Problem $2$: B$-6-61.$ When switch is closed $($ t $<0),$ the current flowing through the capacitor charger it$.$ With time the capacitor will be fully charged and reach a constant value $$ at the steady state. Once the steady state is reached there won$$t be any further charge flow through the capacitor. Therefore, current in the Loop $1$ and that in the Loop $2$ will be equal, and the net current through the capacitor will be zero. Call this steady state through the outer loop as $2.$ Since the resistor R$2$ is parallel to capacitor, the potential drop through them must be equal $($because the voltage b$/$n both the junctions they are connected must be the same$).$ From this relationship you can calculate $3.$ When the switch is opened at t $=0,$ the current will only flow through Loop $1,$ i$.$e$.2=0.$ Note that capacitor will discharge from its initial value $$ The final answer is $1()=(1+2)1$ for $0$