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

In the circuit given below, the current through R is 2 sin 8t. The val

1.	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
Answer» 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).

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