Sinusoidal Steady State Analysis - Introduction to Electrical Engineering | EE 221, Assignments of Electrical and Electronics Engineering

Download Sinusoidal Steady State Analysis - Introduction to Electrical Engineering | EE 221 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! Examples - Chapter 10 - Sinusoidal Steady-State Analysis 1 Chapter 10, Problem 2. (a) If -10 cos ωt + 4 sin ωt = A cos (ωt + φ), where A > 0 and -180° < φ ≤ 180°, find A and φ. (b) If 200 cos (5t +130°) = F cos 5t + G sin 5t, find F and G. (c) Find three values of t, 0 ≤ t ≤ 1 s, at which i(t) = 5 cos 10t - 3 sin 10t = 0. (d) In what time interval between t = 0 and t = 10 ms is 10 cos 100πt ≥ 12 sin 100πt? Chapter 10, Solution 2. (a) (b) (c) (d) 10cos 4sin ACos ( ), A 0, 180 180 A 116 10.770, A cos 10, A sin 4 tan 0.4, 3 quad 21.80 201.8 , too large 201.8 360 158.20 d t t wt− ω + ω + +Φ > − ° < Φ ≤ ° = = Φ = − Φ = − ∴ Φ = ∴Φ = ° = ° ∴Φ = °− ° = − ° 200cos (5 130 ) Fcos5 G sin 5 F 200cos130 128.56 G 200sin130 153.21 + ° = + ∴ = ° = − = − ° = − ° t t t sin10 5( ) 5cos10 3sin10 0, 0 1 , 10 1.0304, cos10 3 0.10304 ; also, 10 1.0304 , 0.4172 ; 2 : 0.7314 = − = ≤ ≤ ∴ = = = = + π = π ti t t t t s t t t s t t s s 0 10ms, 10cos100 12sin100 ; let 10cos100 t =12sin100 t 10tan100 t = , 100 0.6947 2.211ms 0 2.211ms 12 < < π ≥ π π π ∴ π π = ∴ = ∴ < < t t t t t t = Examples - Chapter 10 - Sinusoidal Steady-State Analysis 2 Chapter 10, Problem 6. Compare the following pairs of wave forms, and determine which one is leading: (a) -33 sin (8t - 9°) and 12 cos (8t - 1°) (b) 15 cos (1000t + 66°) and -2 cos (1000t + 450°). (c) sin (t - 13°) and cos (t - 90°). (d) sin t and cos (t -90°). Chapter 10, Solution 6. (a) -33 sin(8t – 9o) → -33∠(-9-90)o = 33∠81o 12 cos (8t – 1o) → 12∠-1o (b) 15 cos (1000t + 66o) → 15 ∠ 66o -2 cos (1000t + 450o) → -2 ∠ 450o = -2 ∠90o = 2 ∠ 270o (c) sin (t – 13o) → 1∠-103o cos (t – 90o) → 1 ∠ -90o (d) sin t → 1 ∠ -90o cos (t – 90o) → 1 ∠ -90o These two waveforms are in phase. Neither leads the other. 33∠81o 12∠-1o -33 sin(8t – 9o) leads 12 cos (8t – 1o) by 81 – (-1) = 82o. 15∠66o 2∠270o 15 cos (1000t + 66o) leads -2 cos (1000t + 450o) by 66 – -90 = 156o. 1∠-103o 1 ∠ -90o cos (t – 90o) leads sin (t – 13o) by 66 – -90 = 156o. Examples - Chapter 10 - Sinusoidal Steady-State Analysis 3 Chapter 10, Problem 18. Assume that the op-amp in Fig. 10.50 is ideal (Ri = ∞ , Ro = 0, and A = ∞ ). Note also that the integrator input has two signals applied to it, -Vm cos ωt and vout. If the product R1C1 is set equal to the ratio L/R in the circuit of Fig. 10.4, show that vout equals the voltage across R in Fig. 10.4. Chapter 10, Solution 18. 1 1 1 1 1 1 1 1 LR , R 0, A , ideal, R C R V cos , R R ( V cos ) C R LV cos R C R For RL circuit, V cos L R LV cos R By c i o m out upper lower c upper lower out m out m out out out out R m r m R R t vi i ii i i v t v t v v v v d vt v dt t v v ω ω ω ω ω = ∞ = = ∞ = = − = ′∴ = + = − = − ′ ′∴ = + = +  = +     ′∴ = + omparison, R outv v= Examples - Chapter 10 - Sinusoidal Steady-State Analysis 4 Chapter 10, Problem 30. Let ω= 4 krad/s, and determine the instantaneous value of xi at 1mst = if Ix equals: (a) 5 80 A∠− ° (b) 4 1.5j− + A Express in polar form the phasor voltage Vx if ( )xv t equals: (c) 50 sin (250t - 40°) (d) 20cos108 30sin108t t− V (e) 33cos (80 50 ) 41cos (80 75 ) Vt t− ° + − ° Chapter 10, Solution 30. 4000, 1mstω = = and ω t = 4 rad (a) (b) (c) (d) (e) rad I 5 80 A 4cos (4 80 ) 4.294A = ∠− ° ∴ = − ° = − x xi rad I 4 1.5 4.272 159.44 A 4.272cos (4 159.44 ) 3.750 A− = − + = ∠ ° ∴ = + ° = x x j i ( ) 50sin (250 40 ) 50cos (250 130 ) V 50 103 V = − ° = − ° → = ∠− ° x x v t t t 20cos108 30sin108 20 30 36.06 56.31 V = − → + = ∠ ° xv t t j 33cos (80 50 ) 41cos (80 75 )! 33 50 41 75 72.27 63.87 V = − ° + − ° → ∠− °+ ∠− ° = ∠− ° xv t t 5 Examples - Chapter 10 - Sinusoidal Steady-State Analysis 9 Chapter 10, Problem 50. Two admittances, Y1 = 3 +j4 mS and Y2 = 5 + j2 mS, are in parallel, and a third admittance, Y3 = 2 - j4 mS, is in series with the parallel combination. If a current I1 = 0.1 A is flowing through Y1, find the magnitude of the voltage across (a) Y1; (b) Y2; (c) Y3; (d) the entire network. Chapter 10, Solution 50. (a) 11 13 1 I 0.1 30V 20 23.13 V 20V Y (3 4)10− ∠ ° = = = ∠− °∴ = + j (b) 2 1 2V V V 20V= ∴ = (c) (d) 3 2 2 2 3 1 2 3 3 33 3 I Y V (5 2)10 20 23.13 0.10770 1.3286 A I I I 0.1 30 0.10770 1.3286 0.2 13.740 A I 0.2 13.740V 44.72 77.18 V V 44.72V Y (2 4)10 j j − − = = + × ∠− ° = ∠− ° ∴ = + = ∠ °+ ∠− ° = ∠ ° ∠ ° ∴ = = = ∠ ° ∴ = − 1 3V V V 20 23.13 44.72 77.18 45.60 51.62 V 45.60V = + + ∠− °+ ∠ ° = ∠ ° ∴ = in in = Examples - Chapter 10 - Sinusoidal Steady-State Analysis 10 Chapter 10, Problem 54. Use phasors and mesh analysis on the circuit of Fig. 10.63 to find IB . Chapter 10, Solution 54. 3I 5(I I ) 0 2I 5I 0 3(I 5) 5(I I ) 6(I 10) 0 5I (9 5) I 60 15 0 5 60 15 9 5 75 300I 2 5 15 18 5 9 5 13.198 154.23 A B B D B D D D B D B D B j j j j j j j j j j j j j j j j j − − = ∴− + = + − − + + = ∴ + − = − − − − − − + = = − − − = ∠ ° -2jIB Examples - Chapter 10 - Sinusoidal Steady-State Analysis 11 Chapter 10, Problem 67. Find the input admittance of the circuit shown in Fig. 10.74, and represent it as the parallel combination of a resistance R and an inductance L , giving values for R and L if ù =1 rad/s. Chapter 10, Solution 67. Note: There is no dependent but one independent source. Therefore, find the input impedance/admittance through an experiment, .e.g., apply an input current of 1 ∠0o A. V 210 0.5V 1V (1 ) 2 11 2 V 1Z 1 2 1 At 1, Z 1 1 2 1 1Y 0.5 0.5 1 1 L L in in in in in j j j j j j j j j j j j j j ω ω ω ω ω ω ω ω ω = ∴ = ∴ = + + = + + ∴ = = + + = = − + = + ∴ = = + + R = 500 mΩ, L = 500 mH. so Yin = ( )12 2 −+ ωω ω j = jω1∠ o, Examples - Chapter 10 - Sinusoidal Steady-State Analysis 12 Chapter 10, Problem 82. In the circuit of Fig. 10.85, find values for (a) I1, I2, and I3.(b) Show Vs, I1, I2, and I3 on a phasor diagram (scales of 50 V/in and 2 A/in work fine). (c) Find Is graphically and give its amplitude and phase angle. Chapter 10, Solution 82. (a) (b) (c) 1 2 3 120I 3 30 A 40 30 120I 2.058 30.96 A 50 30 120I 2.4 53.13 A 30 40 j j = = ∠− ° ∠ ° = = ∠ ° − = = ∠− ° + 1 2 3I I I I 6.265 22.14 A s = + + = ∠− °