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V1=vs1= a) 3.5 b) 2.5 c) -3. 5 d) 0.4 e) -0. 3 f) 1.5 g) -0. 6 h) -1. 5 j) -2. 5 k) -0. 2 l) -0. 1

V1=vs1= a) 3.5 b) 2.5 c) -3. 5 d) 0.4 e) -0. 3 f) 1.5 g) -0. 6 h) -1. 5 j) -2. 5 k) -0. 2 l) -0. 1

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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V1=vs1=
a) 3.5
b) 2.5
c) -3. 5
d) 0.4
e) -0. 3
f) 1.5
g) -0. 6
h) -1. 5
j) -2. 5
k) -0. 2
l) -0. 1
m) non of the listed values

 

 

V2=vs2=
a) 3.5
b) 2.5
c) -3. 5
d) 0.4
e) -0. 3
f) 1.5
g) -0. 6
h) -1. 5
j) -2. 5
k) -0. 2
l) -0. 1
m) non of the

For the circuit shown 1. VI = VS1 = 2. V2 = VS2 = +1V +1V Q1 VT1 = VT2 = 0.4V v1 Knl = Kn2 = 2.5mAV: Q2 V2 4KO -1V
Transcribed Image Text:For the circuit shown 1. VI = VS1 = 2. V2 = VS2 = +1V +1V Q1 VT1 = VT2 = 0.4V v1 Knl = Kn2 = 2.5mAV: Q2 V2 4KO -1V
Expert Solution
Step 1

According to question:

Assuming the saturation condition and is given as

VGS>VTH

 

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