Question

this question please http://i46.tinypic.com/24c9icl.jpg

Answer 1

I;m not sure about this but I'll give it a try

In a steady-state conditions (the current in the various branches are constant), any capacitor acts as an open circuit. It means that the current that contins the capacitor is zero. From this we can deduce the circuit and draw it simply this way:

V=iR = (8)(0.2)= 1.6 V

:) is that right? Anyway, a guess