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Coursehero >> Pennsylvania >> UPenn >> ESE 216

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UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
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UPenn - ESE - 216
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UPenn - ESE - 216
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UPenn - ESE - 216
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION:The correct answer is a.0 1LC11m(10 ) 10,000rads11 Z ( j 0 ) V0 ( j 0 ) 120 c 60 C90 60 10,000(10 )90 60 90V20Chapter 12: Variable Frequency Network PerformanceProblem 12. FE-1
UPenn - ESE - 216
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION:The correct answer is c.0 BW 1LC120m(50 ) 1000radsRrad 200LsR 200(20m) 4Chapter 12: Variable Frequency Network PerformanceProblem 12. FE-2
UPenn - ESE - 216
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION:The correct answer is d.11V0ZCjCRC Gv ( j )11VI Z C R R j jCRCat DC, Gv=1=0dBat 3dB down, 1RCrad1 200s(5k )(1 ) 2ff 31.83 32 Hz2Chapter 12: Variable Frequency Network Pe
UPenn - ESE - 216
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION:The correct answer is b.0 L1LC 1000rads11 100mH2 0 C (1000) 2 (10 )BW Rrad 100LsR 100m(100) 10Chapter 12: Variable Frequency Network PerformanceProblem 12. FE-4
UPenn - ESE - 216
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION:The correct answer is a.f 8 Hz , 1 16rads1 j 2 kjC11j CRCGv ( j 1 ) 11 R j jCRCZC For f=8Hz and 1=16V0ZC j 2000 G v ( j 1 ) 0.707 45VSZ C R 2000 j 2000Gv ( j16 ) 0.707 4
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216
UPenn - ESE - 216