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:

Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter3: Power Transformers
Section: Chapter Questions
Problem 3.30P: Reconsider Problem 3.29. If Va,VbandVc are a negative-sequence set, how would the voltage and...
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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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