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 $\mathrm { R } _ { 1 } = 1.8 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 12 \mathrm { k } \Omega , \mathrm { R } _ { 3 } =$ $2.8 \mathrm { k } \Omega , \mathrm { R } _ { 4 } = 18 \mathrm { k } \Omega , \mathrm { V } _ { \mathrm { s } } = 3 \mathrm {~V} , \mathrm { C } = 0.02 \mu \mathrm { F }$.

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The Answer is

\begin{array} { l } \mathrm { v } ( 0 ) = \mathrm { V } _ { 0 } = \mathrm { Vs } _ { \mathrm { s } } imes \left( \mathrm { R } _ { 2 } \| \mathrm { R } _ { 4 } ight) / \left[ \mathrm { R } _ { 1 } + \left( \mathrm { R } _ { 2 } \| \mathrm { R } _ { 4 } ight) ight] = 2.4 \mathrm {~V} \\mathrm { Req } _ { \mathrm { eq } } = \mathrm { R } _ { 3 } + \left( \mathrm { R } _ { 2 } \| \mathrm { R } _ { 4 } ight) = 10 \mathrm { k } \Omega \ au = \mathrm { Req } _ { \mathrm { eq } } = 0.0002 \mathrm {~s} \\mathrm { v } ( \mathrm { t } ) = 2.4 \mathrm { e } ^ { - 5000 \mathrm { t } } \mathrm { u } ( \mathrm { t } )\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 $= 0$