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

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.