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

Answer 1

Answer: (a) y11 =  \(\frac{D}{B}\)

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

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

y11 =  \(\frac{D}{B}\)

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

y21 = – \(\frac{1}{B}\)

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