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Fill in the blank : For the circuit shown below, solve for rL VDD = 20 V R, = 4.7 k2 IDss = 10 mA Vas(om) = -4 V Cout = 100 µF Cin = 0.1 µF +2.15 V R = 22 kn Re R = 1 kn 300 mV. = 1.5 MN Cs = 100 pF 22kohms O 3.87kohms 1.3kohms O 4.7kohms

Fill in the blank : For the circuit shown below, solve for rL VDD = 20 V R, = 4.7 k2 IDss = 10 mA Vas(om) = -4 V Cout = 100 µF Cin = 0.1 µF +2.15 V R = 22 kn Re R = 1 kn 300 mV. = 1.5 MN Cs = 100 pF 22kohms O 3.87kohms 1.3kohms O 4.7kohms

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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Fill in the blank : For the circuit shown below, solve for rL VDD = 20 V R = 4.7 k Ipss = 10 mA Vas(om = -4 V Cout = 100 µF Cin = 0.1 µF +2.15 V R = 22 k2 Re R = 1 k2 300 mV. = 1.5 MQ Cs = 100 pF 22kohms O 3.87kohms 1.3kohms O 4.7kohms 루
Transcribed Image Text:Fill in the blank : For the circuit shown below, solve for rL VDD = 20 V R = 4.7 k Ipss = 10 mA Vas(om = -4 V Cout = 100 µF Cin = 0.1 µF +2.15 V R = 22 k2 Re R = 1 k2 300 mV. = 1.5 MQ Cs = 100 pF 22kohms O 3.87kohms 1.3kohms O 4.7kohms 루
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