Q3.Calculate the impedance parameter for the circuit below:20Ω100 250 Ω60 212v,Ans:3.70429.88[z] =- 70.37 11.11

Answered: Q3. Calculate the impedance parameter…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Q3.
Calculate the impedance parameter for the circuit below:
20Ω
100 2
50 Ω
60 2
12v,
Ans:
3.704
29.88
[z] =
- 70.37 11.11

Expert Solution

Step 1

Calculate the parameters $Z_{11}$ and $Z_{21}$ ,open circuit the output port $(I_{2} = 0)$ .

$A p p l y K i r c h o f f' s c u r r e n t l a w a t n o d e v_{x} - \frac{v_{x} - V_{1}}{20} + \frac{v_{x}}{50} + \frac{v_{x} - V_{2}}{100} = 0 \frac{5 v_{x} - 5 V_{1}}{100} + \frac{2 v_{x}}{100} + \frac{v_{x} - V_{2}}{100} = 0 8 v_{x} - 5 V_{1} - V_{2} = 0 8 v_{x} = 5 V_{1} + V_{2} . . . . . . (1) A p p l y K i r c h o f f' s c u r r e n t l a w a t n o d e V_{2} \frac{V_{2} - v_{x}}{100} + \frac{V_{2} - (- 12 v_{x})}{60} = 0 \frac{3 V_{2} - 3 v_{x}}{300} + \frac{5 V_{2} + 60 v_{x}}{300} = 0 3 V_{2} - 3 v_{x} + 5 V_{2} + 60 v_{x} = 0 8 V_{2} = - 57 v_{x} V_{2} = - 7.125 v_{x} . . . . . . . (2) R e c a l l i n g e q u a t i o n (1) 8 v_{x} = 5 V_{1} + V_{2} S u b s t i t u t e - 7.125 v_{x} f o r V_{2} 8 v_{x} = 5 V_{1} - 7.125 v_{x} 15.125 v_{x} = 5 V_{1} v_{x} = 0.33 V_{1} T h e c u r r e n t I_{1} i s, I_{1} = \frac{V_{1} - v_{x}}{20} = \frac{V_{1} - 0.33 V_{1}}{20} = \frac{0.67 V_{1}}{20} C a l c u l a t i n g t h e z - p a r a m e t e r Z_{11} Z_{11} = \frac{V_{1}}{I_{1}} S u b s t i t u t e \frac{0.67 V_{1}}{20} f o r I_{1} Z_{11} = \frac{V_{1}}{\frac{0.67 V_{1}}{20}} = 29.9 Ω$

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