The y parameters of the above network are :

A. \(\left[ {\begin{array}{*{20}{c}} { - 1}&1\\ { - 1}&{ - 2} \end{array}} \right]\)

B. \(\left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ 1&{ - 2} \end{array}} \right]\)

C. \(\left[ {\begin{array}{*{20}{c}} {\frac{{ + 1}}{3}}&{\frac{{ - 2}}{3}}\\ {\frac{{ - 1}}{3}}&{\frac{{ - 1}}{3}} \end{array}} \right]\)

D. \(\left[ {\begin{array}{*{20}{c}} {\frac{{ - 2}}{3}}&{\frac{{ - 1}}{3}}\\ {\frac{1}{3}}&{\frac{{ - 1}}{3}} \end{array}} \right]\)

Please scroll down to see the correct answer and solution guide.

Right Answer is: C

SOLUTION

Apply KVL at outer loop

V₁ – 2 I₁ + J₂ (1) = V₂

-2I₁+ I₂ = V₂ – V₁-------(1)

Apply KCL at Node

I₁ + I₂ = -V₂ ------(2)

(2) – (1)

(I₁ + I₂) – (I₂ – 2 I₁) = -V₂ – (V₂– V₁)

3 J₁ = V₁ – 2V₂

\({I_1} = \frac{1}{3}{v_1} - \frac{2}{3}{v_2} \) -------(3)

Substitute (3) in (2)

I₁ + I₂ = -V₂

I₂= V₂ – I₁

\(\begin{array}{l} {I_2} = - {V_2} + \frac{2}{3}{v_2} - \frac{1}{3}\;{v_1}\\ {I_2} = \frac{{ - 1}}{3}{v_2} - \frac{1}{3}{v_1} \end{array}\)

\(\left[ {\begin{array}{*{20}{c}} {{I_1}}\\ {{I_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{3}}&{\frac{{ - 2}}{3}}\\ {\frac{{ - 1}}{3}}&{\frac{{ - 1}}{3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{v_1}}\\ {{v_2}} \end{array}} \right]\)