Question

Consider the following equations:

V₁ = 6 V₂ – 4 I₂

I₁ = 7 V₂ – 2 I₂

A, B, C, D parameters are

1. 6, -4 Ω, 7 mho and -2
2. 6, 4 Ω, 7 mho and 2
3. -6, 4 Ω, -7 mho and 2
4. 6, -4 Ω, -7 mho and -2

Answer 1

Correct Answer - Option 2 : 6, 4 Ω, 7 mho and 2

ABCD Parameters:

It relates the variables at the input port to those at the output port.

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

V1 = AV2 - BI2

I1 = CV2 - DI2

With output open-circuited, i.e. I2 = 0, the two parameters we get are:

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

\({\left. {C = \frac{{{I_1}}}{{{V_2}}}} \right|_{{I_2} = 0}}\)

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

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

\({\left. {D = \frac{{{I_1}}}{{{I_2}}}} \right|_{{V_i} = 0}}\)

Calculation:

Given:

V₁ = 6 V₂ – 4 I₂

I₁ = 7 V₂ – 2 I₂

Comparing this with the standard equations of two-port networks, we get:

A = 6, B = 4 Ω, C = 7 mho, and D = 2