Question

For a T-network if the Short circuit admittance parameters are given as y11, y21, y12, y22, then y22 in terms of Hybrid parameters can be expressed as ________

(a) y22 = \(\left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right)\)

(b) y22 = \(\frac{h_{21}}{h_{11}}\)

(c) y22 = –\(\frac{h_{12}}{h_{11}}\)

(d) y22 = \(\frac{1}{h_{11}} \)

I have been asked this question in unit test.

My question comes from Hybrid (h) Parameter topic in division Two-Port Networks of Network Theory

Answer 1

Correct choice is (a) y22 = \(\left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right)\)

The best explanation: We know that the short circuit admittance parameters can be expressed in terms of voltages and currents as,

I1 = y11 V1 + y12 V2 ……… (1)

I2 = y21 V1 + y22 V2 ………. (2)

And the Hybrid parameters can be expressed in terms of voltages and currents as,

V1 = h11 I1 + h12 V2 ………. (3)

I2 = h21 I1 + h22 V2 ……….. (4)

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

I1 = \(\frac{V_1}{h_{11}}  – \frac{h_{12} V_2}{h_{11}}\)  ………. (5)

And I2 = \(\frac{h_{21} V_1}{h_{11}}  + \left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right) V_2\) ………. (6)

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

y11 = \(\frac{1}{h_{11}} \)

y12 = –\(\frac{h_{12}}{h_{11}}\)

y21 = \(\frac{h_{21}}{h_{11}}\)

y22 = \(\left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right)\).