Question

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

(a) y11 =  \(\frac{D}{B}\)

(b) y11 = \(\frac{C-A}{B}\)

(c) y11 = – \(\frac{1}{B}\)

(d) y11 = \(\frac{A}{B}\)

This question was addressed to me during an interview.

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

Answer 1

Correct option is (a) y11 =  \(\frac{D}{B}\)

For explanation I would say: We know that, V1 = AV2 – BI2 ……… (1)

I1 = CV2 – DI2 …………… (2)

And, I1 = y11 V1 + y12 V2 ……… (3)

I2 = y21 V1 + y22 V2 ………. (4)

Now, (1) and (2) can be rewritten as, I2 = \(\frac{A}{B}V_2 – \frac{1}{B}V_1\) …………. (5)

And I1 = CV2 – D \(\left(\frac{A}{B}V_2 – \frac{1}{B}V_1\right) = \frac{D}{B}V_1 + \left(C-\frac{A}{B}\right) V_2\) …………… (6)

Comparing equations (3), (4) and (5), (6), we get,

y11 =  \(\frac{D}{B}\)

y12 = \(\frac{C-A}{B}\)

y21 = – \(\frac{1}{B}\)

y22 = \(\frac{A}{B}\).