ese12 

For the two-port network shown, the parameter Y12 is equal to

This question was previously asked in
ESE Electronics 2012 Paper 1: Official Paper
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  1. YC
  2. YC + YB
  3. YA + YC
  4. -YC

Answer (Detailed Solution Below)

Option 4 : -YC
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Detailed Solution

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Y (Admittance) Parameters:

They are also called the short circuit parameters, as they are calculated under short circuit conditions, i.e. at V= 0 and V= 0.

Expressed in Matrix Form as:

\(\left[ {\begin{array}{*{20}{c}} {{I_1}}\\ {{I_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{Y_{11}}}&{{Y_{12}}}\\ {{Y_{21}}}&{{Y_{22}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_1}}\\ {{V_2}} \end{array}} \right]\)

I1 = Y11V1 + Y12V2

I2 = Y21V1 + Y22V2

With the output short-circuited, i.e. V2 = 0, the two parameters obtained are:

\({Y_{11}} = {\left. {\frac{{{I_1}}}{{{V_1}}}} \right|_{{V_2} = 0}}\)

\({Y_{21}} = {\left. {\frac{{{I_2}}}{{{V_1}}}} \right|_{{V_2} = 0}}\)

With the input short-circuited, i.e. V1 = 0, the two parameters obtained are:

\({Y_{12}} = {\left. {\frac{{{I_1}}}{{{V_2}}}} \right|_{{V_1} = 0}}\)

\({Y_{22}} = {\left. {\frac{{{I_2}}}{{{V_2}}}} \right|_{{V_1} = 0}}\)

Analysis:

We can see that, 

Y11 = YA + YC

Y12 = Y21 = -YC

Y22 = YB + YC 

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