1.

For a T shaped network, if the Short-circuit admittance parameters are y11, y12, y21, y22, then y12 in terms of Inverse Transmission parameters can be expressed as ________(a) y12 =\(\frac{A’}{B’}\)(b) y12 = – \(\frac{1}{B’}\)(c) y12 = \(\left(C’ – \frac{D’ A’}{B’}\right)\)(d) y12 = \(\frac{D’}{B’}\)The question was posed to me in an internship interview.Asked question is from Relation between Transmission Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters topic in portion Two-Port Networks of Network Theory

Answer»

The correct OPTION is (B) y12 = – \(\frac{1}{B’}\)

The BEST I can explain: 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 =\(\frac{D’}{B’}\).


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