OHW 32 - terry(ect328 – oldhomework 32 – Turner...

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Unformatted text preview: terry (ect328) – oldhomework 32 – Turner – (59130) 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 99 . 8 MHz and 98 . 8 MHz, respectively. Electromagnetic signals from these ra- dio stations simultaneously supply sinusoidal emf s E KUT = V sin( ω KUT t ) and E KMFA = V sin( ω KMFA t ) to the circuit via an antenna. The receiving circuit includes an inductor of 0 . 57 μ H, a variable capacitor C , and a speaker whose resistance you must choose. C R . 57 μ H V sin( ω t ) Speaker K U T K M F A 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 . 46174 pC. Explanation: Let : ν = 99 . 8 MHz , = 9 . 98 × 10 7 Hz , and L = 0 . 57 μ H , = 5 . 7 × 10 − 7 H . The frequency of KUT is ω KUT = 2 π ν KUT = 2 π (9 . 98 × 10 7 Hz) = 6 . 27062 × 10 8 Hz . At resonance ω KUT = 1 √ L C KUT , so C KUT = 1 L ω 2 KUT = 1 (5 . 7 × 10 − 7 H) (6 . 27062 × 10 8 Hz) 2 × 10 12 pC 1 C = 4 . 46174 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 . 55252 pC. Explanation: Let : ν KMFA = 98 . 8 MHz = 9 . 88 × 10 7 Hz The frequency of KMFA is ω KMFA = 2 π ν KMFA = 2 π (9 . 88 × 10 7 Hz) = 6 . 20779 × 10 8 Hz . Thus C KMFA = 1 L ω 2 KMFA = 1 (5 . 7 × 10 − 7 H) (6 . 20779 × 10 8 Hz) 2 × 10 12 pC 1 C = 4 . 55252 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...
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OHW 32 - terry(ect328 – oldhomework 32 – Turner...

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