1.

For the circuit given below, the value of z11 parameter is ____________(a) z11 = 1.775 + j5.739 Ω(b) z11 = 1.775 – j4.26 Ω(c) z11 = -1.775 – j4.26 Ω(d) z11 = 1.775 + j4.26 ΩThis question was addressed to me by my school teacher while I was bunking the class.The question is from Series-Parallel Interconnection of Two Port Network in chapter Two-Port Networks of Network Theory

Answer»

The correct choice is (d) Z11 = 1.775 + j4.26 Ω

To ELABORATE: z1 = \(\frac{12(J10)}{12+j10-j5} = \frac{j120}{12+j5}\)

z2 = \(\frac{j60}{12+j5}\)

Z3 = \(\frac{50}{12+j5}\)

z12 = z21 = z2 = \(\frac{(-j60)(12-j5)}{144+25}\) = -1.775 – j4.26

z11 = z1 + z12 = \(\frac{(j120)(12-j5)}{144+25}\) + z12 = 1.775 + j4.26

z22 = z3 + z21 = \(\frac{(50)(12-j5)}{144+25}\) + z21 = 1.7758 – j5.739

∴ [z] = [1.775 + j4.26; -1.775 – j4.26; -1.775 – j4.26; 1.775 – j5.739] Ω.



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