Question

For a T-network if the Open circuit impedance parameters are given as z11, z21, z12, z22, then z12 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 = \(\left(g_{22} – \frac{g_{21} g_{12}}{g_{11}} \right)\)

Answer 1

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

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