Vrms (220) = = 484 Ω 100 P2 Total resistance across the a.c mains when the two are connected in series R = (96.8 + 484) = 580.8 Ω The root mean square valve of current in the circuit is Resistance of bulb, R2 = I rms = Vrms 220 A = R 580.8 Power consumed by the heater  220  =  × 96.8  580.8  = 13.9 W Power consumed by the bulb  220  =  × 484  580.8  = 69.4 W S Chand & Company Limited