Question

For a T-network if the Open circuit impedance parameters are given as z11, z21, z12, z22, then z11 in terms of Inverse Hybrid parameters can be expressed as ________

(a) z12 = \(\frac{1}{g_{11}} \)

(b) z12 = – \(\frac{g_{12}}{g_{11}} \)

(c) z12 = – \(\frac{g_{21}}{g_{11}} \)

(d) z12 = \(\left(g_{22} – \frac{g_{21} g_{12}}{g_{11}}\right)\)

The question was posed to me in final exam.

The origin of the question is Relation between Hybrid Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters in portion Two-Port Networks of Network Theory

Answer 1

Right choice is (a) z12 = \(\frac{1}{g_{11}} \)

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

V2 = z21 I1 + z22 I2 ………. (2)

And, I1 = g11 V1 + g12 I2 ………. (3)

V2 = g21 V1 + g22 I2 ……….. (4)

Now, (3) and (4) can be rewritten as,

V1 = \(\frac{I_1}{g_{11}}  – \frac{g_{12}}{g_{11}} I_2\) ………… (5)

And V2 = \(\left(g_{22} – \frac{g_{21} g_{12}}{g_{11}}\right) I_2 – \frac{g_{21} I_1}{g_{11}}\)  ……….. (6)

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

z11 = \(\frac{1}{g_{11}} \)

z12 = – \(\frac{g_{12}}{g_{11}} \)

z21 = – \(\frac{g_{21}}{g_{11}} \)

z22 = \(\left(g_{22} – \frac{g_{21} g_{12}}{g_{11}}\right)\).