Question

Resonance: 9.1 Given the network below find 00, 11, (0.) and Q. 100 20mH 6Ccosot o vdo 9.2 Repeat Problem 9.1 if the value of R is changed to 10. 9.3 Determine the resonant frequency, Q. BW. and the average power dissipated by the network at resonance. 20 50mH M SUF 1Ccososto

Answer 1

Resonance. 9.1 Gäven metwork, 22 102 aomh P - Go cosct ( itajwL 14F | Gx. : VCD. @) for Series RLC network, resonant frequency * goodle = 60.=jhe I got a 50 :) w el in = " cat.otik siod fee : V20x10* 1x li) Let vitja Gocos cot V. Roxie 941667 اع . By voltage division aude, Vo(t) = (t & Rtiului at walo, gube so J x G Ohkola Go x 14.142 . .. = 846.536V (ii) Quality factor for a concies R-L-c circuit, Q = load cake om i lod qoxeo 3 14.14 Q = Tolixit Ava Ww=.07 Kaad fee Ivokol = 846. 536V Q. = 14.64

Give M e - ☺ for Seues Rat-c circuit, rejenant frequenter como sic ita f.cift kacole Et) let Villa Go Coswt . By voltage dierson sule, Volts = V, GJ x Rt id-I at Resonance, wh o * Vo(es) = Go coscete wc = Go Gesuk a ROHKEM 194 iſ vocool. 60 x 141.42 = 8465.36 V C) ouaisty factor of Services R-toc Givceüt, Q = = =+ oktober = 141.42 i *** TW.7.07 kradee 1 Volu) =8188.26V 1Q=141.42 ven network, 250 ml 60 5 FT locouet

) for a series Roll Carcedit, resonant frozucies for ansice (uz adata) sofy and Sound Syot = 318.309 Hz e quality factor, Q = a sporto » Q=50 Qi) Bandesidth , B W 318.309_-6-366 Hz 750 Evy At siclonance, wh o In Net impedance imaginary part is zero- Zz Rtiwl-ci But ait Resonance (Free to & Zeer ... Average power distipaited at grejenance, 0 2 2xQ * Pawg - Vm = (0)2 = 100a esit EVg(H) = lo cosuit, Namalov 1 Kafe . fr = 318.2017 Hz, ub= Quoo geodec Q = 50 Bk = 6.366 Hz Paig = 25ki