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

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).