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For a T shaped network, if the Short-circuit admittance parameters are y11, y12, y21, y22, then y12 in terms of Transmission parameters can be expressed as ________(a) y12 = \(\frac{D}{B}\)(b) y12 = \(\frac{C-A}{B}\)(c) y12 = – \(\frac{1}{B}\)(d) y12 = \(\frac{A}{B}\)I have been asked this question in examination.I'm obligated to ask this question of Relation between Transmission Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters topic in division Two-Port Networks of Network Theory

Answer»

Right answer is (b) y12 = \(\frac{C-A}{B}\)

The best 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 = \(\frac{A}{B}\).


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