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300 2 KO. 300 2 = 300 2, v = 2.5 x 108 m/s VL (Rin of receiver) 60 V Zo %3D Vin 100 MHz KO 2 m

300 2 KO. 300 2 = 300 2, v = 2.5 x 108 m/s VL (Rin of receiver) 60 V Zo %3D Vin 100 MHz KO 2 m

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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300 2 KO 300 2 60 V Vin 100 MHz Zo = 300 2, v=2.5 x 108 m/s VL %3D (Rin of receiver) KO 2 m
Transcribed Image Text:300 2 KO 300 2 60 V Vin 100 MHz Zo = 300 2, v=2.5 x 108 m/s VL %3D (Rin of receiver) KO 2 m
b. In a transmission network of AIT Open University, a purely capacitive impedance of -j300 Ohms placed in parallel with two 300 Ohms receivers. The University need you to determine both the input impedance and the power delivered to each receiver. Use the below for your computation. 300 a Z-300 2, v-2.5 10 m 300 2 (Rof receiver) 100 MH
Transcribed Image Text:b. In a transmission network of AIT Open University, a purely capacitive impedance of -j300 Ohms placed in parallel with two 300 Ohms receivers. The University need you to determine both the input impedance and the power delivered to each receiver. Use the below for your computation. 300 a Z-300 2, v-2.5 10 m 300 2 (Rof receiver) 100 MH
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Electrical Engineering homework question answer, step 1, image 1

                                                   Fig:  Given transmission network

It is given that load is two 300 ohm resistors and one -j300 ohm capacitor, all connected in parallel.

Hence,

zL=300∥300∥(-j300)zL=300×300300+300∥(-j300)zL=150∥(-j300)zL=-j300×150150-j300zL=120-j60 ohm

 

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