Download Chapter 11 Part 2-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! February 5, 2006 CHAPTER 11 P.P.11.1 )30t10cos(15)60t10sin(15)t(i °−=°+= )20t10cos(80)t(v °+= )30t10cos()20t10cos()15)(80()t(i)t(v)t(p °−°+== )]30-20cos()3020t20[cos(1580 2 1 )t(p °−+°−°+⋅⋅= =)t(p W7.385)10t20cos(600 +°− =θ−θ= )cos(IV 2 1 P ivmm W7.385 P.P.11.2 °∠== 8200ZIV )cos(IV 2 1 P ivmm θ−θ= =°−°= )308cos()10)(200( 2 1 P W2.927 P.P.11.3 °∠= + °∠ = 57.2653.2 j3 458 I For the resistor, °∠== 57.2653.2R II °∠== 57.2659.73R IV === )59.7)(53.2( 2 1 IV 2 1 P mmR W6.9 3 Ω + − I j1 Ω8∠45° V docsity.com For the inductor, °∠= 57.2653.2LI °∠=°+°∠== 57.11653.2)9057.26(53.2j LL IV =°= )90cos()53.2( 2 1 P 2L W0 The average power supplied is =°−°= )57.2645cos()53.2)(8( 2 1 P W6.9 P.P.11.4 Consider the circuit below. For mesh 1, 0)2j-()2j8(40- 21 =+−+ II 20j)j4( 21 =−− II (1) For mesh 2, 0)2j-()2j4j(20j- 12 =+−+ II 10jjj- 21 =+ II (2) In matrix form, ⎥⎦ ⎤ ⎢⎣ ⎡ =⎥⎦ ⎤ ⎢⎣ ⎡ ⎥⎦ ⎤ ⎢⎣ ⎡ − 10j 20 jj- j-j4 2 1 I I 4j2 +=Δ , j20-101 +=Δ , 60j102 +=Δ °∠= Δ Δ = 14.53511I and °∠=Δ Δ = 11.176.1322I For the 40-V voltage source, °∠= 040sV °∠= 14.5351I =°= )14.53-cos()5)(40( 2 1- Ps W60- j4 Ω -j2 Ω 8 Ω + − 40 V j20 V + − I1 I2 docsity.com == 3 16 Irms A309.2 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ == )9( 3 16 RIP 2rms W48 P.P.11.8 , π=T π<<= t0),tsin(8)t(v ∫∫ π π == 0 2T 0 22 rms dt))tsin(8( 1dtv T 1V [ ] 32dt)t2cos(1 2 164V 0 2 rms =−π = ∫ π =rmsV V657.5 === 6 32 R V P 2 rms W333.5 P.P.11.9 The load impedance is Ω°∠=+= 7.3311.7240j60Z The power factor is =°= )7.33cos(pf (lagging)832.0 Since the load is inductive A7.23-08.2 7.3311.72 10150 °∠= °∠ °∠ == Z V I The apparent power is === )08.2)(150( 2 1 IVS rmsrms VA156 P.P.11.10 The total impedance as seen by the source is 2j8 )6j8)(4j( 10)6j8(||4j10 − − +=−+=Z °∠= 62.2069.12Z The power factor is =°= )62.20cos(pf (lagging)936.0 °∠= °∠ °∠ == 20.62-152.3 62.2069.12 040 Z V I rmsrms docsity.com The average power supplied by the source is equal to the power absorbed by the load. W118)88.11()152.3(RIP 22rms === or === )936.0)(152.3)(40(pfIVP rmsrms W118 P.P.11.11 (a) )15-4.0)(85110(*rmsrms °∠°∠== IVS =S VA7044 °∠ == SS VA44 (b) 35.41j05.157044 +=°∠=S =P W05.15 , =Q VAR35.41 (c) =°= )70cos(pf (lagging)342.0 °∠= °∠ °∠ == 70275 15-4.0 85110 rms rms I V Z =Z Ω+ 4.258j06.94 P.P.11.12 (a) If , °∠= 75-250Z =°= )-75cos(pf (leading)2588.0 (b) = ° = θ =⎯→⎯θ= )75-sin( kVAR10 sin Q SsinSQ 10.35 kVA (c) 8.1608)250)(10353(SV V S rms 2 rms ==⋅=⎯→⎯= Z Z == rmsm V2V kV275.2 P.P.11.13 Consider the circuit below. 20 Ω + − V I + Vo − I2 I1 (60+j20) Ω (30–j10)Ω docsity.com Let be the current through the 60-Ω resistor. 2I 4 60 240 R P IRIP 2222 ===⎯→⎯= (rms)22 =I 40j120)20j60(2o +=+= IV 4.2j2.3 10j30 o 1 +=− = V I 4.2j2.521 +=+= III )40j120()48j104(20 o +++=+= VIV =+= 88j224V 240.7∠21.45˚ Vrms For the 20-Ω resistor, °∠=+== 8.2454.11448j20420 IV °∠=+= 8.24727.54.2j2.5I )24.8-727.5)(8.2454.114(* °∠°∠== IVS =S VA656 For the (30 – j10)-Ω impedance, °∠=+= 43.185.12640j120oV °∠=+= 87.3644.2j2.31I )36.87-4)(18.435.126(*1o1 °∠°∠== IVS =°∠= 18.44-5061S VA160j480 − For the (60 + j20)-Ω impedance, °∠= 022I )0-2)(43.185.126(*2o2 °∠°∠== IVS =°∠= 43.182532S VA80j240 + The overall complex power supplied by the source is )24.8-727.5)(21.4567.240(*T °∠°∠== IVS =°∠= 3.35-3.1378TS VA80j1376 − docsity.com
