the switch in the circuit shown below has been closed for a long time before it is opened at t $= 0$. Find the current $i ( t )$ through the inductor for $t \geq 0$. Assume $R _ { 1 } = 1 \mathrm { k } \Omega , R _ { 2 } = 2 \mathrm { k } \Omega , R _ { 3 } = 3$ $\mathrm { k } \Omega , \mathrm { L } = 250 \mathrm { mH } , \mathrm { C } = 0.025 \mu \mathrm { F } , \mathrm { V } _ { \mathrm { s } } = 2.2 \mathrm {~V}$.

Question

Quizplus · Accepted Answer

The Answer is

\begin{array} { l } \mathrm { R } _ { \mathrm { a } } = \mathrm { R } _ { 2 } \| \mathrm { R } _ { 3 } = 1.2 \mathrm { k } \Omega \\mathrm { v } ( 0 ) = \mathrm { V } _ { 0 } = \mathrm { V } _ { \mathrm { s } } imes \mathrm { R } _ { \mathrm { a } } / \left( \mathrm { R } _ { 1 } + \mathrm { R } _ { \mathrm { a } } ight) = 1.2 \mathrm {~V} \\mathrm { i } ( 0 ) = \mathrm { I } _ { 0 } = \mathrm { V } _ { 0 } / \mathrm { R } _ { 3 } = 0.4 \mathrm {~mA} \\frac { v ( t ) } { R _ { 2 } } + i ( t ) + C \frac { d v ( t ) } { d t } = 0 \v ( t ) = R _ { 3 } i ( t ) + L \frac { d i ( t ) } { d t }\end{array}

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

\frac { R _ { 3 } } { R _ { 2 } } i ( t ) + \frac { L } { R _ { 2 } } \frac { d i ( t ) } { d t } + i ( t ) + R _ { 3 } C \frac { d i ( t ) } { d t } + L C \frac { d ^ { 2 } i ( t ) } { d t ^ { 2 } } = 0

which can be rearranged as

\begin{array}{l}\frac { d ^ { 2 } i ( t ) } { d t ^ { 2 } } + \left( \frac { R _ { 3 } } { L } + \frac { 1 } { R _ { 2 } C } ight) \frac { d i ( t ) } { d t } + \frac { 1 } { L C } \left( 1 + \frac { R _ { 3 } } { R _ { 2 } } ight) i ( t ) = 0\\frac { d i ( t ) } { d t } = \frac { 1 } { L } v ( t ) - \frac { R _ { 3 } } { L } i ( t )\\frac { d i ( 0 ) } { d t } = \frac { 1 } { L } v ( 0 ) - \frac { R _ { 3 } } { L } i ( 0 ) = 0\a _ { 2 } = 1\a _ { 1 } = \frac { R _ { 3 } } { L } + \frac { 1 } { R _ { 2 } C } = 32000\a _ { 0 } = \frac { 1 } { L C } \left( 1 + \frac { R _ { 3 } } { R _ { 2 } } ight) = 4 imes 10 ^ { 8 }\\alpha = a _ { 1 } / 2 = 16000 , \omega _ { 0 } = \sqrt { a _ { 0 } } = 20000\ ext { Since } \alpha < \omega _ { 0 } ext {, this is case } 3 ext { (underdamped). }\\beta = \sqrt { \omega _ { o } ^ { 2 } - \alpha ^ { 2 } } = 12000\s _ { 1 } = - \alpha + j \beta = - 16000 + \mathrm { j } 12000\s _ { 2 } = - \alpha - j \beta = - 16000 - \mathrm { j } 12000\B _ { 1 } = I _ { 0 } = 0.4 \mathrm {~mA}\B _ { 2 } = \frac { D i _ { 0 } + \alpha I _ { 0 } } { \beta } = 0.53333 \mathrm {~mA}\\mathrm { i } ( \mathrm { t } ) = \mathrm { e } ^ { - 1600 \mathrm { t } } [ 0.4 \cos ( 12000 \mathrm { t } ) + 0.53333 \sin ( 12000 \mathrm { t } ) ] \mathrm { u } ( \mathrm { t } ) \mathrm { mA }\end{array}

The Switch in the Circuit Shown Below Has Been Closed =0= 0=0

The Switch in the Circuit Shown Below Has Been Closed $= 0$