Tiêu đề 23. Alternating Current Hard Ans.pdf ✅

1. (d) When key is closed dots are linked with closed loop (i.e. increases from zero to a certain value) so induced current will be clockwise when key is opened dots linked with loop decreases (from a certain value to zero) so induced current will be anticlockwise in direction. 2. (a) e e sint = 0 where e0 = NBA 0 ' e 0 = (9)(4N)BA = 36 e 3. (a) s p p s N N i i = Now, according to the information given in the problem, ip can be calculated by using the formula, V = iR so s p p p s N N R V i =  (This is the secondary current in terms of Vp) Now to rearrange the result obtained above, in terms of secondary voltage, we must replace the term of Vp in the above result by Vs. We know that s p s p N N V V = ; s s p p N V N V = , Substituting this in equation (i) 2 2 s p p s s N N R V i = 4. (a) Comparing the given equation with i i sin t = 0   = 400   2 = 400   = 200 Hz. Also 2 0 i i rms = 2 50 2 = = 50 A. 5. (b) By using i i t i t i t = 0 sin = 0 sin 2 = rms 2 sin 2  i = 20 2 sin(150 t). 6. (b) At t = 0, i 3 A 2 3 2 3 2 sin 0  =  =      = +  . 7. (b) The given equation can be written as follows V = 50 2sin100t cos100t = 50sin 2(100t) = 50sin 200t (  sin 2 = 2 sin cos ) Hence peak voltage V 50 volt 0 = and frequency 100 . 2 200 = = Hz    8. (d) Time difference T.D.   =  2 T  T.D. sec T T 360 1 6 60 1 6 1 2 3 6 =  =  = = =    9. (a) Peak value V (8) (6) 10 volt 2 2 0 = + = so v volt rms 5 2 7.05 2 10 = = = . 10. (c) Phase difference relative to current 6 6 3     = − − = − In degree o  = − 60 i.e. voltage lag behind the current by 60o or current leads the voltage by 60o . 11. (b) r.m.s. value of wattless current sin  2 0 i = In this question i0 = 1 A and 2   = . So r.m.s. value of wattless current A 2 1 = 12. (c) iWL = i rms sin   3 = 2 sin  2 3 sin  =  o  = 60 so p.f. 2 1 = cos = cos 60 = o  . 13. (b) P = Vrms i rms cos  100 =  4  0.5 Vrms  Vrms = 50V so V 2 50 70 volt 0 =  = 14. (b) By using Z R cos =  200 cos 60 o R =  2 200 1 R =  R = 100 . 15. (b) Ammeter reads r.m.s. value so V C X V i rms C rms rms = =    100 (1 10 ) 2 10 20 . 2 200 2 6 2 i rms    =  = mA         = − − 16. (c) A L V X V i rms L rms rms 0.455 2 60 0.7 120 2 =   = = =   . 17. (b)
From the given equation i0 = 1A and V 200 volt 0 = . Hence = = 200  1 200 Z also 2 2 2 Z = R + X L  2 2 2 (200 ) (100 ) = + X L  = 100 3 . XL 18. (a) When dc is applied R V i =  R 100 1 =  R = 100. When ac is applied Z V i =  Z 100 0.5 =  Z = 200. Hence 2 2 2 2 2 2 Z = R + XL = R + 4  L  2 2 2 2 2 (200) = (100) + 4 (50) L  L = 0.55H. 19. (a) Impedance of the circuit = 2 2 2 R +  L Amplitude of voltage = Vo  Amplitude of current = 2 2 2 o R L V +  , Hence (A) is correct. 20. (c) XL = 8 100 , R = 10Ω 10 100 = ; L  100 = 8 100 or L = 8 1 H Z = 2 40 10 10 2 8 1 2 2  + =          I = 5 2A 2 10 10 2 100 Z E = = = Hence (C) is correct. 21. (d) Below resonant frequency the current leads the applied e.m.f., at resonance it is in phase with applied e.m.f. and above resonance frequency it lags the applied e.m.f., Hence (D) is correct. 22. (d) Flux linked with the loop will first increase and then decrease as electron pass by. 23. (b) Vsource = 2 C 2 VR +V  VC = 2 R 2 VSource – V = 2 2 (20) – (12) = 16 V 24. (d) Quantity of heat liberated in the ammeter of resistance R (i) Due to direct current of 3 ampere = [(3)2 R/J] (ii) Due to alternating current of 4 ampere = [(4)2R/J] Total heat produced per second = J (3) R 2 + J (4) R 2 = J 25R Let the equivalent alternating current be I ampere; then J I R 2 = J 25R or I = 5 amp 25. (a) From Kirchoff's current law, i3 = i1 + i2 = 3 sin t + 4 sin (t + 900 ) = 3 4 2(3)(4)cos90o 2 2 + + sin(t+ ) where tan  = 3 4cos90o 4sin 90o + = 3 4  i3 = 5sin(t+530 ) 26. (d) If AC is the square wave then all these three options are possible 27. (d) Volts (V) 8 2 Time 3 t(sec) I = dt dq q = it + a V = c q V = c it + a  V is proportional to time. 28. (b) P = Vrms rms cos P = Vrms Z Vrms Z R = 2 2 rms Z V R Z = 2 2 R + (L) = 2 2 (50) + (0.2250)
= 2 2500 + (50) = 50 2 P = 2 2 200         × 50 2 50 × 50 2 1 = 200 watt 29. (c) 10 = 2 L 2 r + X and r XL = tan 60 10 = 2 2 r + (r 3) or r = 5 , XL = 5 3  Z = 2 2 (5+ 5) + (5+ 3) = 175 = 13.2  tan  = R r XL + = 10 5 3 or  = tan–1         2 3 . 30. (d) When capacitance is removed tan  = R L or L = 100 tan 600...(1) when inductance is removed tan  = ( C)(R) 1  or C 1  = 100 tan 600 ....(2) So Z = R = 100  I = V/R = 200/100 = 2A Power P = I2 R = 4 × 100 = 400 W 31. (b) Equation of straight line from t = 0 to t = 1 f (t) = 4 t – 2 f 2 (t) = 16 t2 – 16 t + 4 rms =  − + T 0 2 (16 16t 4)dt T 1 32. (c) R = 100  XL = 2fL XC = 2 fC 1  Z = 2 L C 2 R + (X − X ) 33. (c) P = Vrms Irms cos  =         2 100         2 100 × 10–3 cos 3  = 2.5 W 34. (c)  = tan–1       R XL = 450  XL = R 35. (d) Use Pav = 2 rms Z V       R 36. (c) Use p s V V = p s N N Frequency in transformer remains same. 37. (b) Let  is the phase difference between V and i tan  = R XL = R L = 100 1001 = 4  From diagram it is clear that i lags with V  Instantaneous current equation will be i0 = sin (100 t –) where  = 4  and i0 = Z V0 = 2 2 0 R ( L) V +  = 2 2 (100) (100 1) 200 +  = 2  i = 2 sin (100 t – 4  ) 38. (b) The circuit is RLC resonant circuit. Xc = 2 A ~ 110 V XL = 2 V R = 55 VL VC VR  Reading of voltmeter = VL – VC = 0 Reading of ammeter = 2A 55 110 R E Z Erms rms = = = 39. (c) Irms = {< I2 >}1/2 = { < I1 2 cos2t + I2 2 sin2t + I1I2sin2t >}1/2 = 2 I I 2 2 2 1 + 40. (d) 2 V0 = V0 sin(t)
T 2 × t = 4   t1 = 8 7 Time taken to reach the peak is 4 T  interval = 8 T 4 T − = 8 T = 8 50 1 8f 1  = = 400 1 = 2.5 × 10–3 sec. 41. (a) P = Vrms Irms cos  =         2 150         2 150 cos /3 = 5625 W 42. (c) 2.5watt. 3 100 (100 10 ) cos 2 1 V i cos 2 1 P 3 0 0  =       =  =     − 43. (d) The given current is a mixture of a dc component of 3A and an alternating current of maximum value 6A Hence r.m.s. value 2 2 = (dc) + (r.m.s. value of ac) 2 2 2 6 (3)         = + (3) (3 2) 3 3 A 2 2 = + = . 44. (c) 2 6 2 2 2 10 2.5 2 50 1 (1000) 2 C 1 Z R             = +       = + −  = + =   2 2 3 Z (3000 ) (4000 ) 8 10 So power factor 0.6 5 10 3000 cos 3 =  = = Z R  and power Z V P V i rms rms rms   cos cos 2 = =  P 4.8W 5 10 (200 ) 0.6 3 2 =   = 45. (a) Capacitance of wire C 0.014 10 200 2.8 10 F 2.8F 6 6 =   =  = − − For impedance of the circuit to be minimum X L = X C  C L   2 1 2 =  0.35 10 H 0.35mH 4(3.14) (5 10 ) 2.8 10 1 4 C 1 L 3 2 2 2 3 2 6 =  =     =   = − −