Question

For the circuit given below, the value of the Transmission parameter B and D are __________ (a) D = 0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω (b) D = 0.6923 + j0.5385 Ω, B = 6.923 + j25.385 Ω (c) D = -0.6923 + j0.5385 Ω, B= 25.385 + j6.923 Ω (d) D = -0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω I had been asked this question in my homework. The doubt is from Advanced Problems Involving Parameters topic in section Two-Port Networks of Network Theory Select the correct answer from above options network theory questions and answers, network theory questions pdf, network theory question bank, network theory gate questions and answers pdf, mcq on network theory pdf, gate network theory questions and solutions, network theory mcq Test , ies network theory questions, Network theory Questions for GATE EC Exam, Network Theory MCQ (Multiple Choice Questions)

Answer 1

The correct choice is (a) D = 0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω Easiest explanation: Z1 = \frac{(-j15)(-j10)}{-j15-j10-j20} = j10 Z2 = \frac{(-j10)(-j20)}{-j15} = -\frac{j40}{3} Z3 = \frac{(j15)(-j20)}{-j15} = j20 -I2 = \frac{20-j40/3}{20-\frac{j40}{3}+j20}I_1 = \frac{3-j2}{3+j} I1 ∴ D = \frac{-I_1}{I_2} = \frac{3+j}{3-j2} = 0.5385 + j0.6923 Ω V1 = [j10 + 2(9+j7)] I1 = jI1 (24 – j18) So, B = –\frac{V_1}{I_2} = \frac{-jI_1 (24-j18)}{-\frac{3-j2}{3+j} I_1} = \frac{6}{13}(-15+j55) ∴ B = -6.923 + j25.385 Ω.