VD=------- Volt1.3313.31330.133VB= -------Volt1414( 0.70 0 0 0ос0.14IB= -------- А.( 12 A12 Micro A12 mAo 12 Mega A0 0 0 0Q/ For the circuit below,VCC=20 Volt, Bd.c.-83, RC=4Kohm, RS= 3K ohm, R1-500K ohm.Find:loss-8 mAV-VVE=0 0 0 0VE-VBVE = VDVE= VS(VE=VC

It is given that: VCC=20 Vβdc=83RC=4kΩRS=3kΩRB=500kΩ

Q/ For the circuit below. VCC-20 Volt, Bd.c.-83, RC=4K ohm, RS= 3K ohm, R1-500K ohm. Find: pss-8 mA --4V ww VD=------ Volt 1.33 13.3 133 0.133 VB= -------Volt 14 1.4 0.7 0.14 IB--------- A. 12 A 12 Micro A 12 mA O 12 Mega A. VE= VE-VB VE = VD VE=VS VE=VC

Q/ For the circuit below. VCC-20 Volt, Bd.c.-83, RC=4K ohm, RS= 3K ohm, R1-500K ohm. Find: pss-8 mA --4V ww VD=------ Volt 1.33 13.3 133 0.133 VB= -------Volt 14 1.4 0.7 0.14 IB--------- A. 12 A 12 Micro A 12 mA O 12 Mega A. VE= VE-VB VE = VD VE=VS VE=VC

Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)

10th Edition

ISBN:9781337399128

Author:Russell E. Smith

Publisher:Russell E. Smith

Chapter6: Reading Schematic Diagrams

Section: Chapter Questions

Problem 4RQ: Schematic diagrams break the wiring of control systems down into a(n) _____ arrangement. a....

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Digital Integrated Circuit Design

Digital IC design is a procedural interaction that involves converting determinations and highlights into computerized blocks, which are then further transformed into basic circuits. A significant number of the imperatives related with advanced IC plan come from the foundry cycle and innovative restrictions.

Question

VD=------- Volt
1.33
13.3
133
0.133
VB= -------Volt
14
14
( 0.7
0 0 0 0
ос
0.14
IB= -------- А.
( 12 A
12 Micro A
12 mA
o 12 Mega A
0 0 0 0
Q/ For the circuit below,
VCC=20 Volt, Bd.c.-83, RC=4K
ohm, RS= 3K ohm, R1-500K ohm.
Find:
loss-8 mA
V-V
VE=
0 0 0 0
VE-VB
VE = VD
VE= VS
(VE=VC

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