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USE APPROXIMATE ANALYSIS USE 2 decimal places, no commas. Given: Voltage-divider biased, npn BJT with 1. VCC=17V 2. R1 33kohms 3. R25kohms 4. RC = 2kohms 5. RE 1 kohms 6. BDC = 144 ** Solve for: 1, VB Blank 1 V: 2. VE Blank 2 V; 3. IE Blank 3 mA; 4. VCE Blank 4 V; 5. VC Blank 5V:

USE APPROXIMATE ANALYSIS USE 2 decimal places, no commas. Given: Voltage-divider biased, npn BJT with 1. VCC=17V 2. R1 33kohms 3. R25kohms 4. RC = 2kohms 5. RE 1 kohms 6. BDC = 144 ** Solve for: 1, VB Blank 1 V: 2. VE Blank 2 V; 3. IE Blank 3 mA; 4. VCE Blank 4 V; 5. VC Blank 5V:

Introductory Circuit Analysis (13th Edition)
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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USE APPROXIMATE ANALYSIS USE 2 decimal places, no commas. Given: Voltage-divider biased, npn BJT with 1. VCC=17V 2. R1 33kohms 3. R25kohms 4. RC = 2kohms 5. RE 1 kohms 6. BDC = 144 ** Solve for: 1, VB Blank 1 V: 2. VE Blank 2 V; 3. IE Blank 3 mA; 4. VCE Blank 4 V; 5. VC Blank 5V:
USE APPROXIMATE ANALYSIS USE 2 decimnal places, no commas. Given: Voltage-divider biased, npn BJT with 1. VCC = 17V 2. R1 = 33kohms 3. R2 = 5kohms 4. RC = 2kohms 5. RE = 1kohms 6. BDC = 144 Solve for: 1. VB = Blank 1 V; 2. VE = Blank 2 V 3. IE = Blank 3 mA; 4. VCE = Blank 4 V; 5. VC = Blank 5 V; Blank 1 pdyour answer
Transcribed Image Text:USE APPROXIMATE ANALYSIS USE 2 decimnal places, no commas. Given: Voltage-divider biased, npn BJT with 1. VCC = 17V 2. R1 = 33kohms 3. R2 = 5kohms 4. RC = 2kohms 5. RE = 1kohms 6. BDC = 144 Solve for: 1. VB = Blank 1 V; 2. VE = Blank 2 V 3. IE = Blank 3 mA; 4. VCE = Blank 4 V; 5. VC = Blank 5 V; Blank 1 pdyour answer
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