his problem analyzes the circuit in below. The capacitor has a preinitial condition of vc$(0-)=10$volts$.$

$($a$)$Redraw the circuit in the Laplace domain, using the parallel version of the Laplace equivalent of the capacitor.

$($b$)$Write the KCL $($Kirchoff$\text{'}$s current law$)$equation for the junction at the top of the capacitor.

$($c$)$Solve your equation from part $($b$)$for VC$($s$).$

$($d$)$Take the inverse Laplace transform of your answer in $($c$)$to find vC$($t$)$for t$0.$

$($e$)$In the remaining parts of this problem, we will find the same solution using superposition. Short$-$circuit the voltage source in your Laplace$-$domain circuit in part $($a$)$to disable it$,$combine the parallel resistors, and then use current division to find the Laplacedomain current through the capacitor. Let's call it IC$(1)($s$).$

$($f$)$Compute VC$(1)($s$)=$IC$(1)$ssC$.$

$($g$)$Take the inverse Laplace transform of your answer in $($f$)$to find vC$(1)($t$)$for t$0.$

$($h$)$Now take your Laplace$-$domain circuit in part $($a$)$and open$-$circuit the current source arising from the initial condition to disable it$.$Combine the parallel resistor and capacitor, and then use voltage division to find the Laplace$-$domain voltage across the capacitor. Let's call it VC$(2)($s$).$

$($i$)$Take the inv