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 ΩThe 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

For the circuit given below, the value of Transmission parameter A and

1.	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 ΩThe 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
Answer» 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.

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