Question

For a T-network if the Open circuit Impedance parameters are z11, z12, z21, z22, then z22 in terms of Transmission parameters can be expressed as ____________

(a) z22 = \(\frac{A}{C}\)

(b) z22 = \(\frac{AD}{C – B}\)

(c) z22 = \(\frac{1}{C}\)

(d) z22 = \(\frac{D}{C}\)

This question was posed to me by my school teacher while I was bunking the class.

My doubt is from Relation between Transmission Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters topic in chapter Two-Port Networks of Network Theory

Answer 1

Correct choice is (d) z22 = \(\frac{D}{C}\)

To explain I would say: We know that, V1 = z11 I1 + z12 I2 …………. (1)

V2 = z21 I1 + z22 I2 ……………. (2)

And   V1 = AV2 – BI2 ……… (3)

I1 = CV2 – DI2 …………… (4)

Rewriting (3) and (4), we get,

V2 = \(\frac{1}{C}I_1 + \frac{D}{C}I_2\) …………… (5)

And V1 = \(A \left(\frac{1}{C}I_1 + \frac{D}{C}I_2\right) – BI_2 = \frac{A}{C}I_1 + \left(\frac{AD}{C} – B\right) I_2\) ………….. (6)

Comparing (1), (2) and (5), (6), we get,

z11 = \(\frac{A}{C}\)

z12 = \(\frac{AD}{C – B}\)

z21 = \(\frac{1}{C}\)

z22 =  \(\frac{D}{C}\).