Question

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

(a) y21 =   \(\frac{A’}{B’}\)

(b) y21 = – \(\frac{1}{B’}\)

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

(d) y21 = \(\frac{D’}{B’}\)

This question was addressed to me during an online interview.

The above asked question is from Relation between Transmission Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters topic in section Two-Port Networks of Network Theory

Answer 1

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

Best explanation: 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 =  \(\frac{D’}{B’}\).