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 $\mathrm { R } _ { 1 } = 8 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 24 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 3$ $\mathrm { k } \Omega , \mathrm { I } _ { \mathrm { s } } = 0.5 \mathrm {~mA} , \mathrm {~V} _ { \mathrm { s } } = 3.2 \mathrm {~V} , \mathrm { C } = 0.05 \mu \mathrm { F }$.

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\begin{array} { l } \mathrm { v } ( 0 ) = \mathrm { V } _ { 0 } = \mathrm { I } _ { \mathrm { s } } imes \left( \mathrm { R } _ { 1 } \| \mathrm { R } _ { 2 } ight) + \mathrm { V } _ { \mathrm { s } } imes \mathrm { R } _ { 2 } / \left( \mathrm { R } _ { 1 } + \mathrm { R } _ { 2 } ight) = 5.4 \mathrm {~V} \\mathrm { v } ( \infty ) = \mathrm { V } _ { \mathrm { inf } } = \mathrm { V } _ { \mathrm { s } } imes \mathrm { R } _ { 2 } / \left( \mathrm { R } _ { 1 } + \mathrm { R } _ { 2 } ight) = 2.4 \mathrm {~V} \\mathrm { Req } _ { \mathrm { eq } } = \mathrm { R } _ { 3 } + \left( \mathrm { R } _ { 1 } \| \mathrm { R } _ { 2 } ight) = 9 \mathrm { k } \Omega \ au = \mathrm { R } _ { \mathrm { eq } } \mathrm { C } = 0.00045 \mathrm {~s} \\mathrm { v } ( \mathrm { t } ) = \left[ \mathrm { V } _ { \mathrm { inf } } + \left( \mathrm { V } _ { 0 } - \mathrm { V } _ { \mathrm { inf } } ight) \mathrm { e } ^ { - 2222.2222 \mathrm { t } } ight] \mathrm { u } ( \mathrm { t } ) = \left[ 2.4 + ( 5.4 - 2.4 ) \mathrm { e } ^ { - 2222.2222 \mathrm { t } } ight] \mathrm { u } ( \mathrm { t } ) \= \left[ 2.4 + 3 \mathrm { e } ^ { - 2222.2222 \mathrm { t } } ight] \mathrm { u } ( \mathrm { t } ) \mathrm { V }\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 $= 0$