Question

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

(a) z22 = 0.0667 Ω

(b) z22 = 2.773 Ω

(c) z22 = 1.667 Ω

(d) z22 = 0.999 Ω

The question was asked in an online interview.

The origin of the question is Advanced Problems on Two Port Network topic in portion Two-Port Networks of Network Theory

Answer 1

The correct option is (b) z22 = 2.773 Ω

To explain I would say: z11 = \(\frac{V_1}{I_1}\)  = 2 + 1 || [2+1 || (2+1)]

z11 = 2 + 1 || (2 + \(\frac{3}{4}\)) = 2 + \(\frac{1×\frac{11}{4}}{1+\frac{11}{4}} = 2 + \frac{11}{15}\) = 2.733

I0 = \(\frac{1}{1+3}\) I’0 = \(\frac{1}{4}\) I’0

And I’0 = 1 + \(\frac{11}{4}\)  I1 = \(\frac{4}{15}\) I1

Or, I0 = \(\frac{1}{4} × \frac{4}{5} I_1 = \frac{1}{15} I_1\)

Or, V2 = I0 = \(\frac{1}{15} I_1\)

z21 = \(\frac{V_2}{I_1} = \frac{1}{15}\) = z12 = 0.0667

z22 = \(\frac{V_2}{I_2}\)  = 2+1 || (2+1||3) = z11 = 2.733

∴ [z] = [2.733:0.0667; 0.0667:2.733] Ω.