Question

AC Electrical Circuit Analysis (ELEC 20001) - Spring - 20- CW (Assignment2) - All - QP Q3. a) An Antenna circuit have range of capacitor (from 40 x10-12 to 360x10-12F) with a 240 x10-6 This antenna coil has a resistance of 12 ohm. Determine the range of frequency for the radio signals make the radio is tunable and find the value of Q for the lowest and highest value of frequency range. (20 marks) b) A current source in a linear circuit has is = 12 cos(450nt - 459) A. Find the value of i, at t=2ms (5 marks)

Answer 1

a Guver :- C = 40 pF to 360 pF 940 цн 12 ). Solo we know that, Б: 2пJoc По» с= 4o pF 2 Jцох 10 "2 x 240 x 10 о В. 16243.8.3Н, Now, for с= 360 PF. а б, - Р 9.11 360 x 1612 x 24o кo6 ev 54 1456. A1 н, 2 * Ronge of frequency is from 541456.11 H₂ to 1694368. Зн,. ie O.Gч\ MH2 So 1.6 24M H2. Now, We Janow that, и -ly R

Foun C = 40PF. - 0,= 240x10-6 40 x 1012 - 2ou Foor c = 360PF 12 240 x 10-6 360 x 1612 68 b] given : is = 12 Cos cason f-45°) Sol? : at t = 2m Sec. In nodian , uso = 0.25 T. 4 :. js = 12 cos (usot so tx 2x10 3 0.250) js = 12 Cos (0.91 0.257 ) is = 12 casco.65T) . ون 11. 9923 A

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