Question

For the circuit given below, the value of Transmission parameter A and C are ____________ (a) A = -0.7692 + j0.3461 Ω, C = 0.03461 + j0.023 Ω (b) A = 0.7692 + j0.3461 Ω, C = 0.03461 + j0.023 Ω (c) A = -0.7692 – j0.3461 Ω, C = -0.03461 + j0.023 Ω (d) A = 0.7692 – j0.3461 Ω, C = 0.023 + j0.03461 Ω The question was asked by my college director while I was bunking the class. This is a very interesting question from Advanced Problems Involving Parameters topic in portion 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

Right option is (b) A = 0.7692 + j0.3461 Ω, C = 0.03461 + j0.023 Ω To explain I would say: V = [20 + (-j10) || (j15 − j20)] I1 V1 = \Big[20 + \frac{(-j10)(-j5)}{-j15}\Big] I1 = [20 – j\frac{10}{3} ] I1 I0 = \left(\frac{-j10}{-j10-j5}\right) I1 = \frac{2}{3} I1 V2 = (-j20) I0 + 20I’0 = –\frac{j40}{3}I_1 + 20I_1 = (20 – \frac{j40}{3}) I_1 ∴ A = \frac{V_1}{V_2} = \frac{(20-\frac{j10}{3})I_1}{20-\frac{j40}{3}) I_1} = 0.7692 + j0.3461 Ω ∴ C = \frac{I_1}{V_2} = \frac{1}{20-j40/3} = 0.03461 + j0.023 Ω.