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hw24-solutions

# hw24-solutions - howard(cah3459 hw24 turner(56705 This...

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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 0 R = 1 2 V 2 0 Z 2 . When the radio is tuned to KUT Z KUT = R , and and P KUT = 1 2 V 2 0 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 0 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 . 004 (part 4 of 6) 10.0 points For this speaker resistance, with the radio re- ceiver tuned resonantly to one of the stations, what amplitude V 0 of emf must be supplied to the circuit in order for the speaker to dissi- pate 2 . 6 W of average power of that station’s sound?
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