Question

In the circuit shown in the figure, if ω = 40 rad/s, then Zin is _____________ (a) 4.77 – j1.15 Ω (b) 2.96 – j0.807 Ω (c) 2.26 – j3.48 Ω (d) 2.26 – j3.48 Ω I got this question during a job interview. The above asked question is from Advanced Problems on Resonance in section Resonance & Magnetically Coupled Circuit 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

Correct choice is (b) 2.96 – j0.807 Ω Easy explanation: Z = Z11 + \(\frac{ω^2 M^2}{Z_{22}} \) Or, Z = 2 + j (40) \((\frac{1}{10}) + \frac{40^2 (\frac{1}{4})^2}{4+j(40)\frac{1}{2}}\) = 2 + j (4) + \(\frac{100}{4+j (20)} \) = 2 + j (4) + \(\frac{100}{4+j (20)} × \frac{4-j (20)}{4-j (20)}\) = 2 + j (4) + \(\frac{400-j (2000)}{416}\) = 2.96 – j0.807 Ω.