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For the circuit given below, the value of g11 and g21 are _________________(a) g11 = 0.0667 – j0.0333 Ω, g21 =0.8 + j0.4 Ω(b) g11 = -0.0667 – j0.0333 Ω, g21 = -0.8 – j0.4 Ω(c) g11 = 0.0667 + j0.0333 Ω, g21 = 0.8 + j0.4 Ω(d) g11 = -0.0667 + j0.0333 Ω, g21 = 0.8 – j0.4 ΩI had been asked this question during an online interview.This intriguing question comes from Advanced Problems Involving Parameters topic in portion Two-Port Networks of Network Theory |
Answer» RIGHT option is (C) g11 = 0.0667 + j0.0333 Ω, G21 = 0.8 + j0.4 Ω For EXPLANATION I would SAY: V1 = (12-j6) I1 Or, g11 = \(\frac{I_1}{V_1} = \frac{1}{12-j6}\) = 0.0667 + j0.0333 Ω g21 = \(\frac{V_2}{V_1}= \frac{12I_1}{(12-j6) I_1}\) = \(\frac{2}{2-j}\) = 0.8 + j0.4 Ω. |
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