947_Physics ProblemsTechnical Physics

947_Physics ProblemsTechnical Physics - 288 P33.72...

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288 Alternating Current Circuits P33.72 ω 0 6 1 100 10 == × LC . r a d s For each angular frequency, we find ZR L C =+ 2 2 1 ωω bg then I Z = 100 V and P = I 2 af . The full width at half maximum is: f f = × = π 2 10005 09995 2 2 159 0 31 .. . s Hz while R L 2 159 3 = × = 21 . 0 01 0 H Hz ej . 0 2 L C ZP I R 1 W 0.9990 0.9991 0.9993 0.9995 0.9997 0.9999 1.0000 1.0001 1.0003 1.0005 1.0007 1.0009 1.0010 999.0 999.1 999.3 999.5 999.7 999.9 1000 1000.1 1000.3 1000.5 1000.7 1000.9 1001 1001.0 1000.9 1000.7 1000.5 1000.3 1000.1 1000.0 999.9 999.7 999.5 999.3 999.1 999.0 2.24 2.06 1.72 1.41 1.17 1.02 1.00 1.02 1.17 1.41 1.72 2.06 2.24 0.19984 0.23569 0.33768 0.49987 0.73524 0.96153 1.00000 0.96154
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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