Question

For the circuit given below, the value of z22 parameter is ____________

(a) z22 = 1.775 + j5.739 Ω

(b) z22 = 1.775 – j4.26 Ω

(c) z22 = -1.775 – j5.739 Ω

(d) z22 = 1.775 + j4.26 Ω

I got this question at a job interview.

Origin of the question is Advanced Problems on Two Port Network in division Two-Port Networks of Network Theory

Answer 1

The correct choice is (c) z22 = -1.775 – j5.739 Ω

Easiest explanation: 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.