RC circuit How does the current vary after the switch is closed?
On simplified level you can see a capacitor in a DC circuit like this:

When it's uncharged (just the moment you close the switch) it's like closed switch.

When it's fully charged (in your quiz called after very long time) it's like a open switch.

Lets call the Resistor closer to the battery \$R_1\$ and the other one \$R_2\$.

So this answers your question:


In the moment the switch is beeing closed you have current flowing through both resistors: 


\$I_1=\frac{U}{R_1}\$ 

and 

\$I_2=\frac{U}{R_2}\$ 

The total current is:

\$I_{Total} = I_1 + I_2\$

And since \$R_1\$ and \$R_2\$ are same the result is

\$I_{Total} = \frac{2*U}{R}\$

Note that this is only valid for exactly the moment the switch is beeing closed (where the capacitor is fully discharged).


When the capacitor is fully charged this part of the circuit has no current flowing anymore so \$I_{Total}\$ come down to beeing \$I_1\$ which is 


\$I_1 = \frac{U}{R}\$