Question

Exercise 31 A series ac circuit contains a 350 - resistor, a 20.0 - mH inductor, a 2.90 - uF capacitor, and an ac power source of voltage amplitude 45.0 V operating at an angular frequency of 360 rad/s. Part A What is the power factor of this circuit? ANSWER: cos - Part B Find the average power delivered to the entire circuit ANSWER: P- w Part C What is the average power delivered to the resistor, to the capacitor, and to the inductor? Enter your answers numerically separated by commas. ANSWER: PR. Pc. PL - w

Answer 1

in series LOR Circuit we have gonem and R = 3son L- 20ml C = 2.9 ME AC power source and anguted frequency w=360 radis. R PARTA power factar is Casa z cohore z is impedance za te rej? R²F *2=WL = 360x20x103 = 7:20 Xe = ㅗ 360x2axi - 957184 360x2:98766 سے نیا so z= Bat (3:2-957-832 2= 1013 n Cost = 350 1013 =0.35 cost=0.35 PART A Pax - Vyms Irms last i Kms Vo - 45 V 2 Irms = Vrms & toth so from cancio

lav = 45 x 45 xais 2x1013 Pava 708.75 2026 0.35W pava 0.25W PARTC average powel delivered to resistor is 2 far= Ver 2 (45/52] - R 45x45 -= 2.99 2x350 350 The average power to inducton and capaction is zero te par for inducter =0 Par for Cepatita o Reason: in case of inducter during the first quarter of each current cycle, as the current increased the magnetic flux tarughe the inclucth boulds up and energy is stored in the inductor from the external source in the next quarter of cylcles as the current decrease the flux deerease and stored energy is returned to the source. There in half cycle hohet power is consumed by the incluctha

in case of copaciter, it absorbed energy from the source for a quarter cycle as it is charged. it returns energy to source in the next quarter cycle as it is discharged. Thus in ce half cycle to net power is consumed by the Capacitor.