InterviewSolution
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InterviewSolution
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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 ΩI 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 |
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Answer» 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. |
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