For the electric circuit shown below, the state variable equations taking the current source as input is

1
\(\left[ {\begin{array}{*{20}{c}} {{{\bar i}_L}}\\ {{{\bar V}_C}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 5/2}&5\\ { - 1}&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{i_L}}\\ {{V_C}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {5/2}&1 \end{array}} \right]i\)
2
\(\left[ {\begin{array}{*{20}{c}} {{{\bar i}_L}}\\ {{{\bar V}_C}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 5}&5\\ { - 1}&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{i_L}}\\ {{V_C}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {5/2}&1 \end{array}} \right]i\)
3
\(\left[ {\begin{array}{*{20}{c}} {{{\bar i}_L}}\\ {{{\bar V}_C}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {5/2}&5\\ { - 1}&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{i_L}}\\ {{V_C}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {5/2}&1 \end{array}} \right]i\)
4
None of these

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