Question

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

(a) y11 =  \(\frac{A’}{B’}\)

(b) y11 = – \(\frac{1}{B’}\)

(c) y11 = \(\left(C’ – \frac{D’ A’}{B’} \right)\)

(d) y11 = \(\frac{D’}{B’}\)

Answer 1

Answer: (a) y11 =  \(\frac{A’}{B’}\)

 We know that, V2 = A’V1 – B’I1 ……… (1)

I2 = C’V1 – D’I1 …………… (2)

And, I1 = y11 V1 + y12 V2 ……… (3)

I2 = y21 V1 + y22 V2 ………. (4)

Now, (1) and (2) can be rewritten as, I1 = – \(\frac{1}{B’} V_2 + \frac{A’}{B’} V_1\) …………. (5)

And I2 = C’V1 – D’ \(\left(- \frac{1}{B’} V_2 + \frac{A’}{B’} V_1 \right) = \left(C’ – \frac{D’ A’}{B’} \right) V_1 + \frac{D’}{B’} V_2\) ………… (6)

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

y11 =  \(\frac{A’}{B’}\)

y12 = – \(\frac{1}{B’}\)

y21 = \(\left(C’ – \frac{D’ A’}{B’} \right)\)

y22 =  \(\frac{D’}{B’}\).