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

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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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Differential amplifier

There are two amplifying stages with common degeneration in a differential amplifier.

Question

ENG
How did he integrate this
step?
Example 2.2: Finding the Performance Parameters of a Battery Charger
The battery voltage in Figure 2.4(a) is E 12 V and its capacity is 100 Wh. The
average 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 2.2
E = 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- a
2R
2л
R
(2.18)
which gives
1
(2V cos a + 2Ea- nE)
2d de
R =
1
(2x84.85x cos 8.13 +2x12x 0.1419 - nx12)= 4.262
2n x5
c) The rms battery current Ims is
1", sin ox – E) d(ox)
%3D
27
a
R²
V2
sin 2a - 4V E cos a
1
+E (7-2a)+.
(2.19)
2 7R?
I = 67.4
I = 8.2 A
The power rating of R is PR = 8.22 x 4.26 = 286.4 W.
ING
CAL ENG

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