Switch in the Circuit Shown Below Has Been Opened for a Long

Question

Examlex · Accepted Answer

The Answer is Change

I _ { s }

and

R _ { 1 }

to a series connection of

\mathrm { V } _ { \mathrm { s } } = \mathrm { R } _ { 1 } \mathrm { I } _ { \mathrm { s } } = 0.5 \mathrm {~V}

and a resistor

\mathrm { R } _ { 1 }

.
Let

\begin{array} { l } \mathrm { R } = \mathrm { R } _ { 1 } + \mathrm { R } _ { 2 } = 300 \Omega \- V _ { s } + R i ( t ) + L \frac { d i ( t ) } { d t } + v ( t ) = 0\end{array}

i ( t ) = C \frac { d v ( t ) } { d t }

Substituting Equation (2) into Equation (1), we obtain

- V _ { s } + R C \frac { d v ( t ) } { d t } + L C \frac { d ^ { 2 } v ( t ) } { d t ^ { 2 } } + v ( t ) = 0

which can be rearranged as

\frac { d ^ { 2 } v ( t ) } { d t ^ { 2 } } + \frac { R } { L } \frac { d v ( t ) } { d t } + \frac { 1 } { L C } v ( t ) = \frac { 1 } { L C } V _ { s }

From Equation (2), we have

\begin{array} { l } \frac { d v ( t ) } { d t } = \frac { i ( t ) } { C } \\frac { d v ( 0 ) } { d t } = \frac { i ( 0 ) } { C } \\mathrm { a } _ { 2 } = 1 \a _ { 1 } = \frac { R } { L } = 3000 \a _ { 0 } = \frac { 1 } { L C } = 5 imes 10 ^ { 8 } \\alpha = \mathrm { a } _ { 1 } / 2 = 1500 , \omega _ { 0 } = \sqrt { a _ { 0 } } = 22360.6798\end{array}

Since

\alpha < \omega _ { 0 }

, this is case 3 (underdamped).

\begin{array} { l } \beta = \sqrt { \omega _ { o } ^ { 2 } - \alpha ^ { 2 } } = 22310.3115 \s _ { 1 } = - \alpha + j \beta = - 1500 + \mathrm { j } 22310.3115\end{array}

s _ { 2 } = - \alpha - j \beta = - 1500 - \mathrm { j } 22310.3115

B _ { 1 } = V _ { 0 } - V _ { \mathrm { s } } = 0 \mathrm {~V}

B _ { 2 } = \frac { D v _ { 0 } + \alpha \left( V _ { 0 } - V _ { s } ight) } { \beta } = 2.241116 \mathrm {~V}

\mathrm { v } ( \mathrm { t } ) = \left[ 0.5 + 2.241116 \mathrm { e } ^ { - 1500 \mathrm { t } } \sin ( 22310.3115 \mathrm { t } ) ight] \mathrm { u } ( \mathrm { t } ) \mathrm { V }

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