Homework Combined

Homework Combined - Kapoor(mk9499 – homework30 – Turner...

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Unformatted text preview: 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 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 . 64 μ H, a variable capacitor C , and a speaker whose resistance you must choose. C R . 64 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 . 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. What resistance R should the speaker have in order for your design to just meet company Kapoor (mk9499) – homework30 – Turner – (60230) 2 specifications? Correct answer: 0 . 812501 Ω. Explanation: In general the average power dissipated in the speaker/resistor is P = 1 2 I 2 R = 1 2 V 2 Z 2 . When the radio is tuned to KUT Z KUT = R , and and P KUT = 1 2 V 2 R . When the radio is tuned for KUT Z KMFA = radicalBigg R 2 + bracketleftbigg ω KMFA L- 1 ω KMFA C KUT bracketrightbigg 2 , and P KMFA = 1 2 V 2 R Z 2 KMFA . To meet the company’s specifications P KUT = 100 P KMFA , or Z 2 KMFA = 100 R 2 = R 2 + parenleftbigg ω KMFA L- 1 ω KMFA C KUT parenrightbigg 2 at KMFA resonance frequency. Solving for R , R = radicalbigg 1 99 bracketleftbigg ω KMFA L- 1 ω KMFA C KUT bracketrightbigg = radicalbigg 1 99 bracketleftbigg ω KMFA L- ω 2 KUT L ω KMFA bracketrightbigg = radicalbigg 1 99 L ω KMFA bracketleftbig ω 2 KMFA- ω 2 KUT bracketrightbig = radicalbigg 1 99 2 π L...
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Homework Combined - Kapoor(mk9499 – homework30 – Turner...

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