the switch in the circuit shown below has been closed for a long time before it is opened at t $= 0$. Find the voltage $\mathrm { v } ( \mathrm { t } )$ across the capacitor for $\mathrm { t } \geq 0$. Assume $\mathrm { R } _ { 1 } = 2 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 3 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 1$ $\mathrm { k } \Omega , \mathrm { L } = 125 \mathrm { mH } , \mathrm { C } = 0.05 \mu \mathrm { F } , \mathrm { V } _ { \mathrm { s } } = 5 \mathrm {~V}$.

Question

Quizplus · Accepted Answer

The Answer is

\begin{array} { l } \mathrm { v } ( 0 ) = \mathrm { V } _ { 0 } = \mathrm { V } _ { \mathrm { s } } imes \mathrm { R } _ { 2 } / \left( \mathrm { R } _ { 1 } + \mathrm { R } _ { 2 } ight) = 3 \mathrm {~V} . \mathrm { i } ( 0 ) = \mathrm { I } _ { 0 } = 0 \\mathrm { R } = \mathrm { R } _ { 2 } + \mathrm { R } _ { 3 } = 4 \mathrm { k } \Omega \\frac { d ^ { 2 } v ( t ) } { d t ^ { 2 } } + \frac { R } { L } \frac { d v ( t ) } { d t } + \frac { 1 } { L C } v ( t ) = 0 \\mathrm { a } _ { 2 } = 1 , \mathrm { a } _ { 1 } = \mathrm { R } / \mathrm { L } = 32000 , \mathrm { a } _ { 0 } = 1 / ( \mathrm { LC } ) = 1.6 imes 10 ^ { 8 } \\alpha = \mathrm { a } _ { 1 } / 2 = 16000 , \omega _ { 0 } = \sqrt { a _ { 0 } } = 12649.11064\end{array}

Since

\alpha > \omega _ { 0 }

, this is case 1 (overdamped).

D v _ { 0 } = \frac { d v ( 0 ) } { d t } = \frac { i ( 0 ) } { C } = \frac { I _ { 0 } } { C } = 0

s _ { 1 } = - \alpha + \sqrt { \alpha ^ { 2 } - \omega _ { 0 } ^ { 2 } } = - 6202.041

s _ { 2 } = - \alpha - \sqrt { \alpha ^ { 2 } - \omega _ { 0 } ^ { 2 } } = - 25797.959

A _ { 1 } = \frac { V _ { 0 } s _ { 2 } - D v _ { 0 } } { s _ { 2 } - s _ { 1 } } = 3.9495

A _ { 2 } = \frac { D v _ { 0 } - V _ { 0 } s _ { 1 } } { s _ { 2 } - s _ { 1 } } = - 0.9495

\mathrm { v } ( \mathrm { t } ) = \left[ 3.9495 \mathrm { e } ^ { - 6202.041 \mathrm { t } } - 0.9495 \mathrm { e } ^ { - 25797.96 \mathrm { t } } ight] \mathrm { u } ( \mathrm { t } )

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

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