Hw24-solutions - howard(cah3459 – hw24 – turner –(56705 1 This print-out should have 16 questions Multiple-choice questions may continue on

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: howard (cah3459) – hw24 – turner – (56705) 1 This print-out should have 16 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 . 4 MHz and 98 . 4 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 receiv- ing circuit includes an inductor of 1 . 2 μ H, a variable capacitor C , and a speaker whose resistance you must choose. C R 1 . 2 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: 2 . 13642 pC. Explanation: Let : ν = 99 . 4 MHz , = 9 . 94 × 10 7 Hz , and L = 1 . 2 μ H , = 1 . 2 × 10- 6 H . The frequency of KUT is ω KUT = 2 π ν KUT = 2 π (9 . 94 × 10 7 Hz) = 6 . 24549 × 10 8 Hz . At resonance ω KUT = 1 √ LC KUT , so C KUT = 1 Lω 2 KUT = 1 (1 . 2 × 10- 6 H) (6 . 24549 × 10 8 Hz) 2 × 10 12 pC 1 C = 2 . 13642 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: 2 . 18006 pC. Explanation: Let : ν KMFA = 98 . 4 MHz = 9 . 84 × 10 7 Hz The frequency of KMFA is ω KMFA = 2 π ν KMFA = 2 π (9 . 84 × 10 7 Hz) = 6 . 18265 × 10 8 Hz . Thus C KMFA = 1 Lω 2 KMFA = 1 (1 . 2 × 10- 6 H) (6 . 18265 × 10 8 Hz) 2 × 10 12 pC 1 C = 2 . 18006 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. howard (cah3459) – hw24 – turner – (56705) 2 What resistance R should the speaker have in order for your design to just meet company specifications? Correct answer: 1 . 52326 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 = R 2 + ω KMFA L- 1 ω KMFA C KUT 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 + ω KMFA L- 1 ω KMFA C KUT 2 at KMFA resonance frequency. Solving for R , R = 1 99 ω KMFA L- 1 ω KMFA C KUT = 1 99 ω KMFA L- ω 2 KUT L ω KMFA = 1 99 L ω KMFA ω 2 KMFA- ω 2 KUT = 1 99 2 π L ν KMFA ν 2 KMFA- ν 2 KUT = 1 99 2 π (1 . 2 × 10- 6 H) (9 . 84 × 10 7 Hz) × (9 . 84 × 10 7 Hz) 2- (9 . 94 × 10 7 Hz) 2 = 1 . 52326 Ω ....
View Full Document

This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.

Page1 / 9

Hw24-solutions - howard(cah3459 – hw24 – turner –(56705 1 This print-out should have 16 questions Multiple-choice questions may continue on

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online