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Question 31 Given: • VDD = 30 V •R, =19 kohms • R2 = 3 kohms • Rp = 629 ohms • R3 = 0 kohms •R = 12 kohms • Iocon) = 18 mA at VGslon) = 7 V • Vasth) = 2 V * Em = 27 ms Sove: Answers should be expressed using 2 decimal places 1. k (transfer constant) = Blank 1 mA/V2 2. Ip = Blank 2 mA 3. V - Blank 3 V %3D

Question 31 Given: • VDD = 30 V •R, =19 kohms • R2 = 3 kohms • Rp = 629 ohms • R3 = 0 kohms •R = 12 kohms • Iocon) = 18 mA at VGslon) = 7 V • Vasth) = 2 V * Em = 27 ms Sove: Answers should be expressed using 2 decimal places 1. k (transfer constant) = Blank 1 mA/V2 2. Ip = Blank 2 mA 3. V - Blank 3 V %3D

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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Question 31 Given: • VDD = 30 V •R, =19 kohms • R2 = 3 kohms • Rp = 629 ohms • R3 = 0 kohms •R = 12 kohms • Iocon) = 18 mA at Veston) = 7 V • Vasth) = 2 V * Em = 27 ms Solve: Answers should be expressed using 2 decimal places 1. k (transfer constant) = Blank 1 mA/V2 2. Ip = Blank 2 mA 3. VG = Blank 3 V 4. Vs = Blank 4 V 5. Ves = Blank 5 v %3D 6. Vos = Blank 6 v 7. Rintot) = Blank 7 kohms 8. Voltage gain, A, = Blank 8 +VpD Rp C3 R1 Vout RL Via R2 Rs
Transcribed Image Text:Question 31 Given: • VDD = 30 V •R, =19 kohms • R2 = 3 kohms • Rp = 629 ohms • R3 = 0 kohms •R = 12 kohms • Iocon) = 18 mA at Veston) = 7 V • Vasth) = 2 V * Em = 27 ms Solve: Answers should be expressed using 2 decimal places 1. k (transfer constant) = Blank 1 mA/V2 2. Ip = Blank 2 mA 3. VG = Blank 3 V 4. Vs = Blank 4 V 5. Ves = Blank 5 v %3D 6. Vos = Blank 6 v 7. Rintot) = Blank 7 kohms 8. Voltage gain, A, = Blank 8 +VpD Rp C3 R1 Vout RL Via R2 Rs
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