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30 - Kapoor(mk9499 homework30 Turner(60230 This print-out...

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Kapoor (mk9499) – homework30 – Turner – (60230) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 6) 10.0 points You are designing a radio receiver based on the idea of a series LRC circuit as shown be- low. The two major radio stations in town are KUT and KMFA, which broadcast at frequen- cies 97 . 2 MHz and 96 . 2 MHz, respectively. Electromagnetic signals from these ra- dio stations simultaneously supply sinusoidal emf s E KUT = V 0 sin( ω KUT t ) and E KMFA = V 0 sin( ω KMFA t ) to the circuit via an antenna. The receiving circuit includes an inductor of 0 . 64 μ H, a variable capacitor C , and a speaker whose resistance you must choose. C R 0 . 64 μ H V 0 sin( ω t ) Speaker KUT KMFA At what value of the variable capacitance C will the sound volume ( i.e. , the power dissi- pated in the speaker) be maximum for KUT? Correct answer: 4 . 18917 pC. Explanation: Let : ν = 97 . 2 MHz , = 9 . 72 × 10 7 Hz , and L = 0 . 64 μ H , = 6 . 4 × 10 7 H . The frequency of KUT is ω KUT = 2 π ν KUT = 2 π (9 . 72 × 10 7 Hz) = 6 . 10726 × 10 8 Hz . At resonance ω KUT = 1 L C KUT , so C KUT = 1 L ω 2 KUT = 1 (6 . 4 × 10 7 H) (6 . 10726 × 10 8 Hz) 2 × 10 10 pC 1 C = 4 . 18917 pC . 002 (part 2 of 6) 10.0 points At what value of the variable capacitance C will the sound volume be maximum for KMFA? Correct answer: 4 . 27671 pC. Explanation: Let : ν KMFA = 96 . 2 MHz = 9 . 62 × 10 7 Hz The frequency of KMFA is ω KMFA = 2 π ν KMFA = 2 π (9 . 62 × 10 7 Hz) = 6 . 04442 × 10 8 Hz . Thus C KMFA = 1 L ω 2 KMFA = 1 (6 . 4 × 10 7 H) (6 . 04442 × 10 8 Hz) 2 × 10 10 pC 1 C = 4 . 27671 pC . 003 (part 3 of 6) 10.0 points In order for the radio receiver to be useful it must discriminate between the two radio stations. Your company’s specifications state that when the radio is tuned to KUT, the sound volume of KMFA signals should be 100 times weaker.
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