Question

Please use the values from the example below

Example 15.2 An RLC Series Circuit The ouput of an ac generator connected to an RLC series combination has a frequency of 200 Hz and an amplitude of 0.100 V. If R-4.00 Ω, L= 3.00x10-3 H, and C= 8.00× 10-4 F. what are (a) the e, (b) the inductive reactance, (c) the difference between the current and the emf of the generator? Strategy The reactances and impedance in (a)-(c) are found by substitutions into Equation 15.3, Equation 15.8, and Equation 15.11, respectively. The current amplitude is calculated from the peak voltage and the impedance. The phase diference between the current and the emf is calculated by the inverse tangent of the difference between the reactances divided by the resistance. Solution a. From Equation 15.3, the capacitive reactance is c-. 2π(200 Hz)(8.00 × 10-4 F) b. From Equation 15.8, the inductive reactance is 2(200 H2(3.00x 10 H) 3.77a c. Substituting the values of R, Xc, and X into Equation 15.11, we obtain for the impedance z-,(4.00 )2 +13.77 Ω-0.995 Ω)2-487 Ω. d. The current amplitude is 4 2.05x10-2A, e. From Equation 15.9, the phase difference between the current and the emf is tan 4.002 Tou = 0.607 rad. Significance The phase angle is positive because the reactance of the inductor is larger than the reactance of the capacitor. This OpenS tacx book is avalable for tree at htt r.orp/contenlool 2074/19 Chapter 151 Alhereaing-Current Cecuits

To solve this problem shown below. Also please show derivations for the problem below since they’re not given in the book. The answers for the values is: 2.00 V; 10.01 V; and 8.01 V.

Answer 1