ENGHow did he integrate thisstep?Example 2.2: Finding the Performance Parameters of a Battery ChargerThe battery voltage in Figure 2.4(a) is E 12 V and its capacity is 100 Wh. Theaverage charging current should be Ide = 5 A. The primary input voltage is Vp =120 V, 60 Hz, and the transformer has a turn ratio of n = 2:1. Calculate(a) the conduction angle & of the diode.(b) the current-limiting resistance R.(c) the power rating PR of R.(d) the charging time ho in hours.(e) the rectifier efficiency n.(f) the PIV of the diode.Solution of Example 2.2E = 12 V, V, = 120 V, V, = V/n= 120/2 = 60 V, and Vm= V2 x Vs, = v2 x 60 =84.85 V.a) From Eq. (2.17), a = sin (12/84.85) = 8.13° or 0.1419 rad. B 180 8.13 =171.87°. The conduction angle is 8 = B- a = 171.87-8.13 = 163.74°.b) The average charging current Ide is.sin ot – E(2V cos a +2Ea- nE), for B = n- a2R2лR(2.18)which gives1(2V cos a + 2Ea- nE)2d deR =1(2x84.85x cos 8.13 +2x12x 0.1419 - nx12)= 4.2622n x5c) The rms battery current Ims is1", sin ox – E) d(ox)%3D27aR²V2sin 2a - 4V E cos a1+E (7-2a)+.(2.19)2 7R?I = 67.4I = 8.2 AThe power rating of R is PR = 8.22 x 4.26 = 286.4 W.INGCAL ENG

Integration of sin(wt) is -coswt. First integrate the sinwt and then substitute lower and upper…

Answered: How did he integrate this step?

Engineering

Electrical Engineering

How did he integrate this step?

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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