Question

In the circuit given below, the current through R is 2 sin 8t. The value of R is ___________ (a) (0.18 + j0.72) (b) (0.46 + j1.90) (c) – (0.18 + j1.90) (d) (0.23 – 0.35 j) The question was posed to me during a job interview. Question is from Relation between Hybrid Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters topic in portion Two-Port Networks 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 (d) (0.23 – 0.35 j) For explanation I would say: Here, Inductor is not given, hence ignoring the inductance. Let I1 and I2 are currents in the loop then, I1 = \frac{2 sin ⁡8t}{3} = 0.66 sin 8t Again, I2 = \frac{-j X 4 X 0.75 I_1}{3.92-2.56j} = (0.23 – 0.35j) sin 8t So, R = (0.23 – 0.35 j).