A two-port network has parameters ABCD. If all the impedances in the network are doubled, then

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  1. A and D remain unchanged, B is doubled and C is halved
  2. A, B, C and D are all doubled
  3. A and D are doubled, C and B remain unchanged
  4. A and D remain unchanged, C is doubled and B is halved

Answer (Detailed Solution Below)

Option 1 : A and D remain unchanged, B is doubled and C is halved
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Detailed Solution

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The ABCD/transmission parameters are given as:

\(\left[ {\begin{array}{*{20}{c}} {{V_1}}\\ {{I_1}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_2}}\\ { - {I_2}} \end{array}} \right]\)

The impedances are:

\(B = \frac{V_1}{I_2}\)

\(C=\frac{I_1}{V_2}\)

Since B is directly proportional to impedance and C is inversely proportional to the impedance. 

∴ Doubling the impedances means, doubling B and also C gets halved, rest A and D are not affected. 

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