Question

A resistor (R = 9.00 x 102.2), a capacitor (C = 0.250 uF), and an inductor (L = 1.70 H) are connected in series across a 2.40 x 102-Hz AC source for which AVmax = 1.50 x 102 V. (a) Calculate the impedance of the circuit. Ο ΚΩ (b) Calculate the maximum current delivered by the source. (c) Calculate the phase angle between the current and voltage. (d) is the current leading or lagging behind the voltage? The current leads the voltage. The current lags behind the voltage. There is no phase difference between the current and voltage. A series RLC circuit has a resistance of 16.0 and an impedance of 75.0 12. If the rms voltage applied to the circuit is 170 V, what average power is delivered to the circuit? An AC voltage of the form Av = (95.0 V)sin(270t) is applied to a series RLC circuit. If R = 46.012, C = 28.0 uF, and L = 0.220 H, find the following. (a) impedance of the circuit (b) rms current in the circuit (c) average power delivered to the circuit

Answer 1

. Given: R = 9 x 10²2 C = 0.250 x 10-6 F Ls 1. To H f = 2.40 X 10 ² Hz A Vmaxe = 1150 X 102 v . (a) Angulare frequency can be given as w=2af . - 21 (2040 X1O? Hz). = 1507.96 Mad/s Impedence of the circuit is z = JR² + (x,-XJ ² = 909)++ (WL-ise) WC = 100)+ (1567.90) * C1.7) – Obras x costat de = 78100 + 3(2563.53) - (2652.5984 5810000 + 793).68 = 904.40 2 = 0.904 kr Maximum current I= AVmaz (6) 2

I = 1.50 X 162 - 9o4.no = 0.17 A (C3 Phase angle, $ tans (*) ah 0 tan | (1507196 x 2) - 1 L- ax60597.9) (6-150x159 - 1567.90 (0.250x156 9x10² 7 $tania [ 943.] ! [ 36 3.33 Goo a tant' (0.404) = 21.9840 (d) The the is current correct lags behind the voltage option. 2014