For the circuit given below, the values of the h parameter is given as follows h = [16, 3; -2, 0.01]. The value of the ratio \(\frac{V_2}{V_1}\)is ______________(a) 0.3299(b) 0.8942(c) 1.6(d) 0.2941This question was posed to me during an internship interview.I'd like to ask this question from Advanced Problems Involving Parameters topic in division Two-Port Networks of Network Theory

For the circuit given below, the values of the h parameter is given as

1.	For the circuit given below, the values of the h parameter is given as follows h = [16, 3; -2, 0.01]. The value of the ratio \(\frac{V_2}{V_1}\)is ______________(a) 0.3299(b) 0.8942(c) 1.6(d) 0.2941This question was posed to me during an internship interview.I'd like to ask this question from Advanced Problems Involving Parameters topic in division Two-Port Networks of Network Theory
Answer» Right option is (d) 0.2941 To elaborate: Replacing the given 2-port circuit by an equivalent circuit and applying nodal ANALYSIS, we get, V2 = (20) (2I1) = 40 I1 Or, -10 + 20I1 + 3V2 = 0 Or, 10 = 20I1 + (3) (40I1) = 140I1 ∴ I1 = \(\frac{1}{14}\) and V2 = \(\frac{40}{14}\) So, V1 = 16I1 + 3V2 = \(\frac{136}{14}\) And I2 = (\(\frac{100}{125}\)) (2I1) = \(\frac{-8}{70}\) ∴ \(\frac{V_2}{V_1}= \frac{40}{136}\) = 0.2941.

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