The y-parameters for the network shown in the figure can be represented by

F1 U.B Deepak 21.11.2019 D 1

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  1. \(\left[ y \right] = \left[ {\begin{array}{*{20}{c}} { - \frac{1}{5}}&{\frac{1}{5}} \\ {\frac{1}{5}}&{ - \frac{1}{5}} \end{array}} \right]\mho \)
  2. \(\left[ y \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{5}}&{ - \frac{1}{5}} \\ { - \frac{1}{5}}&{\frac{1}{5}} \end{array}} \right]\mho \)
  3. \(\left[ y \right] = \left[ {\begin{array}{*{20}{c}} { - 5}&5 \\ 5&{ - 5} \end{array}} \right]\mho \)
  4. \(\left[ y \right] = \left[ {\begin{array}{*{20}{c}} 5&{ - 5} \\ { - 5}&5 \end{array}} \right]\mho \)

Answer (Detailed Solution Below)

Option 2 : \(\left[ y \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{5}}&{ - \frac{1}{5}} \\ { - \frac{1}{5}}&{\frac{1}{5}} \end{array}} \right]\mho \)
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Detailed Solution

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Concept:

Y parameter:

I1 = V1 Y11 + V2 Y12

I2 = V1 Y21 + V2 Y22

\({I_1} = \frac{{{V_1} - {V_2}}}{Z}\)

\({I_1} = \frac{{{V_1}}}{Z} - \frac{1}{Z}\;{V_2}\)     ...1)

\({I_2} = - \frac{1}{Z}\;{V_1} + \frac{1}{Z}\;{V_2}\)     ...2)

\(\left[ y \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{Z}}&{ - \frac{1}{Z}} \\ { - \frac{1}{Z}}&{\frac{1}{Z}} \end{array}} \right]\)

Calculation:

For the given question Z = 5 Ω

\(\therefore \left[ y \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{5}}&{ - \frac{1}{5}} \\ { - \frac{1}{5}}&{\frac{1}{5}} \end{array}} \right]\)

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