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

This is solution to problems related Electrical Circuit Analysis course. It was given by Prof. Gurnam Kanth at Punjab Engineering College. Its main points are: ACB, Sequence, Phase, Circuit, Balanced, Y-connected, Load, Expressions, Voltages, Line

Typology: Exercises

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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