October 26, 2001HV#2-1 ECE209 –01 Network Analysis II Homework #2 (cp-10) October 26, 2001 10.5 750 0 40()150 3000 40 50t for t msitt for t ms≤≤=−≤≤ 201()TrmsIitdtT=∫ ()30.04 0.052200.0410(750 ) (150 3000 )50rmsI t dt t dt=+−∫∫ 323 2 325 60.04 0.0510 750150 9(10 ) 9(10 )0 0.0450 3 2 3rmstttIt=+−+ 323 2 3 2 325 6 25 610 750 (0.04) (0.05) (0.05) (0.04) (0.04)150 (0.05) 9(10 ) 9(10 ) (150 (0.04) 9(10 ) 9(10 )50 3 2 3 2 3rmsI =+−+−+−    ()31012 1125 1125 375 900 720 19250rmsI =+−+−+− ()31015 30050rmsI == 17.32rmsI = t(ms) i(A) 30 10 20 30 40 50 60HV#2-2 10.8 2uF4K10HIg Ig-j5kj1k4K () 30cos100git tmA= 100(10) 1jL j jkω==Ω, 6115100(2)10jkjC jω−==−Ω Using current division 0001 1000 90I3001 4 5 5656.8 45gjkIjk k jk∠==∠+− ∠− I 5.303 135 mA=∠! 2235.303(10 )4000 56.2422IPR mW−== = 2235.303(10 )( 5000) 70.3022CIQX mW−== −=− 56.24 70.30SPjQ j=+ = − 22 2 256.24 70.30 90.03SPQ mVA=+= + = i(t)HV#2-3 10.13 5/9uF480Vg -j360480Vg a. 22240602 2(480)gVPWR== = 22240802 2( 360)gCVQ VARX== =−− 22 2 2max60 60 70 60 100 160pPPQ=+ + = + + = + = b. 22 2 2min60 60 70 60 100 40pPPQ=− + = − + = − =− c. 60PW= d. 80Q VAR=− e. Yes f. ggVV240 0 240 0I 0.5 0.67 0.83 53.13480 360Cjj ARjX∠∠=+ = + =+ = ∠!!! cos(0 53.13 ) 0.6pf =− =! leading g. sin(0 53.13 ) 0.8rf =− =−!HV#2-4 10.20 2400 V(rms) 1 23 0.2Vs 1j1.62 3-+ 118 24SjkVA=+, 236 48SjkVA=−, 318 0SjkVA=+ 12272 24SSS S jkVA=++= − 3*rms0rmsS (72 24)10I3010V 2400 0jjA−== =−∠, I30 10jA=+ 00rmsrmsV2400 075.89 18.9 72 24I3010Zjj∠== = ∠− =−Ω− a. ()VVIsrms rms rms ThZ=+ 0 0()V 2400 0 (30 10)(0.2 1.6) 2390 50 2390 1.2srmsjj j V=∠++ + = +=∠ b. 0()()V2390 1.2I0.2 1.6 72 24 72.2 22.4srmssrmsjj j∠==++− − 00()02390 1.2I 31.61 18.475.59 17.2srmsA∠==∠∠− 22()0.2(31.62) 199.96SSrmsPRI W== = 22()1.6 1.6(31.62) 1600SSrmsQ I VAR== = c. 199.96 72000 72199.96TPW=+= 1600 24000 22400TQ VAR=− =− d. 72000100 100 99.7272199.8STPPη== = IsHV#2-5 10.25 0.14Vg10.050.05Vg2 152SjkVA=+, 23.75 1.5SjkVA=−, 380SjkVA=+ 1( )*11( )52I4016V 125 0rmsrmsSjkVAj+== =+∠!, ! 1( )I 40 16 43.08 21.8rmsjA=− = ∠−! 2( )*22( )3750 1500I3012V 125 0rmsrmsSjj+== =+∠!, 2( )I 30 12 32.31.08 21.8rmsjA=− = ∠−! 3( )*33( )8000 0I32V 250 0rmsrmsSj+== =∠!, ! 1( )I 32 0 32 0rmsjA=+=∠! 1( ) 1( ) 3( )IIIg rms rms rms=+, 1( )I 40 16 32 0 72 16 73.75 12.52grmsjjj A=− ++=− = ∠−! 2( ) 2( ) 3( )IIIg rms rms rms=+, 2( )I 30 12 32 0 62 12 63.15 10.95grmsjjj A=− ++=− = ∠−! 3( ) 1( ) 2( )II-Igrms rms rms=, 3( )I 40 16 30 12 10 4 10.77 21.8grmsjjj A=− −+ =−= ∠−! 11 3V 0.05(I ) 125 0 0.14(I )gg g=+∠+! 1V 0.05(72 16) 125 0 0.14(10 4)gjj j=−+++− 1( )V 130 1.36grmsjV=+ 2( ) 3 2V 0.14(I ) 125 0 0.05(I )grms g g=− + ∠ +! 2V 0.14(10 4 ) 125 0 0.05(62 12)gjj j=− − + + + −! 2( )V 126.6 0.04grmsjV=− Sources complex powers *22()2()V I (126.6 0.04)(62 12)g g rms g rmsSjj==−+ 27849.7 1516.7gSjVA=+ *11()1()V I (130 1.36)(72 16)g g rms g rmsSjj==++ 19338.2 2177.9gSjVA=+ S1 S2 S3 125∠0° V(rms) 125∠0° V(rms) I1 I2 I3 250∠0° V(rms) Ig1 Ig2 Ig3HV#2-6 b. 220.05 10.05(73.75) 271.95gPRI WΩ== = 220.05 20.05(63.15) 199.4gPRI WΩ== = 220.14 30.14(10.77) 16.23gPRI WΩ== = *111V (I ) 125(40 16) 5000 2000ssSjjVA==+=+ *222V (I ) 125(30 12) 3750 1500ssSjjVA==+=+ *333V (I ) 250(32 0) 8000 0ssSjjVA==+=+ 271.95 199.4 16.23 5000 2000 3750 1500 8000 0TSjjj=++++++++ 17237.5 3500TSVA=+ Total source complex power 127849.7 1516.7 9338.2 2177.9gg gSSS j j=+= + + + 1217187.9 3694.6gT g gSSS j VA=+= +HV#2-7 10.31 VgLo300Ro1uF 200Tsπµ=, 10000 /rsω=, ( ) 150 2 cos(10000 )gVt tV=, a. 461110010 10jjjCω−=− =− Ω 4310 (0.6)10 60jL j jω−==Ω Using voltage division () ()100V V 0.1 0.3(150 0 ) 15 45300 100Th rms g rmsjjjVj−==−∠=−−! (300)( 100)30 90300 100ThjZjj−==−Ω− 1030VTh-j90j60 30 90 10 60 40 30TZj j j=− ++ =− Ω (rms)V15 45I 0.78 0.54 0.9487 34.740 30ThjjAZj−== =− = ∠−−! The average power in the 10Ω: 2210 0.9487 9PRI W== = b. *30 90LThZZ j==+Ω, Then Inductance impedance goes max 80jwL j=Ω 222200( ) 30 ( 90 80) 31.62Th ThRRXX=+−+=Ω, We can have R max 20Ω V15 45I 0.46 0.8 0.93 6030 90 20 80ThjjAZj j−== = − =∠−−++! 1-20Ω 1-8mH j10~j80ΩHV#2-8 2220 0.93 17.30PRI W== = c. yes d. *30 90LThZZ j==+Ω, 90jwL j=Ω, 030R =Ω, V15 45I 0.25 0.75 0.79 71.630 90 30 90ThjjAZj j−== =− =∠−−++! 2230 0.79 18.72PRI W== = e. 030R =Ω 90jwL j=Ω, ! , 490910jLmHΩ== f. yesHV#2-9 10.35 -j24j15Vg8R18 j6 ()V 630 0grmsV=∠! a. 24(18 6)815 24724 (18 6)ThjjZj jjj−+=+ + = + Ω−++ 22 2224 7 25Th ThRRX=+=+=Ω b. 24V 630 0 420 2 4524 (18 6)ThjVjj−=∠=∠−−++!! V 420 420ThjV=− 420 2 45 420 2 45I 12 53.1324 7 25 49.49 8.13Aj∠− ∠−===∠−++ ∠!!!! 2225(12) 3600PRI W== = IHV#2-10 10.47 j4010 j40Vg j4010 j40Vg 111110I (40 40)I 30I 30I 360 0jj++− −=∠! 1( )360 0I 7.2 14.410 20rmsjAj∠==−+! 1V ( 40 30)IThjj=− ()10(7.2 14.4) 144 72Th rmsVj j jV=−=+ From the second circuit 1 1112210I (40 40)I 30I 30I 40I 30I 360 0jjjj++− −+ − =∠! 12(10 20)I 10I 360 0jj++=∠! (1) 12( 40 30)I 40I 0jj j−+ + = 1210I 40I 0jj−+ = (2) From (1) and (2) 2I 2.215 3.877jA=− 1I18 0jA=+ 2V144 72Z236I 2.215 3.877ThThjjj+== =+Ω− *Z=Z 2 36LThj=− Ω 144 72I36184jjA+==+ 222(36 18 )2 3240LLPIR W==+= b. *1V I 360(18) 6480ggPW== = j30 360∠0° V(rms) I1 j30 360∠0° V(rms) I1 I2HV#2-11 10.45 (ext) 375100mH40mH400Vg50mH ( ) 248cos10000gvt tV= 12(375 400)I (375 500)I 248 0jj+−+=∠! 12(375 400)I (375 400 1000)I 0jj−+ + ++ = 1I 0.80 0.62 1.012 37.7jA=− = ∠−! 2I 0.4 0.3 05 36.87jA=− =∠−! 2400 21400 502PI WΩ== b. 12I=I -I 0.80 0.62 0.4 0.3 0.4 0.32jjjA=− −+=− I=0.512 38.65 A∠−! 23751I 375 49.202PWΩ== c. 111V I 248(0.8) 99.2022ggPW== = I1