Question

0.500 H, C = 6.00 pF, and 4. In the circuit below, Vo = 120 V, W = 200 rad/s, L R= 325 2. R V = V, sin(ot) C (a) (3 pts) Calculate the rms current in the resistor. (b) (2 pts) Calculate the average power dissipated in the resistor. I (c) (2 pts) You are able to vary the angular frequency w of the AC emf source in this circuit while keeping Vo = 120 V. For what value of w will the average power dissipated in the resistor be largest? (d) 3 pts) Calculate the average power dissipated in the resistor when w has the value

Answer 1

R AAAAM man Vo = 120V 01000 V-VISITO co=200 hadis LEO SOOH C-6x10GE R=3252 @ Empedance of circuit is Za JR+C@L- de)? = (325) + 200X0.500 - 200X6X106 Iling = Veens Z 3252+(-733-33)2 7 = 802.12.12 Rimas value of curbut will current will be same for all components (Vo/J2 Z => 1 Isting = (120/12) 0-106A 0:1064 202-12 O Average fewer dissipated in resistor - P = Toms x R = (0.106) x 325 = P = 3.65 W Z 2 © Polder dissipation will be maximum at Pesonant frequency which is given by de

w = = -> frequency Vic Josoxbx106 W = 577.35 red 16 @ When w is equal to resonant then effective Impedance of circuit becomes equal to Ri.e. - Z=R=325R poder dissipated by Jeesistor - P = Veuns (Vol.2) 온 Vo ZR R R 2 =) p = (1202 2x325 - P = 22.15 W