1.

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»

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


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