Question

The impedance seen by the source in the circuit is given by __________

(a) (0.54 + j0.313) Ω

(b) (4 – j2) Ω

(c) (4.54 – j1.69) Ω

(d) (4 + j2) Ω

I got this question in an international level competition.

I want to ask this question from Fourier Series Analysis using Circuits in chapter Fourier Analysis on Complex Spaces of Signals and Systems

Answer 1

Correct choice is (c) (4.54 – j1.69) Ω

For explanation I would say: Z1 = 10∠30° × \((\frac{1}{4})^2\)

Z1 = (0.54 + j0.31) Ω

Total impedance= (4 – j2) + (0.56 + j0.31)

= (4.54 – j1.69) Ω.