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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 1 00 10 = = × LC . rad s For each angular frequency, we find Z R L C = + 2 2 1 ω ω b g then I Z = 1 00 . V and P = I 2 1 00 . a f . The full width at half maximum is: f f = = = × = ω π ω π π 2 1 000 5 0 999 5 2 1 00 10 2 159 0 3 1 . . . b g s Hz while R L 2 1 00 159 3 π π = × = . 2 1.00 10 H Hz e j . ω ω ω ω 0 2 L C Z P 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
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