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 = \(\frac{D’}{C’}\)

(b) z22 = \(\frac{1}{C’}\)

(c) z22 = \(\left(\frac{A’ D’}{C’}  – B’\right)\)

(d) z22 = \(\frac{A’}{C’}\)

I had been asked this question during an online exam.

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

Answer 1

Right answer is (d) z22 = \(\frac{A’}{C’}\)

Easy explanation: We know that, V1 = z11 I1 + z12 I2 …………. (1)

V2 = z21 I1 + z22 I2 ……………. (2)

And   V2 = A’V1 – B’I1 ……… (3)

I2 = C’V1 – D’I1 …………… (4)

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

V2 = A’ \(\left(\frac{D’}{C’} I_1 + \frac{1}{C’} I_2\right) – B’I_1 = \left(\frac{A’ D’}{C’}  – B’\right) I_1 + \frac{A’}{C’} I_2\) ………… (5)

And V1 = \(\frac{D’}{C’} I_1 + \frac{1}{C’} I_2\) ………….. (6)

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

z11 =  \(\frac{D’}{C’}\)

z12 = \(\frac{1}{C’}\)

z21 = \(\left(\frac{A’ D’}{C’}  – B’\right)\)

z22 =   \(\frac{A’}{C’}\).