In the circuit shown below, let $$\mathrm { R } _ { 1 } = 2 \Omega , \mathrm { R } _ { 2 } = 3 \Omega , \mathrm { C } = 1 \mathrm {~F} , \mathrm {~L} = 0.2 \mathrm { H } , \mathrm { v } \left( 0 ^ { - } \right) = 1 \mathrm {~V} , \mathrm { i } \left( 0 ^ { - } \right) = 2 \mathrm {~A} , \mathrm { i } _ { \mathrm { s } } ( \mathrm { t } ) = 3 \mathrm { u } ( \mathrm { t } )$$
Draw the circuit in the s-domain and find the Norton equivalent current $\mathrm { I } _ { \mathrm { n } } ( \mathrm { s } )$ and the Norton equivalent impedance $Z _ { n } ( s )$.

Question

Quizplus · Accepted Answer

The Answer is $Terminals a and b are short-circuited. - \frac { 3 } { s } + \frac { V _ { 1 } + L i \left( 0 ^ { - } ight) } { s L } + \frac { V _ { 1 } } { R _ { 1 } } = 0 \begin{array} { l } I _ { n } ( s ) = \frac { V _ { 1 } ( s ) } { R _ { 1 } } + \frac { \frac { v \left( 0 ^ { - } ight) } { s } } { \frac { 1 } { s C } } = \frac { s + 11 } { s + 10 } \\ \mathrm { Z } _ { \mathrm { n } } ( \mathrm { s } ) = \mathrm { R } _ { 2 } \| 1 / ( \mathrm { sC } ) \| \left( \mathrm { R } _ { 1 } + \mathrm { sL } ight) = \frac { 1 } { \frac { 1 } { R _ { 2 } } + s C + \frac { 1 } { R _ { 1 } + s L } } = \frac { 3 s + 30 } { 3 s ^ { 2 } + 31 s + 25 } = \frac { s + 10 } { s ^ { 2 } + 10.3333 s + 8.3333 } \end{array}$

Terminals

a

and

b

are short-circuited.

- \frac { 3 } { s } + \frac { V _ { 1 } + L i \left( 0 ^ { - } ight) } { s L } + \frac { V _ { 1 } } { R _ { 1 } } = 0

\begin{array} { l } I _ { n } ( s ) = \frac { V _ { 1 } ( s ) } { R _ { 1 } } + \frac { \frac { v \left( 0 ^ { - } ight) } { s } } { \frac { 1 } { s C } } = \frac { s + 11 } { s + 10 } \\\mathrm { Z } _ { \mathrm { n } } ( \mathrm { s } ) = \mathrm { R } _ { 2 } \| 1 / ( \mathrm { sC } ) \| \left( \mathrm { R } _ { 1 } + \mathrm { sL } ight) = \frac { 1 } { \frac { 1 } { R _ { 2 } } + s C + \frac { 1 } { R _ { 1 } + s L } } = \frac { 3 s + 30 } { 3 s ^ { 2 } + 31 s + 25 } = \frac { s + 10 } { s ^ { 2 } + 10.3333 s + 8.3333 }\end{array}

In the Circuit Shown Below, Let Draw the Circuit in the S-Domain and Find the Norton