Chapter 12 Part 2-Electrical Circuit Analysis-Problem Solutions, Exercises of Electrical Circuit Analysis

Download Chapter 12 Part 2-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! February 5, 2006 CHAPTER 12 P.P.12.1 For the abc sequence, leads by 120° and leads by 120°. anV bnV bnV cnV Hence, =°+°∠= )12030(110anV °∠150110 V =°−°∠= )12030(110cnV 110∠–90˚V P.P.12.2 (a) °∠−°∠=−= 90-12030120bnanab VVV 120j)60j92.103(ab ++=V =abV 207.8∠60˚V Alternatively, using the fact that leads by 30° and has a magnitude of abV anV 3 times that of , anV °∠=°+°∠= 6085.207)3030()120(3abV Following the abc sequence, =bcV 207.8∠–60˚V =caV 207.8∠180˚V (b) Z V I ana = )7.0j6.0()19j24()3.0j4.0( +++++=Z °∠=+= 38.663220j25Z = °∠ °∠ = 38.6632 30120 aI A8.66-75.3 °∠ Following the abc sequence, =°∠= 120-ab II A128.66-75.3 °∠ =°∠= 240-ac II 3.75∠111.34˚A docsity.com P.P.12.3 The phase currents are = °∠ °∠ == Δ 4020 20-180AB AB Z V I A60-9 °∠ =°∠= 120-ABBC II A180-9 °∠ =°∠= 120ABCA II °∠609 The line currents are =°∠=°∠= 90-3930-3ABa II A90-59.15 °∠ =°∠= 120-ab II 15.59∠150˚A =°∠= 120ac II A3059.15 °∠ P.P.12.4 In a delta load, the phase current leads the line current by 30° and has a magnitude 3 1 times that of the line current. Hence, =°∠=°∠= 65 3 5.22 30 3 a AB I I A6513 °∠ Ω°∠=+=Δ 33.6963.2112j18Z )33.6963.21)(6513(ABAB °∠°∠== ΔZIV =ABV V98.692.281 °∠ P.P.12.5 °∠=+= 34.5121.1915j12YZ After converting the Δ-connected source to a Y-connected source, °∠=°−°∠= 15-56.138)30150( 3 240 anV = °∠ °∠ == 51.3421.19 15-56.138 Y an a Z V I A66.34-21.7 °∠ =°∠= 120-ab II A186.34-21.7 °∠ =°∠= 120ac II A53.6621.7 °∠ docsity.com )825.10j25.6-()856.19j392.2(BCCAc −−+=−= III =+= 681.30j642.8cI A74.2787.31 °∠ P.P.12.10 The phase currents are 44j j5- 0220 AB = °∠ =I °∠= °∠ = 3022 10j 0220 BCI °∠= °∠ = 120-22 10 120220 CAI The line currents are )05.19j11-()44j(CAABa −−=−= III =+= 05.63j11aI A80.164 °∠ )44j()11j05.19(ABBCb −+=−= III =−= 33j05.19bI A60-1.38 °∠ )11j05.19()05.19j11-(BCCAc +−−=−= III =−= 05.30j05.30-cI A2255.42 °∠ The real power is absorbed by the resistive load === )10()22()10(P 2 2 CAI kW84.4 P.P.12.11 The schematic is shown below. First, use the AC Sweep option of the Analysis Setup. Choose a Linear sweep type with the following Sweep Parameters : Total Pts = 1, Start Freq = 100, and End Freq = 100. Once the circuit is saved and simulated, we obtain an output file whose contents include the following results. FREQ IM(V_PRINT1) IP(V_PRINT1) 1.000E+02 8.547E+00 -9.127E+01 FREQ VM(A,N) VP(A,N) 1.000E+02 1.009E+02 6.087E+01 From this we obtain, =bBI A91.27-547.8 °∠ , =anV V60.879.100 °∠ docsity.com P.P.12.12 The schematic is shown below. In this case, we may assume that s/rad1=ω , so that Hz1592.021f =π= . Hence, 10XL L =ω= and 1.0X1C c =ω= . docsity.com Use the AC Sweep option of the Analysis Setup. Choose a Linear sweep type with the following Sweep Parameters : Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. Once the circuit is saved and simulated, we obtain an output file whose contents include the following results. FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592E-01 3.724E+01 8.379E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592E-01 1.555E+01 -7.501E+01 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592E-01 2.468E+01 -9.000E+01 From this we obtain, =caI A90-68.24 °∠ =cCI A79.8325.37 °∠ IAB 15.55∠–75.01˚A P.P.12.13 (a) If point o is connected to point B, =2P W0 )(ReP *aAB1 IV= =°+°= )06.410cos()05.18)(200(P1 W2722 )(ReP *cCB3 IV= where °∠=°+°∠== 60200)180-120(200- BCCB VV =°−°= )27.74(60cos)87.31)(200(P3 W6177 (b) Total power is =++=++= 617702722PPPP 321T W8899 P.P.12.14 , V208VL = W-560P1 = , W800P2 = (a) =+=+= 800-560PPP 21T W240 (b) =+=−= )560800(3)PP(3Q 12T 2.356 kVAR (c) °=θ⎯→⎯===θ 18.84815.9 240 6.2355 P Q tan T T docsity.com