Question

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

(a) Y21 =  \(\frac{D}{B}\)

(b) Y21 = \(\frac{C-A}{B}\)

(c) Y21 = – \(\frac{1}{B}\)

(d) Y21 = \(\frac{A}{B}\)

I had been asked this question in my homework.

My question is based upon Relation between Transmission Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters topic in chapter Two-Port Networks of Network Theory

Answer 1

Correct answer is (c) Y21 = – \(\frac{1}{B}\)

Easy explanation: 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 = \(\frac{A}{B}\).