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

For a T-network if the Open circuit Impedance parameters are z11, z12,

1.	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» 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) AndV1 = 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 =\(\frac{D}{C}\).

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