Question

A 2-port network is represented by the following equations:

V₁ = 60 I₁ + 20 I₂

V₂ = 20 I₁ + 40 I₂

The ABCD parameters of the above network would be
1. \(\left[ {\begin{array}{*{20}{c}} 2&{\frac{1}{{20}}}\\ 3&{100} \end{array}} \right]\)
2. \(\left[ {\begin{array}{*{20}{c}} {100}&3\\ 2&{\frac{1}{{20}}} \end{array}} \right]\)
3. \(\left[ {\begin{array}{*{20}{c}} {100}&{20}\\ 6&3 \end{array}} \right]\)
4. \(\left[ {\begin{array}{*{20}{c}} 3&{100}\\ {\frac{1}{{20}}}&2 \end{array}} \right]\)

Answer 1

Correct Answer - Option 4 : \(\left[ {\begin{array}{*{20}{c}} 3&{100}\\ {\frac{1}{{20}}}&2 \end{array}} \right]\)

2 port network is represented by-

V₁ = 60 I₁ + 20 I₂     …1)

V₂ = 20 I₁ + 40 I₂      …2)

Now, to find ABCD parameters-

V₁ = AV₂ – BI₂

I₁ = CV₂ – DI₂

Hence, from equation 2)-

20I₁ = V₂ – 40 I₂

\( \Rightarrow {I_1} = \frac{1}{{20}}{V_2} - 2{I_2}\)     …3)

From equation 1) and 3)-

\({V_1} = 60\left[ {\frac{1}{{20}}{V_2} - 2{I_2}} \right] + 20{I_2}\)

V₁ = 3V₂ – 100 I₂      …4)

By comparing equation 3) and 4) with the standard equation,

A = 3, B = 100 Ω

\(C = \frac{1}{{20}},℧{\rm{\;D}} = 2\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3&{100}\\ {\frac{1}{{20}}}&2 \end{array}} \right]\)